Area in complex plane

let $z=x+iy$

$S_{1}$ denotes the interior of circle of radius $4$ units

$S_{3}$ denotes $x>0$

$S_{2}=Im(\frac{(x-1+i(y+\sqrt{3}))(1+i\sqrt{3})}{4})>0$

$S_{2}=i(y+\sqrt{3}x)>0$

Now the shaded region represents the required area

Required area= $\text{Area of quarter of circle}+\text{Area of sector}$

$\large{=\frac{\pi r^{2}}{4}+\frac{\pi r^{2} \theta}{360^{o}}}$

$\large{=4 \pi+\frac{8}{3}\pi}$

$\large{\boxed{\frac{20}{3}\pi}}$

Let's represent $z$ in polar form, since it turns out to be more beneficial. Let $z = r e^{i \theta}$ . Then the conditions become:

• From $S_1$ : $r < 4$
• From $S_3$ : $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$
• From $S_2$ : ...???

Well, let's break that down.

$\begin{aligned} \displaystyle \frac{z - 1 + i \sqrt{3}}{1 - i \sqrt{3}} &= \frac{z}{1 - i \sqrt{3}} - 1 \\ &= \frac{z (1 + i \sqrt{3})}{4} - 1 \end{aligned}$

Since we're only concerned about the imaginary part, the -1 doesn't do anything. Moreover, since we only need it to be positive, the 4 doesn't do anything too; any $z$ such that $\Im (z (1 + i \sqrt{3})) > 0$ will be in $S_2$ and vice versa.

Now convert to polar form:

$\begin{aligned} \displaystyle z (1 + i \sqrt{3}) &= r e^{i \theta} \cdot 2 e^{i \pi / 3} \\ &= 2r e^{i (\theta + \pi / 3)} \end{aligned}$

We need $0 < \theta + \frac{\pi}{3} < \pi$ to have $z \in S_2$ , or $- \frac{\pi}{3} < \theta < \frac{2\pi}{3}$ . Intersecting with $S_3$ gives $- \frac{\pi}{3} < \theta < \frac{\pi}{2}$ . (Since angle only works modulo $2\pi$ , we need to make sure we don't accidentally skip any angle here; it can be easily proven that we have all the solutions.)

Thus $S$ looks like a sector of a circle. The radius is 4, from $S_1$ , and the angle is $\frac{5\pi}{6}$ , from the above. Plugging into the formula $\frac{\theta}{2} r^2$ for the area of a sector of a circle gives an area of $\boxed{\frac{20 \pi}{3}}$ .