Complex Numbers for JEE: Notes, Modulus, Argument, Roots of Unity and PYQs

The equation $x^2+1=0$ has no real root, because the square of a real number is never negative. Mathematicians added one new number $i$ with $i^2=-1$ and found that every polynomial equation then has a root. The resulting numbers $a+ib$ add, multiply and divide like real numbers, carry a length and a direction, and turn rotation in the plane into a single multiplication. That is why complex numbers appear in quadratics, trigonometric sums, roots of unity and coordinate geometry.

This note covers the powers of $i$, the algebra of $a+ib$ and equality, real and imaginary parts, the conjugate and its properties, the modulus with the triangle inequality and the parallelogram law, the argument and the principal argument, polar and Euler form, square roots, solving equations in $z$, De Moivre's theorem, the cube roots of unity, the $n$th roots of unity, and quadratic equations with complex roots or coefficients. Geometry in the Argand plane (rotation, lines, circles and loci) is summarised in one short section and developed fully in the Complex Numbers in Coordinate Geometry note. Each section ends with worked examples from JEE Main, JEE Advanced and IIT JEE papers, graded from easy to hard.

The number $i$ and its powers

The imaginary unit $i$ is defined by

$$i^2=-1$$

so $i=\sqrt{-1}$. For a positive real $a$, the square root of $-a$ is

$$\sqrt{-a}=i\sqrt a$$

A number of the form $ib$ with $b$ real and $b\neq0$ is called purely imaginary.

Powers of $i$

Multiplying by $i$ again and again gives a cycle of length $4$:

$$i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1$$

So for any integer $n$, write $n=4q+r$ with $r\in\{0,1,2,3\}$. Then

$$i^n=i^{4q+r}=\left(i^4\right)^q i^r=i^r$$

The same rule works for negative $n$: $i^{-1}=\frac1i=-i$ because $i\cdot(-i)=1$.

The four powers of i on the unit circle, each a quarter turn from the previous one

Four consecutive powers always add to zero:

$$i^n+i^{n+1}+i^{n+2}+i^{n+3}$$ $$=i^n(1+i-1-i)=0$$

Examples

Example 1. The value of $i^{57}+i^{70}+i^{-3}+i^{-102}$ is

▸ SHOW SOLUTION

Step 1. Reduce each power modulo 4

$57=4\cdot14+1$, so $i^{57}=i$. $70=4\cdot17+2$, so $i^{70}=-1$.

$$i^{-3}=i^{-4}\cdot i=i$$$$i^{-102}=i^{-104}\cdot i^{2}=-1$$

Step 2. Add

$$i-1+i-1=-2+2i$$

Answer: option (A)

Example 2. Which of the following is correct?

▸ SHOW SOLUTION

Step 1. Write each root with $i$

$$\sqrt{-4}=2i,\qquad \sqrt{-9}=3i$$

Step 2. Multiply

$$\sqrt{-4}\,\sqrt{-9}=6i^2=-6$$

$\sqrt{(-4)(-9)}=\sqrt{36}=6$, the positive root. So (A), (C) and (D) are false.

Answer: option (B)

Example 3 (IITianForum). The value of $\displaystyle\sum_{k=1}^{40} i^{\,k!}$ is

▸ SHOW SOLUTION

Step 1. The first three terms

$1!=1$, $2!=2$, $3!=6$:

$$i^{1}+i^{2}+i^{6}=i-1-1=i-2$$

Step 2. The remaining terms

For $k\ge4$, $k!$ contains the factor $4$, so $i^{k!}=1$. There are $40-3=37$ such terms.

Step 3. Add

$$(i-2)+37=35+i$$

Answer: option (A)

Complex numbers: algebra and equality

A complex number is $z=x+iy$ with $x,y$ real. The real number $x$ is the real part, $\mathrm{Re}(z)$, and the real number $y$ is the imaginary part, $\mathrm{Im}(z)$. Note that $\mathrm{Im}(z)$ is $y$, not $iy$.

Formally, $z$ is the ordered pair $(x,y)$ of real numbers, and the set of all such pairs is $\mathbb C$. The real numbers are the pairs $(x,0)$, and $i$ is the pair $(0,1)$.

Equality

Two complex numbers are equal exactly when both parts are equal:

$$a+ib=c+id\iff a=c\ \text{and}\ b=d$$

This is the most used fact of the chapter: one complex equation gives two real equations.

Operations

For $z_1=a+ib$ and $z_2=c+id$:

$$z_1\pm z_2=(a\pm c)+i(b\pm d)$$ $$z_1z_2=(ac-bd)+i(ad+bc)$$

The product follows from expanding and using $i^2=-1$. To divide, multiply the numerator and the denominator by $c-id$:

$$\frac{a+ib}{c+id}=\frac{(a+ib)(c-id)}{c^2+d^2}$$

The multiplicative inverse of $z=a+ib\neq0$ is

$$\frac1z=\frac{a}{a^2+b^2}-i\,\frac{b}{a^2+b^2}$$

Addition and multiplication are commutative and associative, multiplication distributes over addition, $0$ and $1$ are the identities, and every non-zero $z$ has an inverse.

Purely real and purely imaginary

$z$ is purely real when $\mathrm{Im}(z)=0$ and purely imaginary when $\mathrm{Re}(z)=0$, $z\neq0$. For a quotient, first put it in the form $a+ib$ and then set the required part to zero.

Examples

Example 4. If $\dfrac{3+2i}{1-i}+(2-i)^2=a+ib$ with $a,b$ real, then $2(a-b)$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The quotient

$$\frac{3+2i}{1-i}=\frac{(3+2i)(1+i)}{2}=\frac{1+5i}{2}$$

Step 2. The square

$$(2-i)^2=4-4i+i^2=3-4i$$

Step 3. Add and read off $a$, $b$

$$a+ib=\frac72-\frac32\,i$$$$2(a-b)=2\left(\frac72+\frac32\right)=10$$

Answer: $10$

Example 5 (JEE Main 2026, 2 April Shift 1). Let $x$ and $y$ be real numbers such that $50\left(\dfrac{2x}{1+3i}-\dfrac{y}{1-2i}\right)=31+17i$, $i=\sqrt{-1}$. Then the value of $10(x-3y)$ is

▸ SHOW SOLUTION

Step 1. Real denominators

$(1+3i)(1-3i)=10$ and $(1-2i)(1+2i)=5$:

$$\frac{2x}{1+3i}=\frac{x(1-3i)}{5}$$$$\frac{y}{1-2i}=\frac{y(1+2i)}{5}$$

Step 2. The equation

$$10x(1-3i)-10y(1+2i)=31+17i$$

Step 3. Equate parts

$$10x-10y=31,\qquad -30x-20y=17$$

So $x-y=3.1$. Putting $x=3.1+y$ in the second: $-93-50y=17$, so $y=-2.2$ and $x=0.9$.

$$10(x-3y)=10(0.9+6.6)=75$$

Answer: option (4)

Example 6 (JEE Main 2016). A value of $\theta$ for which $\dfrac{2+3i\sin\theta}{1-2i\sin\theta}$ is purely imaginary, is

▸ SHOW SOLUTION

Step 1. Real denominator

Multiply by $1+2i\sin\theta$. Write $s=\sin\theta$.

$$\frac{(2+3is)(1+2is)}{1+4s^2}=\frac{(2-6s^2)+7is}{1+4s^2}$$

Step 2. Real part zero

$$2-6s^2=0\Rightarrow s^2=\frac13$$

The imaginary part $\frac{7s}{1+4s^2}$ is then non-zero, so the number is purely imaginary. $\sin\theta=\frac1{\sqrt3}$ gives $\theta=\sin^{-1}\frac1{\sqrt3}$.

Answer: option (C)

Example 7 (JEE Main 2025, 4 April Shift 2). Let the product of $\omega_1=(8+i)\sin\theta+(7+4i)\cos\theta$ and $\omega_2=(1+8i)\sin\theta+(4+7i)\cos\theta$ be $\alpha+i\beta$, $i=\sqrt{-1}$. Let $p$ and $q$ be the maximum and minimum values of $\alpha+\beta$ respectively. Then $p+q$ is equal to

▸ SHOW SOLUTION

Step 1. Separate real and imaginary parts

Let $a=8\sin\theta+7\cos\theta$ and $b=\sin\theta+4\cos\theta$. Then

$$\omega_1=a+ib,\qquad \omega_2=b+ia$$

Step 2. The product

$$\omega_1\omega_2=ab+ia^2+ib^2+i^2ab$$$$=i\left(a^2+b^2\right)$$

So $\alpha=0$ and $\beta=a^2+b^2$.

Step 3. Simplify $a^2+b^2$

$$a^2+b^2=65\sin^2\theta+65\cos^2\theta$$$$\quad+120\sin\theta\cos\theta$$$$\alpha+\beta=65+60\sin2\theta$$

Step 4. Extremes

$p=65+60=125$ and $q=65-60=5$, so $p+q=130$.

Answer: option (2)

The conjugate

The conjugate of $z=x+iy$ is

$$\bar z=x-iy$$

In the Argand plane (next section) $\bar z$ is the mirror image of $z$ in the real axis.

Point z with its conjugate, its negative and the negative of its conjugate in the four quadrants

Properties

Each follows by writing $z=x+iy$ and expanding.

The product rule: if $z_1=a+ib$ and $z_2=c+id$, both sides equal $(ac-bd)-i(ad+bc)$. The quotient rule follows from the product rule applied to $z_1=\frac{z_1}{z_2}\cdot z_2$.

