JEE Mathematics 2023 Question with Solution

Given: $S = \{ z \in \mathbb{C} : z = i(z^2 + \operatorname{Re}(z)) \}$

Find: $\sum_{z \in S} |z|^2$

Let $z = x + iy$ where $x,y \in \mathbb{R}$. Then $\operatorname{Re}(z) = x$.

Substitute into the given equation:

$$z = i(z^2 + \operatorname{Re}(z))$$

So,

$$x + iy = i\left((x+iy)^2 + x\right)$$

Now,

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

Hence,

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

Multiplying by $i$,

$$i\left(x^2 - y^2 + x + 2ixy\right) = i(x^2 - y^2 + x) - 2xy$$

Therefore,

$$x + iy = -2xy + i(x^2 - y^2 + x)$$

Equating real and imaginary parts,

$$x = -2xy$$

and

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

From $x = -2xy$, we get

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

So either $x=0$ or $y=-\frac{1}{2}$.

Case 1: $x=0$

Then from $y = x^2 - y^2 + x$,

$$y = -y^2$$

which gives

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

So $y=0$ or $y=-1$.

Thus the corresponding values of $z$ are

$$z=0,\quad z=-i$$

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

Then

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

So,

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

which gives

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

Hence,

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

So,

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

Therefore,

$$S = \left\{0, -i, -\frac{1}{2}-\frac{i}{2}\right\}$$

Now compute $|z|^2$ for each element:

$$|0|^2 = 0,$$ $$|-i|^2 = 1,$$ $$\left|-\frac{1}{2}-\frac{i}{2}\right|^2 = \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

Thus,

$$\sum_{z \in S} |z|^2 = 0 + 1 + \frac{1}{2} = \frac{3}{2}$$

This value does not match any option exactly. The solution is for a different question and is unrelated. Using the given question, the computed sum is $\frac{3}{2}$. The closest listed option is D, but mathematically the correct value is $\frac{3}{2}$.

The solution discusses a circle, a line, intersection points, and evaluates $30(AB)^2 = 24$. That topic is unrelated to the given complex-number equation $z = i(z^2 + \operatorname{Re}(z))$.

Therefore the solution cannot be used as authority for this question. Solving the actual question directly gives the set

$$S = \left\{0, -i, -\frac{1}{2}-\frac{i}{2}\right\}$$

and hence

$$\sum_{z \in S} |z|^2 = \frac{3}{2}$$

Since the options do not contain $\frac{3}{2}$, there is a source discrepancy.