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Area of Triangle from Roots of z³+iz²+2i=0 | JEE

JEE Maths question with a full step-by-step solution.

Question

If the roots of

z^3 + iz^2 + 2i = 0

represent the vertices of

\triangle ABC

in the Argand plane, then the area of the triangle is (in square units)

3

1

4

2

correct

Solution

Step 1: Find one root by inspection. Test

z = i

i^3 + i\cdot i^2 + 2i = -i + i(-1) + 2i = -i - i + 2i = 0

(z - i)

is a factor. Step 2: Divide to obtain the quadratic factor:

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

(Check by expansion:

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

. ) Step 3: Solve

z^2 + 2iz - 2 = 0

with the quadratic formula. The discriminant is:

(2i)^2 - 4(1)(-2) = -4 + 8 = 4 \implies \sqrt{4} = 2

z = \frac{-2i \pm 2}{2} = -i \pm 1

So the other two roots are

z = 1 - i

and

z = -1 - i

. Step 4: List the three vertices as points in the plane:

A = i = (0, 1), \quad B = 1 - i = (1, -1), \quad C = -1 - i = (-1, -1)

Step 5: Apply the area formula

\text{Area} = \dfrac{1}{2}\,|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|

= \frac{1}{2}\,\big|\,0(-1 - (-1)) + 1(-1 - 1) + (-1)(1 - (-1))\,\big| = \frac{1}{2}\,|0 - 2 - 2| = \frac{1}{2}\cdot 4 = 2

Correct answer: (4)

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