So the conjugate of any expression built from $z$'s by $+,-,\times,\div$ and real constants is the same expression in the $\bar z$'s. For example, if $p(x)$ has real coefficients then $\overline{p(z)}=p(\bar z)$. This is why the non-real roots of a real polynomial come in conjugate pairs.

Examples

Example 8. Let $z$ be a non-real complex number such that $z+\dfrac1z$ is real. Then

▸ SHOW SOLUTION

Step 1. Real means equal to its conjugate

$$z+\frac1z=\bar z+\frac1{\bar z}$$

Step 2. Factor

$$(z-\bar z)-\frac{z-\bar z}{z\bar z}=0$$$$(z-\bar z)\left(1-\frac1{|z|^2}\right)=0$$

Step 3. Use the non-real condition

$z$ is non-real, so $z-\bar z\neq0$. Hence $|z|^2=1$.

Answer: option (A)

Example 9 (IITianForum). For non-zero complex numbers $z_1,z_2$, the minimum value of $\dfrac{z_1\bar z_2+\bar z_1z_2+z_1z_2+\bar z_1\bar z_2}{|z_1z_2|}$ is ______.

▸ SHOW SOLUTION

Step 1. Factor the numerator

$$z_1\bar z_2+\bar z_1z_2+z_1z_2+\bar z_1\bar z_2$$$$=(z_1+\bar z_1)(z_2+\bar z_2)$$

Step 2. Use $z+\bar z=2\,\mathrm{Re}(z)$

With $z_k=r_k(\cos\theta_k+i\sin\theta_k)$:

$$(z_1+\bar z_1)(z_2+\bar z_2)=4r_1r_2\cos\theta_1\cos\theta_2$$

Step 3. The ratio

$$\frac{4r_1r_2\cos\theta_1\cos\theta_2}{r_1r_2}=4\cos\theta_1\cos\theta_2$$

This lies in $[-4,4]$ and equals $-4$ for $\theta_1=0$, $\theta_2=\pi$, for example $z_1=1$, $z_2=-1$.

Answer: $-4$

Example 10 (JEE Main 2025, 7 April Shift 1). Among the statements

(S1): The set $\left\{z\in\mathbb C-\{-i\}:|z|=1\text{ and }\dfrac{z-i}{z+i}\text{ is purely real}\right\}$ contains exactly two elements, and

(S2): The set $\left\{z\in\mathbb C-\{-1\}:|z|=1\text{ and }\dfrac{z-1}{z+1}\text{ is purely imaginary}\right\}$ contains infinitely many elements,

▸ SHOW SOLUTION

Step 1. Statement S1

Let $w=\frac{z-i}{z+i}$ be real. Then $w\neq1$, and solving for $z$:

$$z=\frac{i(1+w)}{1-w}$$

which is purely imaginary or zero. On $|z|=1$ this leaves $z=\pm i$. $z=-i$ is excluded, so only $z=i$ remains. S1 is false.

Step 2. Statement S2

For $|z|=1$, $\bar z=\frac1z$. Let $u=\frac{z-1}{z+1}$:

$$\bar u=\frac{\frac1z-1}{\frac1z+1}=\frac{1-z}{1+z}=-u$$

So $u$ is purely imaginary (or $0$ at $z=1$) for every $z\neq-1$ on the circle. There are infinitely many such $z$. S2 is true.

Answer: option (3)

Example 11 (JEE Advanced 2016, Paper 2). Let $a,b\in\mathbb R$ and $a^2+b^2\neq0$. Suppose $S=\left\{z\in\mathbb C:z=\dfrac{1}{a+ibt},\ t\in\mathbb R,\ t\neq0\right\}$, where $i=\sqrt{-1}$. If $z=x+iy$ and $z\in S$, then $(x,y)$ lies on

▸ SHOW SOLUTION

Step 1. Real and imaginary parts

$$z=\frac{a-ibt}{a^2+b^2t^2}$$$$x=\frac{a}{a^2+b^2t^2},\qquad y=\frac{-bt}{a^2+b^2t^2}$$

Step 2. Case $a\neq0$, $b\neq0$

$$x^2+y^2=\frac{1}{a^2+b^2t^2}=\frac xa$$

This is the circle with centre $\left(\frac1{2a},0\right)$ and radius $\frac1{2|a|}$. For $a>0$ this is (A). For $a<0$ the centre is $\left(\frac1{2a},0\right)$, not $\left(-\frac1{2a},0\right)$, so (B) is false.

Step 3. The two special cases

$b=0$: $y=0$ and $x=\frac1a$, a point of the $x$-axis, so (C) holds. $a=0$: $x=0$, so $z$ is on the $y$-axis, and (D) holds.

Answer: (A), (C) and (D)

Example 12 (JEE Advanced 2026, Paper 2). Let $\mathbb R$ denote the set of all real numbers and let $i=\sqrt{-1}$. Consider the matrices $S=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and $T=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. Let $a,b,c,d$ be real numbers such that $ST=\begin{bmatrix}a&b\\c&d\end{bmatrix}$. Let $H=\{x+iy:x,y\in\mathbb R\text{ and }y>0\}$. Then which of the following statements is (are) TRUE?

▸ SHOW SOLUTION

Step 1. The matrix

$$ST=\begin{bmatrix}0&-1\\1&1\end{bmatrix}$$

So $a=0$, $b=-1$, $c=1$, $d=1$.

Step 2. Statement (1)

$$\frac{b+ia}{d+ic}=\frac{-1}{1+i}=\frac{-1+i}{2}\neq i$$

False.

Step 3. Statement (2)

$\omega+1=-\omega^2$, so

$$\frac{a\omega+b}{c\omega+d}=\frac{-1}{\omega+1}=\frac{1}{\omega^2}=\omega$$

True.

Step 4. Statement (3)

Let $A=ST$. Then $A^2=\begin{bmatrix}-1&-1\\1&0\end{bmatrix}$ and $A^3=-I$, so $A^6=I$. $A^m=A^2$ holds exactly when $m-2$ is a multiple of $6$, for example $m=14$, which is not a multiple of $8$. False.

Step 5. Statement (4)

$\frac{az+b}{cz+d}=\frac{-1}{z+1}$. Multiply by $\overline{z+1}$:

$$\frac{-1}{z+1}=\frac{-(\bar z+1)}{|z+1|^2}$$$$\mathrm{Im}\left(\frac{-1}{z+1}\right)=\frac{\mathrm{Im}(z)}{|z+1|^2}>0$$

True.

Answer: (2) and (4)

Modulus

The Argand plane shows $z=x+iy$ as the point $(x,y)$. The modulus of $z$ is its distance from the origin:

$$|z|=\sqrt{x^2+y^2}$$

$|z_1-z_2|$ is the distance between the points $z_1$ and $z_2$.

The point z equals x plus iy at distance r from O, making angle theta with the real axis

Properties

The product rule comes from $|z_1z_2|^2=z_1z_2\,\bar z_1\bar z_2=|z_1|^2|z_2|^2$. The sum rule comes from

$$|z_1+z_2|^2=(z_1+z_2)(\bar z_1+\bar z_2)$$ $$=|z_1|^2+|z_2|^2+z_1\bar z_2+\bar z_1z_2$$

and $z_1\bar z_2+\bar z_1z_2=2\,\mathrm{Re}(z_1\bar z_2)$. Adding the $+$ and $-$ forms gives the parallelogram law: the sum of the squares of the diagonals equals the sum of the squares of the four sides.

Parallelogram with sides z1 and z2 and diagonals z1 plus z2 and z1 minus z2

Triangle inequality

$$|z_1+z_2|\le|z_1|+|z_2|$$ $$\big||z_1|-|z_2|\big|\le|z_1-z_2|$$

Reason: $\mathrm{Re}(z_1\bar z_2)\le|z_1\bar z_2|=|z_1||z_2|$, so

$$|z_1+z_2|^2\le|z_1|^2+|z_2|^2+2|z_1||z_2|$$

In the figure, the sides of triangle $O,z_1,z_1+z_2$ have lengths $|z_1|$, $|z_2|$, $|z_1+z_2|$. The general form is $|z_1+z_2+\dots+z_n|\le|z_1|+|z_2|+\dots+|z_n|$.

Equality cases. $|z_1+z_2|=|z_1|+|z_2|$ holds exactly when $z_1$ and $z_2$ point the same way, that is $z_1\bar z_2$ is real and non-negative ($\arg z_1=\arg z_2$ for non-zero numbers). $|z_1-z_2|=\big||z_1|-|z_2|\big|$ holds under the same condition. $|z_1+z_2|=|z_1-z_2|$ holds exactly when $\mathrm{Re}(z_1\bar z_2)=0$, that is $\arg z_1-\arg z_2=\pm\frac{\pi}{2}$.

Examples

Example 13 (JEE Main 2026, 24 January Shift 2). Let $z=(1+i)(1+2i)(1+3i)\cdots(1+ni)$, where $i=\sqrt{-1}$. If $|z|^2=44200$, then $n$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Modulus of a product

$$|z|^2=\prod_{r=1}^{n}|1+ri|^2=\prod_{r=1}^{n}\left(1+r^2\right)$$

Step 2. Multiply until 44200

The factors are $2,5,10,17,26,\dots$:

$$2\cdot5\cdot10\cdot17\cdot26=44200$$

So $n=5$.

