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Complex Numbers question

2025 · 3 Apr · Shift 1 · Q42
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Complex Numbers question

2025 · 3 Apr · Shift 1 · Q42

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let z∈Cz \in Cz∈C be such that z2+3iz−2+i=2+3i\frac{z^2+3 i}{z-2+i}=2+3 iz−2+iz2+3i​=2+3i. Then the sum of all possible values of z2z^2z2 is :
  1. A
    −19+2i-19+2 i−19+2i
  2. B
    −19−2i-19-2 i−19−2i
  3. C
    19−2i19-2 i19−2i
  4. D
    19+2i19+2 i19+2i
View written solutionFree

Correct answer: B

  1. Start with the given equation:
z2+3iz−2+i=2+3i\frac{z^2+3i}{z-2+i}=2+3iz−2+iz2+3i​=2+3i

We must have z−2+i≠0z-2+i \neq 0z−2+i=0, i.e. z≠2−iz \neq 2-iz=2−i.

  1. Cross-multiply:
z2+3i=(2+3i)(z−2+i)z^2+3i=(2+3i)(z-2+i)z2+3i=(2+3i)(z−2+i)

Now simplify the right-hand side.

First,

(2+3i)(−2+i)=−4+2i−6i+3i2=−4−4i−3=−7−4i(2+3i)(-2+i) = -4+2i-6i+3i^2 = -4-4i-3 = -7-4i(2+3i)(−2+i)=−4+2i−6i+3i2=−4−4i−3=−7−4i

So,

(2+3i)(z−2+i)=(2+3i)z+(−7−4i)(2+3i)(z-2+i)=(2+3i)z+(-7-4i)(2+3i)(z−2+i)=(2+3i)z+(−7−4i)

Hence,

z2+3i=(2+3i)z−7−4iz^2+3i=(2+3i)z-7-4iz2+3i=(2+3i)z−7−4i

Bring all terms to one side:

z2−(2+3i)z+7+7i=0z^2-(2+3i)z+7+7i=0z2−(2+3i)z+7+7i=0

So the possible values of zzz are the roots of the quadratic

z2−(2+3i)z+7+7i=0z^2-(2+3i)z+7+7i=0z2−(2+3i)z+7+7i=0
  1. We need the sum of all possible values of z2z^2z2.

Let the roots be z1,z2z_1,z_2z1​,z2​. Then by Vieta:

z1+z2=2+3i,z1z2=7+7iz_1+z_2=2+3i, \qquad z_1z_2=7+7iz1​+z2​=2+3i,z1​z2​=7+7i

Now,

z12+z22=(z1+z2)2−2z1z2z_1^2+z_2^2=(z_1+z_2)^2-2z_1z_2z12​+z22​=(z1​+z2​)2−2z1​z2​

Substitute the Vieta values:

z12+z22=(2+3i)2−2(7+7i)z_1^2+z_2^2=(2+3i)^2-2(7+7i)z12​+z22​=(2+3i)2−2(7+7i)

Compute:

(2+3i)2=4+12i+9i2=4+12i−9=−5+12i(2+3i)^2=4+12i+9i^2=4+12i-9=-5+12i(2+3i)2=4+12i+9i2=4+12i−9=−5+12i

and

2(7+7i)=14+14i2(7+7i)=14+14i2(7+7i)=14+14i

Therefore,

z12+z22=(−5+12i)−(14+14i)=−19−2iz_1^2+z_2^2=(-5+12i)-(14+14i)=-19-2iz12​+z22​=(−5+12i)−(14+14i)=−19−2i
  1. Check the restriction z≠2−iz \neq 2-iz=2−i.

Substitute z=2−iz=2-iz=2−i into the quadratic:

(2−i)2−(2+3i)(2−i)+7+7i(2-i)^2-(2+3i)(2-i)+7+7i(2−i)2−(2+3i)(2−i)+7+7i =(3−4i)−(7+4i)+7+7i=3−i≠0= (3-4i)-(7+4i)+7+7i=3-i \neq 0=(3−4i)−(7+4i)+7+7i=3−i=0

So z=2−iz=2-iz=2−i is not a root, and both quadratic roots are valid.

Therefore the sum of all possible values of z2z^2z2 is

−19−2i\boxed{-19-2i}−19−2i​

Thus the correct option is B.

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