Complex NumbershardComprehensionFree

Complex Numbers: Let Complex Numbers Satisfying Non Real Cube Root

JEE Maths reading comprehension with full step-by-step solutions.

Passage
Let z1z_1 and z2z_2 be two complex numbers satisfying
(z−ω)2+(z−ω2)2=0,\left(z-\omega\right)^{2} + \left(z-\omega^{2}\right)^{2} = 0 ,
 where ω\omega is a non-real cube root of unity. Also let α\alpha be a variable point on the circle
∣z+12∣=32,\left|z + \frac12\right| = \frac{\sqrt3}{2} ,
 and let α1\alpha_1, α2\alpha_2 be the values of α\alpha for which ∣α∣|\alpha| is maximum and minimum respectively.
Question 1 · Single correct
The value of expression
∣z1−α∣2+∣z2−α∣2+∣ω−α∣2+∣ω2−α∣2\left|z_1-\alpha\right|^{2} + \left|z_2-\alpha\right|^{2} + \left|\omega-\alpha\right|^{2} + \left|\omega^{2}-\alpha\right|^{2}
 is
A33
B3\sqrt3
C66correct
D11
Solution
Question attachment Step 1: Solve the quadratic for zz. Expanding,
2z2−2z(ω+ω2)+ω2+ω4=0.2z^{2} - 2z\left(\omega+\omega^{2}\right) + \omega^{2} + \omega^{4} = 0 .
 Step 2: Simplify using 1+ω+ω2=01 + \omega + \omega^{2} = 0 and ω3=1\omega^{3} = 1:
ω+ω2=−1,ω4=ω,ω2+ω4=ω2+ω=−1.\omega + \omega^{2} = -1, \qquad \omega^{4} = \omega, \qquad \omega^{2} + \omega^{4} = \omega^{2} + \omega = -1 .
 So the equation becomes
2z2+2z−1=0.2z^{2} + 2z - 1 = 0 .
 Step 3: Solve it.
z=−2±4+84=−1±32,z = \frac{-2 \pm \sqrt{4+8}}{4} = \frac{-1 \pm \sqrt3}{2},
 so
z1=−1−32,z2=−1+32,z_1 = \frac{-1-\sqrt3}{2}, \qquad z_2 = \frac{-1+\sqrt3}{2},
 both real. Step 4: Show these are ends of a diameter of the given circle. The circle has centre −12-\dfrac12 and radius 32\dfrac{\sqrt3}{2}, and
z1=−12−32,z2=−12+32,z_1 = -\frac12 - \frac{\sqrt3}{2}, \qquad z_2 = -\frac12 + \frac{\sqrt3}{2},
 i.e. the centre displaced by ∓\mp the radius along the real axis. So z1z2z_1 z_2 is a diameter. Step 5: Show ω\omega and ω2\omega^{2} are also ends of a diameter of the same circle. With ω=−12+32i\omega = -\dfrac12 + \dfrac{\sqrt3}{2}i and ω2=−12−32i\omega^{2} = -\dfrac12 - \dfrac{\sqrt3}{2}i,
∣ω+12∣=∣32i∣=32✓,\left|\omega + \frac12\right| = \left|\frac{\sqrt3}{2}i\right| = \frac{\sqrt3}{2} \checkmark ,
 and likewise for ω2\omega^{2}; they are the centre displaced by ±\pm the radius along the imaginary axis. Step 6: Use the semicircle right-angle property. If AA and BB are ends of a diameter and α\alpha is any other point of the circle, then ∠AαB=90∘\angle A\alpha B = 90^\circ, so by Pythagoras
∣A−α∣2+∣B−α∣2=∣AB∣2=(2r)2.\left|A-\alpha\right|^{2} + \left|B-\alpha\right|^{2} = |AB|^{2} = \left(2r\right)^{2} .
 Step 7: Apply it to each diameter, with 2r=32r = \sqrt3:
∣z1−α∣2+∣z2−α∣2=3,∣ω−α∣2+∣ω2−α∣2=3.\left|z_1-\alpha\right|^{2} + \left|z_2-\alpha\right|^{2} = 3, \qquad \left|\omega-\alpha\right|^{2} + \left|\omega^{2}-\alpha\right|^{2} = 3 .
 Step 8: Add.
3+3=6.3 + 3 = 6 .
 (Notice the answer does not depend on where α\alpha sits on the circle.) Answer: (3).
Question 2 · Single correct
Which of the following is incorrect?
A∣α1∣+∣α2∣=∣α1−α2∣\left|\alpha_1\right| + \left|\alpha_2\right| = \left|\alpha_1 - \alpha_2\right|
Bα1=z2\alpha_1 = z_2 and α2=z1\alpha_2 = z_1, where z1<0z_1 < 0, z2>0z_2 > 0correct
C∣2z1+1∣=∣2z2+1∣\left|2z_1 + 1\right| = \left|2z_2 + 1\right|
Damp⁡(z1z2)=amp⁡(z1z2)\operatorname{amp}\left(\dfrac{z_1}{z_2}\right) = \operatorname{amp}\left(z_1 z_2\right)
Solution
Step 1: Get z1z_1 and z2z_2 (as in the previous question). The equation reduces to 2z2+2z−1=02z^{2} + 2z - 1 = 0, so
z1=−1−32≈−1.366,z2=−1+32≈0.366,z_1 = \frac{-1-\sqrt3}{2} \approx -1.366, \qquad z_2 = \frac{-1+\sqrt3}{2} \approx 0.366 ,
