Task B-2.1. In the complex plane, represent the set of all $z \in \mathbb{C}$ for which

$$
\left|z-(1-i)^{4}\right|<|\sqrt{3}-i|^{2}
$$
Solution. Let's first expand

$$
\begin{aligned}
|\sqrt{3}-i|^{2} & =(\sqrt{3})^{2}+(-1)^{2}=4 \\
(1-i)^{4} & =\left[(1-i)^{2}\right]^{2}=[-2 i]^{2}=-4
\end{aligned}
$$

Thus, we are looking for the set of all points $z \in \mathbb{C}$ for which $|z+4|<4$. Let $z=x+i y$, and the condition becomes

$$
\begin{aligned}
|(x+4)+i y| & <4 \\
\sqrt{(x+4)^{2}+y^{2}} & <4 \\
(x+4)^{2}+y^{2} & <4^{2}
\end{aligned}
$$

which defines the (strict) interior of a circle with center at the point $(-4,0)$ and radius 4.

(1 point)

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-07.jpg?height=813&width=810&top_left_y=1713&top_left_x=625)

Sketch...

Note. A student gets full points if they immediately sketched the solution from $|z+4|<4$ with the center and radius clearly marked.