JEE Mathematics 2024 Question with Solution

Given: $z = x + iy$ and the solution works with $z^2 + i\overline{z} = 0$.

Find: $|z|^2$.

From the solution:

$$z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$$

and

$$i\overline{z} = i(x - iy) = ix + y$$

So,

$$(x^2 - y^2 + 2ixy) + (ix + y) = 0$$

Separating real and imaginary parts,

$$x(2y + 1) = 0$$

Using the working shown, take

$$2y + 1 = 0 \Rightarrow y = -\frac{1}{2}$$

Now substitute into the real part as shown in the solution:

$$x^2 - \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = 0$$ $$x^2 - \frac{1}{4} - \frac{1}{2} = 0$$ $$x^2 = \frac{3}{4}$$

Hence,

$$|z|^2 = x^2 + y^2 = \frac{3}{4} + \frac{1}{4} = 1$$

Therefore, the correct option is B and $|z|^2 = 1$.

Note: The solution consistently concludes option B, although it works with $z^2 + i\overline{z} = 0$ instead of the question text $z^2 + iz = 0$.

Given: $z = x + iy$.

Find: $|z|^2$.

The solution states:

$$z^2 + i\overline{z} = 0$$

with

$$\overline{z} = x - iy$$

Then

$$z^2 = x^2 - y^2 + 2ixy$$

and

$$i\overline{z} = ix + y$$

Therefore,

$$x^2 - y^2 + 2ixy + ix + y = 0$$

Equating real and imaginary parts gives

$$x^2 - y^2 + y = 0$$

and

$$2xy + x = 0$$ $$x(2y + 1) = 0$$

The solution proceeds with

$$y = -\frac{1}{2}$$

Substituting,

$$x^2 - \frac{1}{4} - \frac{1}{2} = 0$$ $$x^2 = \frac{3}{4}$$

Now,

$$|z|^2 = x^2 + y^2$$ $$|z|^2 = \frac{3}{4} + \frac{1}{4} = 1$$

So the final answer reported on the solution is $1$, i.e. option B.