 matching the stated signs z1<0z_1 < 0 and z2>0z_2 > 0. Step 2: Find α1\alpha_1 and α2\alpha_2. On a circle, the points nearest to and farthest from an external point lie on the line through that point and the centre. Here the origin and the centre −12-\dfrac12 both lie on the real axis, so α1,α2\alpha_1, \alpha_2 are the two points where the circle meets the real axis:
−12±32,i.e. exactly z1 and z2.-\frac12 \pm \frac{\sqrt3}{2}, \quad\text{i.e. exactly } z_1 \text{ and } z_2 .
 Step 3: Decide which is which.
∣z1∣=1+32≈1.366,∣z2∣=3−12≈0.366.\left|z_1\right| = \frac{1+\sqrt3}{2} \approx 1.366, \qquad \left|z_2\right| = \frac{\sqrt3-1}{2} \approx 0.366 .
 So the maximum occurs at z1z_1 and the minimum at z2z_2:
α1=z1,α2=z2.\alpha_1 = z_1, \qquad \alpha_2 = z_2 .
 Step 4: Test (B). It claims α1=z2\alpha_1 = z_2 and α2=z1\alpha_2 = z_1 - the two are swapped. So (B) is incorrect, and that is what the question asks for. Confirm the other three are correct. Step 5: Test (A). Since z1<0<z2z_1 < 0 < z_2, the origin lies between them, so
∣α1−α2∣=∣z1−z2∣=3=1+32+3−12=∣α1∣+∣α2∣.\left|\alpha_1 - \alpha_2\right| = \left|z_1 - z_2\right| = \sqrt3 = \frac{1+\sqrt3}{2} + \frac{\sqrt3-1}{2} = \left|\alpha_1\right| + \left|\alpha_2\right| .
 Step 6: Test (C).
∣2z1+1∣=∣−1−3+1∣=3,∣2z2+1∣=∣−1+3+1∣=3.\left|2z_1+1\right| = \left|-1-\sqrt3+1\right| = \sqrt3, \qquad \left|2z_2+1\right| = \left|-1+\sqrt3+1\right| = \sqrt3 .
 (Equivalently, both say the distance from the centre is one radius.) Step 7: Test (D). As z1<0z_1 < 0 and z2>0z_2 > 0, both z1z2\dfrac{z_1}{z_2} and z1z2z_1z_2 are negative reals, so both amplitudes equal π\pi. Answer: (2).
Question 3 · Single correct
The number of complex numbers α\alpha such that
∣α−α1∣+∣α−α2∣=6\left|\alpha - \alpha_1\right| + \left|\alpha - \alpha_2\right| = \sqrt6
 is
A44
B33
C22correct
D00
Solution
Step 1: Recall from the paragraph that
α1=−12−32,α2=−12+32,\alpha_1 = -\frac12 - \frac{\sqrt3}{2}, \qquad \alpha_2 = -\frac12 + \frac{\sqrt3}{2},
 so the distance between them is ∣α1−α2∣=3\left|\alpha_1 - \alpha_2\right| = \sqrt3. Step 2: Recognise the condition. "Sum of distances from two fixed points is constant" is the definition of an ellipse with α1,α2\alpha_1, \alpha_2 as foci, provided the constant exceeds the distance between the foci. Here 6>3\sqrt6 > \sqrt3 ✓, so it is a genuine ellipse. Step 3: Extract its dimensions. With 2a=62a = \sqrt6 and 2c=32c = \sqrt3,
a=62,c=32,b2=a2−c2=64−34=34  ⟹  b=32.a = \frac{\sqrt6}{2}, \qquad c = \frac{\sqrt3}{2}, \qquad b^{2} = a^{2} - c^{2} = \frac64 - \frac34 = \frac34 \implies b = \frac{\sqrt3}{2} .
 Step 4: Compare with the circle. The ellipse is centred at −12-\dfrac12 - the same centre as the circle - and its semi-minor axis b=32b = \dfrac{\sqrt3}{2} is exactly the circle's radius. Step 5: So the circle is the "minor-axis circle" of the ellipse: it lies inside the ellipse and touches it only at the two ends of the minor axis. Since α\alpha is restricted to the circle, we want how many points lie on both curves. Step 6: Verify algebraically. Writing X=x+12X = x + \dfrac12, the circle is X2+y2=34X^{2}+y^{2} = \dfrac34 and the ellipse is X23/2+y23/4=1\dfrac{X^{2}}{3/2} + \dfrac{y^{2}}{3/4} = 1. Substituting X2=34−y2X^{2} = \dfrac34 - y^{2} into the ellipse:
34−y232+y234=1  ⟹  12−23y2+43y2=1  ⟹  23y2=12  ⟹  y2=34.\frac{\frac34 - y^{2}}{\frac32} + \frac{y^{2}}{\frac34} = 1 \implies \frac12 - \frac23 y^{2} + \frac43 y^{2} = 1 \implies \frac23 y^{2} = \frac12 \implies y^{2} = \frac34 .
 Step 7: Then X2=34−34=0X^{2} = \dfrac34 - \dfrac34 = 0, so x=−12x = -\dfrac12 and y=±32y = \pm\dfrac{\sqrt3}{2} - that is, α=ω\alpha = \omega and α=ω2\alpha = \omega^{2}. Step 8: Exactly 22 such points. Answer: (3).
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