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Sum of Imaginary Parts of the Roots of a Complex Cubic | JEE Advanced

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

Question
If z1z_1, z2z_2, z3z_3 are the roots of the equation
z3−z2(1+3i)+z(3i−2)+2=0,z^{3} - z^{2}(1+3i) + z(3i-2) + 2 = 0 ,
 then Im⁡(z1)+Im⁡(z2)+Im⁡(z3)\operatorname{Im}(z_1) + \operatorname{Im}(z_2) + \operatorname{Im}(z_3) is
A00
B11
C22
D33correct
Solution
Step 1: Note that taking imaginary parts is additive: for any complex numbers,
Im⁡(z1)+Im⁡(z2)+Im⁡(z3)=Im⁡(z1+z2+z3).\operatorname{Im}(z_1) + \operatorname{Im}(z_2) + \operatorname{Im}(z_3) = \operatorname{Im}\left(z_1 + z_2 + z_3\right).
 So we only need the sum of the roots - no need to find them individually. Step 2: Apply the sum of roots of a cubic. For z3+Az2+Bz+C=0z^{3} + Az^{2} + Bz + C = 0 the sum of roots is −A-A. Here A=−(1+3i)A = -(1+3i), so
z1+z2+z3=1+3i.z_1 + z_2 + z_3 = 1 + 3i .
 Step 3: Take the imaginary part.
Im⁡(1+3i)=3.\operatorname{Im}(1 + 3i) = 3 .
 Step 4: (Check, by actually factorising.) Grouping,
z3−z2−2z+2−3iz2+3iz=(z−1)(z2−2)−3iz(z−1)=(z−1)(z2−3iz−2),z^{3} - z^{2} - 2z + 2 - 3i z^{2} + 3iz = \left(z-1\right)\left(z^{2}-2\right) - 3iz\left(z - 1\right) = (z-1)\left(z^{2} - 3iz - 2\right),
 and z2−3iz−2=0z^{2} - 3iz - 2 = 0 gives z=2iz = 2i or z=iz = i. So the roots are 1, 2i, i1,\ 2i,\ i and the imaginary parts are 0+2+1=30 + 2 + 1 = 3. Answer: (4).
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