Answer: $5$

Example 14 (JEE Main 2019, 11 January Shift 2). Let $z$ be a complex number such that $|z|+z=3+i$, where $i=\sqrt{-1}$. Then $|z|$ is equal to

▸ SHOW SOLUTION

Step 1. Equate parts

Let $z=x+iy$. $|z|$ is real, so the imaginary part gives $y=1$, and the real part gives

$$\sqrt{x^2+1}+x=3$$

Step 2. Solve

$$x^2+1=(3-x)^2=9-6x+x^2$$$$x=\frac43$$$$|z|=3-x=\frac53$$

Answer: option (C)

Example 15 (IITianForum). Complex numbers $z_1,z_2$ satisfy $|z_1|=|z_1+z_2|=2$ and $|z_1-z_2|=2\sqrt3$. Then

▸ SHOW SOLUTION

Step 1. Parallelogram law

$$|z_1+z_2|^2+|z_1-z_2|^2=2\left(|z_1|^2+|z_2|^2\right)$$$$4+12=2\left(4+|z_2|^2\right)\Rightarrow|z_2|=2$$

Step 2. The real part

$$|z_1+z_2|^2=|z_1|^2+|z_2|^2+2\,\mathrm{Re}(z_1\bar z_2)$$$$4=8+2\,\mathrm{Re}(z_1\bar z_2)\Rightarrow\mathrm{Re}(z_1\bar z_2)=-2$$

Step 3. The angle

$\mathrm{Re}(z_1\bar z_2)=|z_1||z_2|\cos\phi$, where $\phi$ is the angle between them:

$$\cos\phi=\frac{-2}{4}=-\frac12$$

So $\phi=\frac{2\pi}{3}$ and $\arg\frac{z_2}{z_1}=\pm\frac{2\pi}{3}$.

Answer: (A), (C) and (D)

Example 16 (AIEEE 2009). If $\left|z-\dfrac4z\right|=2$, then the maximum value of $|z|$ is equal to

▸ SHOW SOLUTION

Step 1. Triangle inequality

$$|z|=\left|\left(z-\frac4z\right)+\frac4z\right|\le2+\frac4{|z|}$$

Step 2. Solve for $r=|z|$

$$r^2-2r-4\le0\Rightarrow r\le1+\sqrt5$$

Step 3. The bound is reached

Equality needs $z-\frac4z$ and $\frac4z$ in the same direction. Take $z=i(1+\sqrt5)$: then $\frac4z=-i(\sqrt5-1)$ and $z-\frac4z=2i$, which has modulus $2$.

Answer: option (A)

Example 17 (IITianForum). If $7z^8+6iz^7+6iz-7=0$, then

▸ SHOW SOLUTION

Step 1. Isolate $z^7$

$$z^7(7z+6i)=7-6iz\Rightarrow z^7=\frac{7-6iz}{7z+6i}$$

Step 2. Compare the two moduli

Let $z=x+iy$. Expanding both:

$$|7-6iz|^2=49+36|z|^2+84y$$$$|7z+6i|^2=49|z|^2+36+84y$$

The difference is $13\left(1-|z|^2\right)$.

Step 3. Rule out the other cases

If $|z|>1$: then $|z^7|>1$, but the difference is negative, so the quotient has modulus less than $1$. Contradiction. If $|z|<1$: then $|z^7|<1$, but the difference is positive, so the quotient has modulus greater than $1$. Contradiction again. So $|z|=1$.

Answer: option (A)

Example 18 (JEE Advanced 2013, Paper 1). Let complex numbers $\alpha$ and $\dfrac1{\bar\alpha}$ lie on the circles $(x-x_0)^2+(y-y_0)^2=r^2$ and $(x-x_0)^2+(y-y_0)^2=4r^2$ respectively. If $z_0=x_0+iy_0$ satisfies the equation $2|z_0|^2=r^2+2$, then $|\alpha|=$

▸ SHOW SOLUTION

Step 1. Write both conditions with $z\bar z$

Let $a=|\alpha|^2$, $p=\mathrm{Re}(\alpha\bar z_0)$ and $c=|z_0|^2$. Since $\frac1{\bar\alpha}=\frac{\alpha}{a}$:

$$a-2p+c=r^2$$$$\frac1a-\frac{2p}{a}+c=4r^2$$

Step 2. Remove $p$

From the first, $2p=a+c-r^2$. Put it in the second (multiplied by $a$):

$$1-a-c+r^2=4ar^2-ac$$

Step 3. Use $c=\frac{r^2+2}{2}$

$$\frac{r^2}{2}-a=\frac{7ar^2}{2}-a\Rightarrow a=\frac17$$

So $|\alpha|=\frac1{\sqrt7}$.

Answer: option (C)

Argument and principal argument

For $z\neq0$, an argument of $z$ is an angle $\theta$ that the line from $O$ to $z$ makes with the positive real axis. In the figure of the last section,

$$\cos\theta=\frac{x}{|z|},\qquad \sin\theta=\frac{y}{|z|}$$

If $\theta$ is an argument, so is $\theta+2k\pi$ for every integer $k$. The one value in $(-\pi,\pi]$ is the principal argument, written $\arg z$ or $\mathrm{Arg}\,z$. The argument of $0$ is not defined.

Finding the principal argument

First find the acute angle $\alpha=\tan^{-1}\left|\frac yx\right|$, then place it by quadrant.

Principal argument in the four quadrants built from the acute angle alpha

Some books use the interval $[0,2\pi)$. Then the third and fourth quadrant values become $\pi+\alpha$ and $2\pi-\alpha$. A JEE question always states its convention; the default is $(-\pi,\pi]$.

Rules for arguments

$$\arg(z_1z_2)=\arg z_1+\arg z_2+2k\pi$$ $$\arg\frac{z_1}{z_2}=\arg z_1-\arg z_2+2k\pi$$ $$\arg\bar z=-\arg z$$ $$\arg(-z)=\arg z\pm\pi$$ $$\arg\left(z^n\right)=n\arg z+2k\pi$$

The integer $k$ is chosen to bring the answer back into $(-\pi,\pi]$. The first rule follows from polar form in the next section: moduli multiply and angles add. The rule for $\bar z$ needs $z$ not on the negative real axis (there $\arg\bar z=\arg z=\pi$).

Examples

Example 19. Let $z_1=-1+i$ and $z_2=-1+i\sqrt3$. Then the principal argument of $z_1z_2$ is

▸ SHOW SOLUTION

Step 1. Each argument

$z_1$ is in the second quadrant with $\alpha=\frac{\pi}{4}$, so $\arg z_1=\frac{3\pi}{4}$. $z_2$ is in the second quadrant with $\alpha=\frac{\pi}{3}$, so $\arg z_2=\frac{2\pi}{3}$.

Step 2. Add and reduce

$$\frac{3\pi}{4}+\frac{2\pi}{3}=\frac{17\pi}{12}>\pi$$$$\arg(z_1z_2)=\frac{17\pi}{12}-2\pi=-\frac{7\pi}{12}$$

Check: $z_1z_2=(1-\sqrt3)-i(1+\sqrt3)$ is in the third quadrant.

Answer: option (B)

Example 20 (JEE Main 2013). If $z$ is a complex number of unit modulus and argument $\theta$, then $\arg\left(\dfrac{1+z}{1+\bar z}\right)$ equals

▸ SHOW SOLUTION

Step 1. Use $|z|=1$

$z\bar z=1$, so $\bar z=\frac1z$:

$$1+\bar z=1+\frac1z=\frac{z+1}{z}$$

Step 2. Simplify

$$\frac{1+z}{1+\bar z}=\frac{(1+z)z}{1+z}=z$$

So the argument is $\theta$.

Answer: option (C)

Example 21 (JEE Advanced 2018, Paper 1). For a non-zero complex number $z$, let $\arg(z)$ denote the principal argument with $-\pi<\arg(z)\le\pi$. Then, which of the following statement(s) is (are) FALSE?

▸ SHOW SOLUTION

Step 1. Statement (A)

$-1-i$ is in the third quadrant with $\alpha=\frac{\pi}{4}$, so $\arg(-1-i)=-\frac{3\pi}{4}$. (A) is false.

Step 2. Statement (B)

$-1+it$ moves along the line $x=-1$. For $t>0$ it is in the second quadrant and $f(t)\to\pi$ as $t\to0^+$. For $t<0$ it is in the third quadrant and $f(t)\to-\pi$ as $t\to0^-$. So $f$ jumps at $t=0$. (B) is false.

Step 3. Statement (C)

$\arg\frac{z_1}{z_2}=\arg z_1-\arg z_2+2k\pi$, so the expression equals $2k\pi$. (C) is true.

Step 4. Statement (D)

The argument $\pi$ means the expression is a negative real number. Then $\frac{z-z_1}{z-z_3}$ and $\frac{z_2-z_1}{z_2-z_3}$ have arguments differing by $\pi$, so $z$ and $z_2$ see the segment $z_1z_3$ under supplementary angles from opposite sides. The locus is an arc of the circle through $z_1,z_2,z_3$, not a line. (D) is false.

Answer: (A), (B) and (D)

Polar and Euler form

With $r=|z|$ and $\theta$ an argument, $x=r\cos\theta$ and $y=r\sin\theta$, so

$$z=r(\cos\theta+i\sin\theta)$$

This is the polar form. Writing $\cos\theta+i\sin\theta=e^{i\theta}$ (Euler's formula) gives the Euler form

$$z=re^{i\theta}$$

From Euler's formula: $e^{i\pi}=-1$, $e^{i\pi/2}=i$, $\left|e^{i\theta}\right|=1$, and

$$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2},\qquad \sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$

Products and quotients

For $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$:

$$z_1z_2=r_1r_2\,e^{i(\theta_1+\theta_2)}$$ $$\frac{z_1}{z_2}=\frac{r_1}{r_2}\,e^{i(\theta_1-\theta_2)}$$

Reason: expand $(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)$ and use the addition formulas for $\cos$ and $\sin$. So multiplying by $z_2$ scales by $r_2$ and rotates by $\theta_2$. In particular, multiplying by $i=e^{i\pi/2}$ is a quarter turn, and multiplying by $e^{i\alpha}$ is a rotation through $\alpha$.

