Draw an area on the complex plane

Question

On the complex plane draw the area: $$ \begin{equation} \begin{cases} |z+4i| < 3 \\ |\arg(z-5-5i)|<\frac{\pi}{3} \end{cases} \end{equation} $$ Where $ \arg(z) \in (-\pi, \pi ]$

I can draw $|z+4i| < 3$:

$|x + iy + 4i|<3 \Rightarrow \sqrt{x^2 + (y + 4)^2}<3 \Rightarrow x^2 + (y + 4)^2<9$:

However, I have no idea how to draw and intersect with $|arg(z-5-5i)|<\frac{\pi}{3}$

Answer 1

You must intersect the circle with:

$$|arg(z-5-5i)|<\frac{\pi}{3}$$

which means:

$$\left| \arctan \left(\frac{\Im(z - 5 - 5i)}{\Re(z - 5 - 5i)} \right) \right | < \frac{\pi}{3}$$

$$\Longleftrightarrow \Re \left\{\arctan \left(\frac{\Im(z) - 5}{\Re(z) - 5} \right)^2 \right\} + \Im \left\{\arctan \left(\frac{\Im(z) - 5}{\Re(z) - 5} \right) ^2\right\} < \frac{\pi^2}{9}$$

You know the argument is an angle in the complex plane. It is thus always real:

$$\arctan \left(\frac{\Im(z) - 5}{\Re(z) - 5} \right) < \pm \frac{\pi}{3}$$

$$\Longrightarrow \frac{\Im(z) - 5}{\Re(z) - 5} < \pm \tan\frac{\pi}{3} = \pm \sqrt 3$$

$$\Longleftrightarrow \Im (z) < \pm \sqrt 3(\Re (z)-5) + 5$$

You can now draw in the complex plane:

$$y = \pm \sqrt 3 (x - 5) + 5$$

Finally, considering $<$, you obtain your domain:

There is thus no connection at all with your circle.

Answer 2

To draw the region $|\text{arg}(z-5-5i)| < \frac \pi 3$, you should first draw the region $|\text{arg}(z)| < \frac \pi 3$ and then translate that shape by $5+5i$.

The region $|\text{arg}(z)| < \frac \pi 3$ is just all points $r e^{i \theta}$ with $-\pi/3 < \theta < \pi/3$. It's the region between two rays that start at $z=0$.

Therefore, the region $|\text{arg}(z-5-5i)| < \frac \pi 3$ is basically the same shape except the rays start at $z=5+5i$ instead.

Combining that with the disc you already drew, we actually find that there is no overlap at all! Thus the region you want is actually empty.