Polar Form of a Complex Number

Let the complex number be written in polar form as $$z = re^{i\pi/4}$$, where $$r = |z| \ge 0$$ because $$\text{Arg}(z)=\frac{\pi}{4}$$ places it on the line making an angle $$45^{\circ}$$ with the positive real axis.

Expressing $$z$$ in Cartesian form, we get
$$z = r\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = r\left(\frac{\sqrt2}{2} + i\,\frac{\sqrt2}{2}\right).$$
Hence its coordinates are
$$x = r\frac{\sqrt2}{2}, \qquad y = r\frac{\sqrt2}{2}.$$

The condition $$|z-2| = 2$$ means the point $$z$$ lies on the circle centred at $$2+0i$$ with radius $$2$$:
$$(x-2)^2 + y^2 = 2^2 = 4.$$

Because $$y=x$$ (the line $$\theta=\pi/4$$), substitute $$y=x$$ into the circle equation:
$$(x-2)^2 + x^2 = 4.$$

Simplify:
$$x^2 - 4x + 4 + x^2 = 4$$
$$2x^2 - 4x = 0$$
$$2x(x-2)=0.$$

This gives $$x=0$$ or $$x=2$$.
• $$x=0$$ would imply $$r=0$$, but $$\text{Arg}(0)$$ is undefined, so this root is discarded.
• Therefore $$x = 2$$.

Back-substitute to find $$r$$:
$$x = r\frac{\sqrt2}{2} = 2 \;\;\Longrightarrow\;\; r = 2\cdot\frac{2}{\sqrt2} = \frac{4}{\sqrt2} = 2\sqrt2.$$

Finally,
$$|z|^2 = r^2 = (2\sqrt2)^2 = 8.$$

Option C which is: $$8$$