Rays to z1, z2 and z1 z2; the angle of the product is the sum of the two angles

For a unit complex number, $|z|=1$:

$$\bar z=\frac1z=e^{-i\theta},\qquad \frac{z}{\bar z}=e^{2i\theta}$$

Square roots of a complex number

Let $z=a+ib$ with $b\neq0$. Its square roots are $\pm(x+iy)$ where

$$x=\sqrt{\frac{|z|+a}{2}},\qquad y=\sqrt{\frac{|z|-a}{2}}$$

with $x$ and $y$ of the same sign when $b>0$ and of opposite signs when $b<0$.

Reason: $(x+iy)^2=a+ib$ gives $x^2-y^2=a$ and $2xy=b$. Also $|x+iy|^2=|z|$, so $x^2+y^2=|z|$. Adding and subtracting give $x^2$ and $y^2$, and $2xy=b$ fixes the signs. When $b=0$: $\sqrt a$ for $a>0$ is real, and for $a<0$ the roots are $\pm i\sqrt{|a|}$.

In polar form the two roots of $re^{i\theta}$ are

$$\pm\sqrt r\,e^{i\theta/2}$$ Complex number z and its two square roots w and minus w at half the angle

Examples

Example 22. For $\pi<\alpha<2\pi$, the principal argument of $1+\cos\alpha+i\sin\alpha$ is

▸ SHOW SOLUTION

Step 1. Half-angle formulas

$$1+\cos\alpha=2\cos^2\frac{\alpha}{2}$$$$\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$$$$1+\cos\alpha+i\sin\alpha$$$$=2\cos\frac{\alpha}{2}\left(\cos\frac{\alpha}{2}+i\sin\frac{\alpha}{2}\right)$$

Step 2. Make the modulus positive

$\frac{\alpha}{2}\in\left(\frac{\pi}{2},\pi\right)$, so $\cos\frac{\alpha}{2}<0$. Use $-e^{iu}=e^{i(u-\pi)}$:

$$=-2\cos\frac{\alpha}{2}\,e^{i\left(\frac{\alpha}{2}-\pi\right)}$$

Now $-2\cos\frac{\alpha}{2}>0$ and $\frac{\alpha}{2}-\pi\in\left(-\frac{\pi}{2},0\right)$, inside $(-\pi,\pi]$.

Answer: option (B)

Example 23. The square roots of $-15-8i$ are

▸ SHOW SOLUTION

Step 1. Modulus

$$|z|=\sqrt{225+64}=17$$

Step 2. The formula

$$x=\sqrt{\frac{17-15}{2}}=1$$$$y=\sqrt{\frac{17+15}{2}}=4$$

Step 3. Signs

$b=-8<0$, so $x$ and $y$ have opposite signs. The roots are $\pm(1-4i)$.

Check: $(1-4i)^2=1-8i-16=-15-8i$.

Answer: option (A)

Example 24 (IITianForum). Let $E=\left(\sqrt3+1+i\left(\sqrt3-1\right)\right)^n$, $n\in\mathbb N$. For the least $n$ for which $E$ is a positive real number, $\log_2E$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Modulus

$$\left(\sqrt3+1\right)^2+\left(\sqrt3-1\right)^2=8$$$$r=2\sqrt2$$

Step 2. Argument

The number is in the first quadrant with

$$\tan\theta=\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3\Rightarrow\theta=\frac{\pi}{12}$$

Step 3. Positive real power

$$E=\left(2\sqrt2\right)^n e^{in\pi/12}$$

$E$ is a positive real exactly when $\frac{n\pi}{12}$ is a multiple of $2\pi$, that is $24\mid n$. Least $n=24$:

$$E=\left(2^{3/2}\right)^{24}=2^{36}$$

Answer: $36$

Example 25 (JEE Main 2025, 22 January Shift 1). Let $z_1,z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $\arg(z_1)=-\dfrac{\pi}{4}$, $\arg(z_2)=0$ and $\arg(z_3)=\dfrac{\pi}{4}$. If $\left|z_1\bar z_2+z_2\bar z_3+z_3\bar z_1\right|^2=\alpha+\beta\sqrt2$, $\alpha,\beta\in\mathbb Z$, then the value of $\alpha^2+\beta^2$ is

▸ SHOW SOLUTION

Step 1. Euler form

$z_1=e^{-i\pi/4}$, $z_2=1$, $z_3=e^{i\pi/4}$, and $\bar z=e^{-i\theta}$ on the unit circle.

$$z_1\bar z_2=e^{-i\pi/4},\qquad z_2\bar z_3=e^{-i\pi/4}$$$$z_3\bar z_1=e^{i\pi/2}=i$$

Step 2. The sum

$$2e^{-i\pi/4}+i=\sqrt2-i\sqrt2+i$$$$=\sqrt2+i\left(1-\sqrt2\right)$$

Step 3. Modulus squared

$$2+\left(1-\sqrt2\right)^2=5-2\sqrt2$$

So $\alpha=5$, $\beta=-2$, and $\alpha^2+\beta^2=29$.

Answer: option (4)

Example 26 (IITianForum). A complex number $z$ satisfies $|z|^2+\dfrac{9}{|z|^2}-3\left(\dfrac{z}{\bar z}+\dfrac{\bar z}{z}\right)-8=0$. If the maximum value of $|z|$ is $\sqrt a+\sqrt b$, where $a,b\in\mathbb N$, then $a+b$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Euler form

Let $z=re^{i\theta}$. Then $\frac{z}{\bar z}=e^{2i\theta}$ and

$$\frac{z}{\bar z}+\frac{\bar z}{z}=2\cos2\theta$$

Step 2. The equation

$$r^2+\frac{9}{r^2}=8+6\cos2\theta\le14$$

Step 3. Largest $r$

$r^2+\frac9{r^2}$ increases for $r>\sqrt3$, so $r$ is largest when the right side is $14$:

$$r^4-14r^2+9=0\Rightarrow r^2=7+2\sqrt{10}$$$$r=\sqrt5+\sqrt2$$

because $\left(\sqrt5+\sqrt2\right)^2=7+2\sqrt{10}$. So $a+b=7$.

Answer: $7$

Solving equations in $z$

Three standard methods:

1. Put $z=x+iy$ and equate real and imaginary parts. This gives two real equations.
2. Take the modulus of both sides first. For example, $4z^2=-\bar z$ gives $4|z|^2=|z|$.
3. Take the conjugate of the equation and solve the pair as two linear equations in $z$ and $\bar z$.

For method 3, the equation $sz+t\bar z+r=0$ with its conjugate $\bar t z+\bar s\bar z+\bar r=0$ gives, after removing $\bar z$,

$$z\left(|s|^2-|t|^2\right)=t\bar r-\bar s r$$

So if $|s|\neq|t|$ there is exactly one solution. If $|s|=|t|$ there is either no solution or a whole line of solutions.

Examples

Example 27 (JEE Main 2026, 22 January Shift 2). Let $S=\{z\in\mathbb C:4z^2+\bar z=0\}$. Then $\displaystyle\sum_{z\in S}|z|^2$ is equal to

▸ SHOW SOLUTION

Step 1. Take the modulus

$$4|z|^2=|z|\Rightarrow|z|=0\ \text{or}\ |z|=\frac14$$

Step 2. The non-zero solutions

Let $z=\frac14e^{i\theta}$. Then $4z^2=\frac14e^{2i\theta}$ and $\bar z=\frac14e^{-i\theta}$:

$$e^{2i\theta}=-e^{-i\theta}\Rightarrow e^{3i\theta}=-1$$

So $\theta=\frac{\pi}{3},\pi,\frac{5\pi}{3}$, three solutions, each with $|z|^2=\frac1{16}$.

The solution 0 and three solutions on the circle of radius one quarter

Step 3. Add

$$0+3\cdot\frac{1}{16}=\frac{3}{16}$$

Answer: option (1)

Example 28 (IIT JEE 2009, Paper 1). Let $z=x+iy$ be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation $z\bar z^3+\bar z z^3=350$ is

▸ SHOW SOLUTION

Step 1. Factor with $z\bar z$

$$z\bar z\left(\bar z^2+z^2\right)=350$$

$z\bar z=x^2+y^2$ and $z^2+\bar z^2=2\left(x^2-y^2\right)$:

$$\left(x^2+y^2\right)\left(x^2-y^2\right)=175$$

Step 2. Integer factor pairs

$x^2+y^2>x^2-y^2>0$ and both have the same parity. From $175=175\cdot1=35\cdot5=25\cdot7$, only $25$ and $7$ give squares:

$$x^2=16,\qquad y^2=9$$

Step 3. The rectangle

The roots are $\pm4\pm3i$, a rectangle with sides $8$ and $6$. Area $=48$.

Answer: option (A)

Example 29 (JEE Main 2026, 6 April Shift 1). Let the set of all values of $k\in\mathbb R$ such that the equation $z(\bar z+2+i)+k(2+3i)=0$, $z\in\mathbb C$, has at least one solution, be the interval $[\alpha,\beta]$. Then $9(\alpha+\beta)$ is equal to

▸ SHOW SOLUTION

Step 1. Put $z=x+iy$

$$z\bar z+z(2+i)$$$$=(x^2+y^2+2x-y)+i(x+2y)$$

Step 2. Equate parts

$$x^2+y^2+2x-y+2k=0$$$$x+2y+3k=0$$

A circle and a line.

