Find Square of Modulus of Complex Number satisfying Conditions

To find the value of $|z|^2$, let the complex number $z$ be represented as $z = x + iy$, where $x, y \in \mathbb{R}$.

Step 1: Simplify the first condition The first condition is given as: $|z + 2| = |z - 2|$

Substituting $z = x + iy$: $|x + 2 + iy| = |x - 2 + iy|$

Squaring both sides: $(x + 2)^2 + y^2 = (x - 2)^2 + y^2$ $x^2 + 4x + 4 + y^2 = x^2 - 4x + 4 + y^2$ $8x = 0 \implies x = 0$

Thus, $z$ is purely imaginary, i.e., $z = iy$.

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Step 2: Simplify the second condition The second condition is: $\arg\left(\frac{z + 3}{z - i}\right) = \frac{\pi}{4}$

Substitute $z = iy$ into the expression: $\frac{z + 3}{z - i} = \frac{3 + iy}{iy - i} = \frac{3 + iy}{i(y - 1)}$

Multiply the numerator and the denominator by $-i$: $\frac{3 + iy}{i(y - 1)} = \frac{-i(3 + iy)}{y - 1} = \frac{-3i + y}{y - 1} = \left(\frac{y}{y - 1}\right) + i\left(\frac{-3}{y - 1}\right)$

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Step 3: Apply the argument condition For a complex number $w = u + iv$ to have $\arg(w) = \frac{\pi}{4}$, both its real and imaginary parts must be strictly positive ($u > 0$ and $v > 0$), and $\frac{v}{u} = \tan\left(\frac{\pi}{4}\right) = 1$.

1. Imaginary part condition: $\text{Im}\left(\frac{z + 3}{z - i}\right) = \frac{-3}{y - 1} > 0 \implies y - 1 < 0 \implies y < 1$

2. Real part condition: $\text{Re}\left(\frac{z + 3}{z - i}\right) = \frac{y}{y - 1} > 0$ Since $y - 1 < 0$, for the fraction to be positive, we must have $y < 0$.

3. Ratio condition: $\frac{\text{Im}\left(\frac{z + 3}{z - i}\right)}{\text{Re}\left(\frac{z + 3}{z - i}\right)} = \tan\left(\frac{\pi}{4}\right) = 1$ $\frac{\frac{-3}{y - 1}}{\frac{y}{y - 1}} = 1 \implies \frac{-3}{y} = 1 \implies y = -3$

This satisfies $y < 0$.

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Step 4: Calculate $|z|^2$ Since $x = 0$ and $y = -3$, we have $z = -3i$.

Now, calculate $|z|^2$: $|z|^2 = | -3i |^2 = 0^2 + (-3)^2 = 9$

Correct Answer: Option A (9)