Let \alpha and \beta be the sum and the product of all the non-zero solutions of the equation (\barz)^2+|z|=0, z \in C . Then 4(\alpha^2+\beta^2) is equal to :

Complex Numbers PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q49 with Solution

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Correct answer: A

(\bar z)^2 + |z| = 0.

z = x+iy \quad \Rightarrow \quad \bar z = x-iy,

and let

|z| = r \ge 0.

A better form is to use polar representation:

z = re^{i\theta}, \quad r=|z|.

Then

\bar z = re^{-i\theta}.

(\bar z)^2 = r^2 e^{-i2\theta}.

The equation becomes

r^2 e^{-i2\theta} + r = 0.

Since we want non-zero solutions, $r\neq 0$ . Hence divide by $r$ :

r e^{-i2\theta} + 1 = 0

r e^{-i2\theta} = -1.

|r e^{-i2\theta}| = |-1| \Rightarrow r=1.

e^{-i2\theta} = -1 = e^{i(\pi+2k\pi)}.

Thus

-2\theta = \pi + 2k\pi

which gives

2\theta = -\pi - 2k\pi.

Hence

\theta = \frac{\pi}{2} + n\pi

(for integer $n$ ). Therefore the distinct solutions are

z = e^{i\pi/2} = i,

and

z = e^{i3\pi/2} = -i.

So the non-zero solutions are $i$ and $-i$ .

\alpha = i + (-i) = 0.

Their product is

\beta = i(-i) = 1.

4(\alpha^2 + \beta^2) = 4(0^2 + 1^2) = 4.

\boxed{4}

which is Option A.