Step 3. Line must meet the circle

Centre $\left(-1,\frac12\right)$, radius squared $\frac54-2k$. Distance from the centre to the line:

$$\frac{|-1+1+3k|}{\sqrt5}=\frac{3|k|}{\sqrt5}$$$$\frac{9k^2}{5}\le\frac54-2k\Rightarrow36k^2+40k-25\le0$$

Step 4. Sum of the end points

$\alpha,\beta$ are the roots of $36k^2+40k-25=0$, so $\alpha+\beta=-\frac{10}{9}$ and $9(\alpha+\beta)=-10$.

Answer: option (1)

Example 30 (JEE Main 2025, 3 April Shift 1). Let $z\in\mathbb C$ be such that $\dfrac{z^2+3i}{z-2+i}=2+3i$. Then the sum of all possible values of $z^2$ is

▸ SHOW SOLUTION

Step 1. Clear the denominator

$$z^2+3i=(2+3i)(z-2+i)$$

$(2+3i)(-2+i)=-7-4i$, so

$$z^2-(2+3i)z+7+7i=0$$

$z=2-i$ is not a root, so both roots are allowed.

Step 2. Sum of squares of roots

$z_1+z_2=2+3i$ and $z_1z_2=7+7i$:

$$z_1^2+z_2^2=(2+3i)^2-2(7+7i)$$$$=-5+12i-14-14i=-19-2i$$

Answer: option (2)

Example 31 (IITianForum). Let $(x,y)$ be a pair of real numbers satisfying $5x-12y=\dfrac{x}{x^2+y^2}$ and $12x+5y=-\dfrac{y}{x^2+y^2}$. Then $13\left(|x|+|y|\right)$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Combine into one complex equation

Let $z=x+iy$. Take the first equation plus $i$ times the second:

$$(5x-12y)+i(12x+5y)$$$$=(5+12i)(x+iy)$$$$\frac{x-iy}{x^2+y^2}=\frac{\bar z}{|z|^2}=\frac1z$$

Step 2. Solve

$$(5+12i)z=\frac1z$$$$z^2=\frac{1}{5+12i}=\frac{1}{(3+2i)^2}$$$$z=\pm\frac{1}{3+2i}=\pm\frac{3-2i}{13}$$

Step 3. Read off

$|x|+|y|=\frac{5}{13}$, so $13\left(|x|+|y|\right)=5$.

Answer: $5$

Example 32 (JEE Advanced 2017, Paper 1). Let $a,b,x$ and $y$ be real numbers such that $a-b=1$ and $y\neq0$. If the complex number $z=x+iy$ satisfies $\mathrm{Im}\left(\dfrac{az+b}{z+1}\right)=y$, then which of the following is(are) possible value(s) of $x$?

▸ SHOW SOLUTION

Step 1. Imaginary part of the quotient

Multiply by $\overline{z+1}=(x+1)-iy$:

$$\mathrm{Im}\big((az+b)(\bar z+1)\big)$$$$=ay(x+1)-y(ax+b)=(a-b)y=y$$$$\mathrm{Im}\left(\frac{az+b}{z+1}\right)=\frac{y}{(x+1)^2+y^2}$$

Step 2. The condition

$y\neq0$, so

$$(x+1)^2+y^2=1\Rightarrow x=-1\pm\sqrt{1-y^2}$$

Answer: (B) and (D)

Example 33 (JEE Advanced 2018, Paper 2). Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+iy$ ($x,y\in\mathbb R$, $i=\sqrt{-1}$) of the equation $sz+t\bar z+r=0$, where $\bar z=x-iy$. Then, which of the following statement(s) is (are) TRUE?

▸ SHOW SOLUTION

Step 1. Conjugate equation

$$sz+t\bar z+r=0,\qquad \bar tz+\bar s\bar z+\bar r=0$$

Multiply the first by $\bar s$, the second by $t$, and subtract:

$$z\left(|s|^2-|t|^2\right)=t\bar r-\bar sr$$

Step 2. The two cases

If $|s|\neq|t|$, exactly one solution. If $|s|=|t|$, then with $z=x+iy$ the equation is one real linear equation (a line) or has no solution. For example, $s=t=1$, $r=i$ gives $2x+i=0$, no solution.

Step 3. Check each statement

(A)True: one element forces $|s|\neq|t|$.

(B)False, by the example.

(C)True: $L$ is a point, a line or empty, and a line meets a circle in at most $2$ points.

(D)True: more than one element means the line case.

Answer: (A), (C) and (D)

De Moivre's theorem

For every integer $n$,

$$(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$$

In Euler form this is $\left(e^{i\theta}\right)^n=e^{in\theta}$. Reason: for $n\ge1$ apply the product rule $n$ times, adding the angle $\theta$ each time; for negative $n$ use $\frac1{e^{i\theta}}=e^{-i\theta}$. So

$$\left(re^{i\theta}\right)^n=r^ne^{in\theta}$$

Uses

Powers. To find $(a+ib)^n$, write $a+ib$ in polar form first. Common values:

$$1+i=\sqrt2\,e^{i\pi/4},\qquad 1+i\sqrt3=2e^{i\pi/3}$$ $$\sqrt3+i=2e^{i\pi/6}$$

Multiple angles. Expanding $(\cos\theta+i\sin\theta)^n$ by the binomial theorem and comparing parts gives $\cos n\theta$ and $\sin n\theta$ as polynomials. For $n=3$:

$$\cos3\theta=4\cos^3\theta-3\cos\theta$$ $$\sin3\theta=3\sin\theta-4\sin^3\theta$$

Trigonometric sums. $\sum\cos k\theta$ and $\sum\sin k\theta$ are the real and imaginary parts of the geometric series $\sum e^{ik\theta}$.

Binomial sums. Sums like $\sum{}^nC_k\,3^{k/2}$ with alternating signs are the real or imaginary part of $\left(1+i\sqrt3\right)^n$.

Examples

Example 34 (JEE Main 2026, 23 January Shift 2). If $z=\dfrac{\sqrt3}{2}+\dfrac i2$, $i=\sqrt{-1}$, then $\left(z^{201}-i\right)^8$ is equal to

▸ SHOW SOLUTION

Step 1. Polar form

$$z=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}=e^{i\pi/6}$$

Step 2. The power

$$\frac{201\pi}{6}=32\pi+\frac{3\pi}{2}$$$$z^{201}=e^{i3\pi/2}=-i$$

Step 3. Finish

$$\left(-i-i\right)^8=(-2i)^8=2^8\,i^8=256$$

Answer: option (4)

Example 35 (IIT JEE 2009, Paper 1). Let $z=\cos\theta+i\sin\theta$. Then the value of $\displaystyle\sum_{m=1}^{15}\mathrm{Im}\left(z^{2m-1}\right)$ at $\theta=2^\circ$ is

▸ SHOW SOLUTION

Step 1. Imaginary parts

By De Moivre, $\mathrm{Im}\left(z^{2m-1}\right)=\sin(2m-1)\theta$:

$$S=\sin\theta+\sin3\theta+\dots+\sin29\theta$$

Step 2. Telescope

$2\sin\theta\sin(2m-1)\theta=\cos(2m-2)\theta-\cos2m\theta$. Adding for $m=1$ to $15$:

$$2\sin\theta\cdot S=1-\cos30\theta$$

Step 3. Put $\theta=2^\circ$

$$S=\frac{1-\cos60^\circ}{2\sin2^\circ}=\frac{1}{4\sin2^\circ}$$

Answer: option (D)

Example 36 (JEE Main 2025, 28 January Shift 1). If $\alpha=1+\displaystyle\sum_{r=1}^{6}(-3)^{r-1}\,{}^{12}C_{2r-1}$, then the distance of the point $\left(12,\sqrt3\right)$ from the line $\alpha x-\sqrt3\,y+1=0$ is ______.

▸ SHOW SOLUTION

Step 1. Recognise the sum

In $\left(1+i\sqrt3\right)^{12}=\sum{}^{12}C_k\left(i\sqrt3\right)^k$, the odd terms $k=2r-1$ give

$$\left(i\sqrt3\right)^{2r-1}=i\sqrt3\,(-3)^{r-1}$$$$\mathrm{Im}\left(1+i\sqrt3\right)^{12}$$$$=\sqrt3\sum_{r=1}^{6}(-3)^{r-1}\,{}^{12}C_{2r-1}$$

Step 2. De Moivre

$$\left(1+i\sqrt3\right)^{12}=2^{12}e^{i4\pi}=4096$$

The imaginary part is $0$, so the sum is $0$ and $\alpha=1$.

Step 3. Distance

$$\frac{\left|12-3+1\right|}{\sqrt{1+3}}=5$$

Answer: $5$

Example 37 (IITianForum). If $z$ is a non-real complex number, then the least value of $\dfrac{\mathrm{Im}\left(z^5\right)}{\left(\mathrm{Im}\,z\right)^5}$ is ______.

▸ SHOW SOLUTION

Step 1. Polar form

Let $z=r(\cos\theta+i\sin\theta)$ with $\sin\theta\neq0$. By De Moivre:

$$\frac{\mathrm{Im}\left(z^5\right)}{\left(\mathrm{Im}\,z\right)^5}=\frac{r^5\sin5\theta}{r^5\sin^5\theta}=\frac{\sin5\theta}{\sin^5\theta}$$

Step 2. Expand $\sin5\theta$

The imaginary part of $(c+is)^5$ is $5c^4s-10c^2s^3+s^5$. With $c^2=1-s^2$:

$$\sin5\theta=16s^5-20s^3+5s$$

Step 3. Minimise

With $t=\frac{1}{s^2}\ge1$:

$$\frac{\sin5\theta}{\sin^5\theta}=5t^2-20t+16=5(t-2)^2-4$$

Least value $-4$ at $t=2$, that is $\theta=\frac{\pi}{4}$.

