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JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Complex Numbers

Let

S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}

. Then

\sum_{z \in S} (Re (z) + Im (z))

is equal to ______.

Answer: 0

Solution

Let

z = x + i y

where

x, y \in R

.

Then

z^{2} = x^{2} - y^{2} + 2 i x y

and

\overset{z}{ˉ} = x - i y

.

The equation

z^{2} + \overset{z}{ˉ} = 0

becomes:

(x^{2} - y^{2} + x) + i (2 x y - y) = 0

Equating real and imaginary parts:

x^{2} - y^{2} + x = 0 and y (2 x - 1) = 0

From the second equation:

y = 0

or

x = 1/2

.

Case 1:

y = 0

x^{2} + x = 0 \Rightarrow x (x + 1) = 0 \Rightarrow x = 0 or x = - 1

This gives

z = 0

and

z = - 1

.

Case 2:

x = 1/2

1/4 - y^{2} + 1/2 = 0 \Rightarrow y^{2} = 3/4 \Rightarrow y = \pm \frac{3}{2}

This gives

z = \frac{1}{2} + i \frac{3}{2}

and

z = \frac{1}{2} - i \frac{3}{2}

.

Now compute the sum:

z \in S \sum (Re (z) + Im (z)) = (0 + 0) + (- 1 + 0) + (\frac{1}{2} + \frac{3}{2}) + (\frac{1}{2} - \frac{3}{2})

= 0 - 1 + \frac{1}{2} + \frac{1}{2} = 0

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Complex Numbers →

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