Draw an area on the complex plane 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 $|arg(z-5-5i)|<\frac{\pi}{3}$
 A: 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.
A: 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.