Answer: $-4$

Example 38 (JEE Advanced 2013, Paper 2). Let $w=\dfrac{\sqrt3+i}{2}$ and $P=\{w^n:n=1,2,3,\dots\}$. Further $H_1=\left\{z\in\mathbb C:\mathrm{Re}\,z>\dfrac12\right\}$ and $H_2=\left\{z\in\mathbb C:\mathrm{Re}\,z<-\dfrac12\right\}$, where $\mathbb C$ is the set of all complex numbers. If $z_1\in P\cap H_1$, $z_2\in P\cap H_2$ and $O$ represents the origin, then $\angle z_1Oz_2=$

▸ SHOW SOLUTION

Step 1. The set $P$

$w=e^{i\pi/6}$, so $P$ is the $12$ points $e^{ik\pi/6}$, $k=0,1,\dots,11$, on the unit circle.

Step 2. Points in each half-plane

$\mathrm{Re}=\cos\frac{k\pi}{6}$. $\cos>\frac12$ only for angles $0,\pm\frac{\pi}{6}$. $\cos<-\frac12$ only for $\pi,\pm\frac{5\pi}{6}$. The points at $\pm\frac{\pi}{3}$ and $\pm\frac{2\pi}{3}$ lie on the boundary lines and are not included.

Twelve powers of w on the unit circle with the half-planes Re z greater than one half and less than minus one half

Step 3. Possible angles

The angle between a point at $a\in\left\{0,\pm\frac{\pi}{6}\right\}$ and a point at $b\in\left\{\pi,\pm\frac{5\pi}{6}\right\}$ takes the values $\frac{2\pi}{3}$, $\frac{5\pi}{6}$ and $\pi$. Among the options, $\frac{2\pi}{3}$ and $\frac{5\pi}{6}$ occur.

Answer: (C) and (D)

Cube roots of unity

The equation $z^3=1$ factors as $(z-1)\left(z^2+z+1\right)=0$. Its roots are

$$1,\qquad \omega=\frac{-1+i\sqrt3}{2}$$ $$\omega^2=\frac{-1-i\sqrt3}{2}$$

In Euler form $\omega=e^{2\pi i/3}$ and $\omega^2=e^{4\pi i/3}=e^{-2\pi i/3}$. They form an equilateral triangle on the unit circle.

The cube roots of unity 1, omega and omega squared at the vertices of an equilateral triangle on the unit circle

Properties

The sum property: $\omega$ is a root of $z^2+z+1=0$. The power sum: if $3\nmid n$, then $\omega^n$ is $\omega$ or $\omega^2$ and the sum is again $1+\omega+\omega^2$.

Factorisations

$$x^2+x+1=(x-\omega)\left(x-\omega^2\right)$$ $$x^2+xy+y^2=(x-y\omega)\left(x-y\omega^2\right)$$ $$x^3+y^3=(x+y)(x+y\omega)\left(x+y\omega^2\right)$$ $$a^3+b^3+c^3-3abc$$ $$=(a+b+c)\left(a+b\omega+c\omega^2\right)$$ $$\quad\times\left(a+b\omega^2+c\omega\right)$$

Examples

Example 39 (JEE Main 2026, 21 January Shift 1). If $x^2+x+1=0$, then the value of $\left(x+\dfrac1x\right)^4+\left(x^2+\dfrac1{x^2}\right)^4+\left(x^3+\dfrac1{x^3}\right)^4+\cdots+\left(x^{25}+\dfrac1{x^{25}}\right)^4$ is

▸ SHOW SOLUTION

Step 1. Each bracket

$x=\omega$ (or $\omega^2$; the answer is the same) and $\frac1{x^n}=\omega^{2n}$:

$$x^n+\frac1{x^n}=\omega^n+\omega^{2n}=\begin{cases}2,&3\mid n\\-1,&3\nmid n\end{cases}$$

Step 2. Count

From $1$ to $25$ there are $8$ multiples of $3$ and $17$ other numbers.

$$8\cdot2^4+17\cdot(-1)^4=128+17=145$$

Answer: option (4)

Example 40 (JEE Main 2026, 22 January Shift 1). Let $\alpha=\dfrac{-1+i\sqrt3}{2}$ and $\beta=\dfrac{-1-i\sqrt3}{2}$. If $(7-7\alpha+9\beta)^{20}+(9+7\alpha-7\beta)^{20}+(-7+9\alpha+7\beta)^{20}+(14+7\alpha+7\beta)^{20}=m^{10}$, then $m$ is ______.

▸ SHOW SOLUTION

Step 1. Rename

$\alpha=\omega$, $\beta=\omega^2$. Let $B=9+7\omega-7\omega^2$.

Step 2. First and third brackets are multiples of $B$

$$\omega^2B=9\omega^2+7-7\omega=7-7\omega+9\omega^2$$$$\omega B=9\omega+7\omega^2-7=-7+9\omega+7\omega^2$$

Step 3. Last bracket

$$14+7\left(\omega+\omega^2\right)=14-7=7$$

Step 4. Add

$$B^{20}\left(\omega^{40}+1+\omega^{20}\right)+7^{20}$$$$=B^{20}\left(\omega+1+\omega^2\right)+7^{20}=7^{20}$$

$7^{20}=49^{10}$, so $m=49$.

Answer: $49$

Example 41 (IITianForum). Let $\omega$ be a non-real cube root of unity, $p=2+3\omega+5\omega^2$, $q=3+5\omega+2\omega^2$ and $r=5+2\omega+3\omega^2$. Then

▸ SHOW SOLUTION

Step 1. Relate $q$ and $r$ to $p$

$$\omega^2p=2\omega^2+3\omega^3+5\omega^4$$$$=3+5\omega+2\omega^2=q$$$$\omega p=2\omega+3\omega^2+5\omega^3$$$$=5+2\omega+3\omega^2=r$$

So $q=\omega^2p$, not $\omega p$; (D) is false.

Step 2. The sums

$$p+q+r=p\left(1+\omega^2+\omega\right)=0$$$$p^2+q^2+r^2=p^2\left(1+\omega^4+\omega^2\right)=0$$

(A) is false and (C) is true.

Step 3. The product

$$pqr=p\cdot\omega^2p\cdot\omega p=\omega^3p^3=p^3$$

(B) is true.

Answer: (B) and (C)

Example 42 (IIT JEE 2010, Paper 1). Let $\omega$ be the complex number $\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}$. Then the number of distinct complex numbers $z$ satisfying $\begin{vmatrix}z+1&\omega&\omega^2\\\omega&z+\omega^2&1\\\omega^2&1&z+\omega\end{vmatrix}=0$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Add all rows to the first

Each column adds to $z+1+\omega+\omega^2=z$:

$$\Delta=z\begin{vmatrix}1&1&1\\\omega&z+\omega^2&1\\\omega^2&1&z+\omega\end{vmatrix}$$

Step 2. Subtract columns

Use $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$:

$$\Delta=z\begin{vmatrix}z+\omega^2-\omega&1-\omega\\1-\omega^2&z+\omega-\omega^2\end{vmatrix}$$

Step 3. Expand

$$\Delta=z\Big(z^2-\left(\omega^2-\omega\right)^2$$$$\quad-(1-\omega)\left(1-\omega^2\right)\Big)$$

$\left(\omega^2-\omega\right)^2=\omega^4-2\omega^3+\omega^2=-3$ and $(1-\omega)\left(1-\omega^2\right)=3$, so

$$\Delta=z^3$$

$\Delta=0$ only for $z=0$: one value.

Answer: $1$

Example 43 (JEE Advanced 2016, Paper 1). Let $z=\dfrac{-1+\sqrt3\,i}{2}$, where $i=\sqrt{-1}$, and $r,s\in\{1,2,3\}$. Let $P=\begin{bmatrix}(-z)^r&z^{2s}\\z^{2s}&z^r\end{bmatrix}$ and $I$ be the identity matrix of order $2$. Then the total number of ordered pairs $(r,s)$ for which $P^2=-I$ is ______.

▸ SHOW SOLUTION

Step 1. Square the matrix

$z=\omega$. The off-diagonal entries of $P^2$ are

$$z^{2s}\left((-z)^r+z^r\right)=\omega^{2s}\omega^r\left((-1)^r+1\right)$$

These are $0$ only when $r$ is odd: $r=1$ or $3$.

Step 2. Diagonal entries

$$z^{2r}+z^{4s}=\omega^{2r}+\omega^{4s}=-1$$

$r=1$: $\omega^2+\omega^{s}=-1$ needs $\omega^s=\omega$, so $s=1$. $r=3$: $1+\omega^s=-1$ is impossible because $|\omega^s|=1$.

So only $(r,s)=(1,1)$.

Answer: $1$

Example 44 (IITianForum). The sum of the squares of the moduli of the non-real roots of $z^5-2z^4+z^3-2z^2+z-2=0$ is ______.

▸ SHOW SOLUTION

Step 1. Group in pairs

$$z^4(z-2)+z^2(z-2)+(z-2)$$$$=(z-2)\left(z^4+z^2+1\right)$$

Step 2. Factor the quartic

$$z^4+z^2+1=\left(z^2+1\right)^2-z^2$$$$=\left(z^2+z+1\right)\left(z^2-z+1\right)$$

The non-real roots are $\omega,\omega^2,-\omega,-\omega^2$.

