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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 z3+iz2+2i=0z^3 + iz^2 + 2i = 0 represent the vertices of △ABC\triangle ABC in the Argand plane, then the area of the triangle is (in square units)
A33
B11
C44
D22correct
Solution
Step 1: Find one root by inspection. Test z=iz = i:
i3+i⋅i2+2i=−i+i(−1)+2i=−i−i+2i=0i^3 + i\cdot i^2 + 2i = -i + i(-1) + 2i = -i - i + 2i = 0
 So (z−i)(z - i) is a factor. Step 2: Divide to obtain the quadratic factor:
z3+iz2+2i=(z−i)(z2+2iz−2)z^3 + iz^2 + 2i = (z - i)(z^2 + 2iz - 2)
 (Check by expansion: (z−i)(z2+2iz−2)=z3+2iz2−2z−iz2−2i2z+2i=z3+iz2−2z+2z+2i=z3+iz2+2i(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 z2+2iz−2=0z^2 + 2iz - 2 = 0 with the quadratic formula. The discriminant is:
(2i)2−4(1)(−2)=−4+8=4  ⟹  4=2(2i)^2 - 4(1)(-2) = -4 + 8 = 4 \implies \sqrt{4} = 2
z=−2i±22=−i±1z = \frac{-2i \pm 2}{2} = -i \pm 1
 So the other two roots are z=1−iz = 1 - i and z=−1−iz = -1 - i. Step 4: List the three vertices as points in the plane:
A=i=(0,1),B=1−i=(1,−1),C=−1−i=(−1,−1)A = i = (0, 1), \quad B = 1 - i = (1, -1), \quad C = -1 - i = (-1, -1)
 Step 5: Apply the area formula Area=12 ∣xA(yB−yC)+xB(yC−yA)+xC(yA−yB)∣\text{Area} = \dfrac{1}{2}\,|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|:
=12 ∣ 0(−1−(−1))+1(−1−1)+(−1)(1−(−1)) ∣=12 ∣0−2−2∣=12⋅4=2= \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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