Step 3. Add

Each has modulus $1$, so the sum is $4$.

Answer: $4$

Example 45 (JEE Advanced 2026, Paper 1). Match each entry in List-I to the correct entry in List-II.

The correct option is

▸ SHOW SOLUTION

Step 1. Item P

$\alpha=\omega$, $\beta=\omega^2$, so $\alpha+1=-\omega^2$ and $\beta+1=-\omega$. Even power: the signs go.

$$\frac{1}{\omega^{4052}}=\frac1{\omega^2}=\omega,\qquad \frac{1}{\omega^{2026}}=\frac1\omega=\omega^2$$

Roots $\omega,\omega^2$: $x^2+x+1=0$. P→(1).

Step 2. Item Q

Odd power: each term changes sign, giving $-\omega^2$ and $-\omega$. Sum $1$, product $1$: $x^2-x+1=0$. Q→(2).

Step 3. Item R

$\gamma=-\omega$, $\delta=-\omega^2$. Then $\gamma-1=-(1+\omega)=\omega^2$ and $\delta-1=\omega$:

$$\frac{1}{\omega^{4052}}+\frac{1}{\omega^{2026}}=\omega+\omega^2=-1$$

R→(4).

Step 4. Item S

$p^2+p=1$ gives $\frac1{p+1}=p$. So the sum is $p^3+r^3$ with $p+r=-1$, $pr=-1$:

$$p^3+r^3=(p+r)^3-3pr(p+r)$$$$=-1-3=-4$$

S→(5).

Answer: option (3)

Example 46 (JEE Advanced 2025, Paper 2). For a non-zero complex number $z$, let $\arg(z)$ denote the principal argument of $z$, with $-\pi<\arg(z)\le\pi$. Let $\omega$ be the cube root of unity for which $0<\arg(\omega)<\pi$. Let $\alpha=\arg\left(\displaystyle\sum_{n=1}^{2025}(-\omega)^n\right)$. Then the value of $\dfrac{3\alpha}{\pi}$ is ______.

▸ SHOW SOLUTION

Step 1. The base

$$-\omega=\frac12-\frac{\sqrt3}{2}i=e^{-i\pi/3}$$

Its powers repeat with period $6$, and any six consecutive powers add to $0$.

Step 2. Reduce the sum

$2025=6\cdot337+3$, so only the last three terms remain:

$$e^{-i\pi/3}+e^{-2i\pi/3}+e^{-i\pi}$$$$=\left(\frac12-\frac{\sqrt3}{2}i\right)+\left(-\frac12-\frac{\sqrt3}{2}i\right)-1$$$$=-1-i\sqrt3$$

Step 3. Argument

$-1-i\sqrt3$ is in the third quadrant with $\alpha=-\left(\pi-\frac{\pi}{3}\right)=-\frac{2\pi}{3}$. So $\frac{3\alpha}{\pi}=-2$.

Answer: $-2$

$n$th roots of unity

The equation $z^n=1$ has exactly $n$ roots. Writing $1=e^{2k\pi i}$:

$$z_k=e^{2k\pi i/n},\qquad k=0,1,\dots,n-1$$

With $\alpha=e^{2\pi i/n}$ they are $1,\alpha,\alpha^2,\dots,\alpha^{n-1}$, a geometric progression. They are the vertices of a regular $n$-gon inscribed in the unit circle, one vertex at $1$, adjacent vertices $\frac{2\pi}{n}$ apart.

The seventh roots of unity at the vertices of a regular heptagon on the unit circle

Properties

For $n\ge2$:

Reasons. The sum is a GP: $\frac{1-\alpha^n}{1-\alpha}=0$. The product is $(-1)^{n-1}$ times the constant term of $z^n-1$, by Vieta. Dividing the factorisation by $z-1$:

$$1+z+\dots+z^{n-1}$$ $$=(z-\alpha)\cdots\left(z-\alpha^{n-1}\right)$$

and putting $z=1$ gives the product of differences.

$n$th roots of any complex number

The $n$ roots of $z^n=re^{i\theta}$ are

$$\sqrt[n]{r}\,e^{i(\theta+2k\pi)/n},\qquad k=0,1,\dots,n-1$$

They also form a regular $n$-gon, of radius $\sqrt[n]r$. Equivalently: one root times each $n$th root of unity.

Examples

Example 47. Let $\alpha=e^{2\pi i/7}$. Then $\displaystyle\prod_{k=1}^{6}\left(2-\alpha^k\right)$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Factorisation

$$z^7-1=(z-1)\prod_{k=1}^{6}\left(z-\alpha^k\right)$$$$\prod_{k=1}^{6}\left(z-\alpha^k\right)=1+z+z^2+\dots+z^6$$

Step 2. Put $z=2$

$$1+2+4+\dots+64=2^7-1=127$$

Answer: $127$

Example 48 (JEE Advanced 2014, Paper 2). Let $z_k=\cos\dfrac{2k\pi}{10}+i\sin\dfrac{2k\pi}{10}$, $k=1,2,\dots,9$. Match the lists.

▸ SHOW SOLUTION

Step 1. P

$z_k$ are the $10$th roots of unity other than $1$. $z_kz_{10-k}=1$ and $z_{10-k}$ is also in the list. P is true.

Step 2. Q

$z_1\neq0$, so $z=\frac{z_k}{z_1}$ always exists. Q is false.

Step 3. R

$(1-z_1)(1-z_2)\cdots(1-z_9)=10$, so the expression is $1$.

Step 4. S

The real parts of all ten roots add to $0$, so $\sum_{k=1}^{9}\cos\frac{2k\pi}{10}=-1$ and S $=2$.

Answer: option (C)

Example 49 (JEE Advanced 2015, Paper 2). For any integer $k$, let $\alpha_k=\cos\left(\dfrac{k\pi}{7}\right)+i\sin\left(\dfrac{k\pi}{7}\right)$, where $i=\sqrt{-1}$. The value of the expression $\dfrac{\displaystyle\sum_{k=1}^{12}\left|\alpha_{k+1}-\alpha_k\right|}{\displaystyle\sum_{k=1}^{3}\left|\alpha_{4k-1}-\alpha_{4k-2}\right|}$ is ______.

▸ SHOW SOLUTION

Step 1. Every term is the same

$\alpha_k=e^{ik\pi/7}$, so

$$\left|\alpha_{k+1}-\alpha_k\right|=\left|\alpha_k\right|\left|e^{i\pi/7}-1\right|$$$$=\left|e^{i\pi/7}-1\right|$$

These are the equal sides of a regular $14$-gon. Call the common length $d$.

Step 2. Count

The numerator has $12$ terms and the denominator $3$, all equal to $d$:

$$\frac{12d}{3d}=4$$

Answer: $4$

Example 50 (IITianForum). Let $\alpha_1,\alpha_2,\dots,\alpha_{12}$ be all the $12$th roots of unity, and $S=\displaystyle\sum_{1\le i<j\le12}\left(\alpha_i\alpha_j\right)^P$. Then

▸ SHOW SOLUTION

Step 1. Write $S$ with power sums

Let $T_m=\sum_{i=1}^{12}\alpha_i^m$. Then

$$S=\frac{T_P^2-T_{2P}}{2}$$

$T_m=12$ if $12\mid m$ and $0$ otherwise.

Step 2. Each value of $P$

$P=1$: $T_1=T_2=0$, $S=0$. $P=4$: $T_4=T_8=0$, $S=0$, so (B) is false. $P=6$: $T_6=0$, $T_{12}=12$, $S=-6$. $P=12$: $T_{12}=T_{24}=12$:

$$S=\frac{144-12}{2}=66$$

Answer: (A), (C) and (D)

Example 51 (IITianForum). The number of complex numbers $z$ that satisfy both $z^{60}=1$ and $z^{84}=1$ is ______.

▸ SHOW SOLUTION

Step 1. A common root has order dividing both

$z^{60}=1$ and $z^{84}=1$. Since $\gcd(60,84)=12$, there are integers $a,b$ with $60a+84b=12$, so

$$z^{12}=\left(z^{60}\right)^a\left(z^{84}\right)^b=1$$

Step 2. Conversely

Every $12$th root of unity satisfies both, because $12\mid60$ and $12\mid84$. So there are $12$ such $z$.

Answer: $12$

Example 52 (IITianForum). Let $P=\cos\dfrac{2\pi}{7}+i\sin\dfrac{2\pi}{7}$, $a=P+P^2+P^4$ and $b=P^3+P^5+P^6$. Then $|a-b|^2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Sum

$a+b=P+P^2+\dots+P^6=-1$.

Step 2. Product

$$ab=P^4+P^6+P^7+P^5+P^7$$$$\quad+P^8+P^7+P^9+P^{10}$$

$P^7=1$, $P^8=P$, $P^9=P^2$, $P^{10}=P^3$:

$$ab=3+\left(P+P^2+\dots+P^6\right)$$$$=3-1=2$$

Step 3. Difference

$$(a-b)^2=(a+b)^2-4ab=1-8=-7$$

So $a-b=\pm i\sqrt7$ and $|a-b|^2=7$.

Answer: $7$

Quadratic equations with complex roots

For $az^2+bz+c=0$ with $a\neq0$, the roots are

$$z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

This formula holds for complex $a,b,c$ too, with $\sqrt{b^2-4ac}$ meaning either square root (found by the method of the polar form section). Vieta's relations also hold:

$$z_1+z_2=-\frac ba,\qquad z_1z_2=\frac ca$$

Real coefficients

If $a,b,c$ are real and $D=b^2-4ac<0$, the roots are the conjugate pair

$$\frac{-b\pm i\sqrt{-D}}{2a}$$

More generally, a polynomial with real coefficients has its non-real roots in conjugate pairs, because $p(\bar z)=\overline{p(z)}=0$. Then

$$z_1+z_2=2\,\mathrm{Re}(z_1),\qquad z_1z_2=|z_1|^2$$

Complex coefficients

When the coefficients are not all real, conjugate pairing fails. Use the formula with the square root of a complex number, or Vieta's relations. The sign test $D<0$ has no meaning, because $D$ is complex.

Examples

Example 53 (JEE Main 2026, 4 April Shift 2). Let $S=\{z\in\mathbb C:z^2+4z+16=0\}$. Then $\displaystyle\sum_{z\in S}\left|z+\sqrt3\,i\right|^2$ is equal to

▸ SHOW SOLUTION

Step 1. Roots

$$(z+2)^2=-12\Rightarrow z=-2\pm2\sqrt3\,i$$

Step 2. Shift by $\sqrt3\,i$

$$z+\sqrt3\,i=-2+3\sqrt3\,i\quad\text{or}\quad-2-\sqrt3\,i$$

Step 3. Moduli

$$(4+27)+(4+3)=38$$

Answer: option (4)

Example 54 (JEE Main 2026, 5 April Shift 2). Let $z_1,z_2\in\mathbb C$ be the distinct solutions of $z^2+4z-(1+12i)=0$. Then $|z_1|^2+|z_2|^2$ is equal to

▸ SHOW SOLUTION

Step 1. Discriminant

$$b^2-4ac=16+4(1+12i)=20+48i$$

Step 2. Its square root

$|20+48i|=52$. Then $x=\sqrt{\frac{52+20}{2}}=6$ and $y=\sqrt{\frac{52-20}{2}}=4$, same sign:

$$\sqrt{20+48i}=\pm(6+4i)$$

Step 3. Roots

$$z=\frac{-4\pm(6+4i)}{2}=1+2i\ \text{or}\ -5-2i$$

Step 4. Moduli

$$|z_1|^2+|z_2|^2=5+29=34$$

Answer: option (4)

Example 55 (JEE Main 2026, 5 April Shift 1). Let $a,b\in\mathbb C$. Let $\alpha,\beta$ be the roots of $x^2+ax+b=0$. If $\beta-\alpha=\sqrt{11}$ and $\beta^2-\alpha^2=3i\sqrt{11}$, then $\left(\beta^3-\alpha^3\right)^2$ is equal to

▸ SHOW SOLUTION

Step 1. Sum of the roots

$$(\beta+\alpha)(\beta-\alpha)=3i\sqrt{11}\Rightarrow\beta+\alpha=3i$$

Step 2. Product of the roots

$$4\alpha\beta=(\alpha+\beta)^2-(\beta-\alpha)^2$$$$=-9-11=-20$$

So $\alpha\beta=-5$.

Step 3. Difference of cubes

$$\beta^3-\alpha^3=(\beta-\alpha)\left((\alpha+\beta)^2-\alpha\beta\right)$$$$=\sqrt{11}\,(-9+5)=-4\sqrt{11}$$$$\left(\beta^3-\alpha^3\right)^2=16\cdot11=176$$

Answer: option (2)

Example 56 (JEE Main 2026, 4 April Shift 2). If $\alpha=1$ and $\beta=1+i\sqrt2$, where $i=\sqrt{-1}$, are two roots of the equation $x^3+ax^2+bx+c=0$, $a,b,c\in\mathbb R$, then $\displaystyle\int_{-1}^{1}\left(x^3+ax^2+bx+c\right)dx$ is equal to

▸ SHOW SOLUTION

Step 1. Third root

Real coefficients, so $1-i\sqrt2$ is also a root.

$$\left(x-1-i\sqrt2\right)\left(x-1+i\sqrt2\right)$$$$=x^2-2x+3$$

Step 2. The cubic

$$(x-1)\left(x^2-2x+3\right)$$$$=x^3-3x^2+5x-3$$

Step 3. Integrate

Odd powers give $0$ on $[-1,1]$:

$$\int_{-1}^{1}\left(-3x^2-3\right)dx=-2-6=-8$$

Answer: option (3)

Example 57 (JEE Main 2026, 6 April Shift 2). Let $S=\left\{z\in\mathbb C:z^2+\sqrt6\,iz-3=0\right\}$. Then $\displaystyle\sum_{z\in S}z^8$ is equal to

▸ SHOW SOLUTION

Step 1. Roots

$$z=\frac{-\sqrt6\,i\pm\sqrt{-6+12}}{2}=\frac{\sqrt6}{2}(\pm1-i)$$

Step 2. Eighth powers

$(1-i)^2=-2i$ and $(-1-i)^2=2i$, so both eighth powers are $16$:

$$(1-i)^8=(-2i)^4=16$$$$(-1-i)^8=(2i)^4=16$$

Step 3. Add

$$\left(\frac{\sqrt6}{2}\right)^8=\frac{6^4}{2^8}=\frac{81}{16}$$$$\sum z^8=\frac{81}{16}(16+16)=162$$

Answer: option (1)

Example 58 (AIEEE 2011). Let $\alpha,\beta$ be real and $z$ be a complex number. If $z^2+\alpha z+\beta=0$ has two distinct roots on the line $\mathrm{Re}\,z=1$, then it is necessary that

▸ SHOW SOLUTION

Step 1. The roots are a conjugate pair

Two distinct roots with real part $1$ cannot both be real (they would both equal $1$). Real coefficients, so the roots are $1\pm iy$ with $y\neq0$.

Step 2. Product

$$\beta=(1+iy)(1-iy)=1+y^2>1$$

Answer: option (D)

Complex numbers in geometry: a short summary

Every result of this section is developed with figures and many more examples in the Complex Numbers in Coordinate Geometry note. The list below is for quick recall.

For $|z|$ over a region: the least and greatest distance from a point $P$ to a circle with centre $C$ and radius $r$ are $|PC-r|$ and $PC+r$.

Examples

Example 59 (JEE Main 2026, 21 January Shift 2). Let $z$ be the complex number satisfying $|z-5|\le3$ and having maximum positive principal argument. Then $34\left|\dfrac{5z-12}{5iz+16}\right|^2$ is equal to

▸ SHOW SOLUTION

Step 1. The extreme point

The disc has centre $5$ and radius $3$. The argument is greatest at the upper tangent point $z$ from $O$.

Disc of radius 3 centred at 5 with the tangent from the origin touching it at z

Step 2. Locate it

$|z|=\sqrt{25-9}=4$ and $\sin(\arg z)=\frac35$, so $\cos(\arg z)=\frac45$:

$$z=4\left(\frac45+\frac35i\right)=\frac{16+12i}{5}$$

Step 3. Evaluate

$5z-12=4+12i$ and $5iz+16=4+16i$:

$$34\cdot\frac{16+144}{16+256}=34\cdot\frac{160}{272}=20$$

Answer: option (4)

Example 60 (JEE Advanced 2025, Paper 1). Let $\mathbb R$ denote the set of all real numbers. Let $z_1=1+2i$ and $z_2=3i$ be two complex numbers, where $i=\sqrt{-1}$. Let $S=\{(x,y)\in\mathbb R\times\mathbb R:|x+iy-z_1|=2|x+iy-z_2|\}$. Then which of the following statements is (are) TRUE?

▸ SHOW SOLUTION

Step 1. Square both sides

$$(x-1)^2+(y-2)^2=4\left(x^2+(y-3)^2\right)$$$$3x^2+3y^2+2x-20y+31=0$$

Step 2. Centre and radius

Divide by $3$: centre $\left(-\frac13,\frac{10}{3}\right)$ and

$$r^2=\frac19+\frac{100}{9}-\frac{31}{3}=\frac89\Rightarrow r=\frac{2\sqrt2}{3}$$Apollonius circle for the points z1 and z2 with ratio 2, and its centre C

Answer: (A) and (D)

Which method to use

What JEE asks

Across the JEE Main papers from 2009 to 2026 and the JEE Advanced and IIT JEE papers, the chapter appears in these forms.

1. Algebra, equality and real or imaginary parts (Examples 4 to 7, 10, 12 and 32).
2. Modulus, conjugate and inequalities (Examples 8, 9, 11 and 13 to 18).
3. Argument and polar form (Examples 19 to 22, 24 to 26 and 46).
4. Equations in $z$ and $\bar z$ (Examples 27 to 31 and 33).
5. De Moivre, trigonometric and binomial sums (Examples 34 to 38).
6. Cube roots of unity (Examples 39 to 46). This is the most common form in JEE Advanced, often inside a matrix or a determinant.
7. $n$th roots of unity (Examples 47 to 52).
8. Quadratic equations with complex roots or coefficients (Examples 53 to 58). Five of the thirteen 2026 JEE Main questions in this note are of this type.
9. Geometry in the Argand plane (Examples 59 and 60, and the coordinate geometry note).

JEE Advanced 2026 asked for $\omega$ inside a list match of quadratics (Example 45) and for the map $z\mapsto\frac{az+b}{cz+d}$ on the upper half-plane (Example 12). JEE Advanced 2025 asked for the argument of a sum of powers of $-\omega$ (Example 46) and an Apollonius circle (Example 60).

Quick recall

Practise these methods on real exam questions in the doMath Complex Numbers chapter. Rotation, lines, circles and loci in the Argand plane are developed in the Complex Numbers in Coordinate Geometry note, and quadratic equations with real coefficients in the Quadratic Equations note.