## Task B-2.1.

Determine the real parameter $k$ so that the system of equations

$$
\left[\operatorname{Im}\left(z+\frac{1}{2} i\right)\right]^{2}=|z+2|^{2}+\frac{5}{4} \quad \text { and } \quad \operatorname{Im}(z)-2 \operatorname{Re}(z)=k, z \in \mathbb{C}
$$

has only one solution?
## Solution.

Let $z=x+y i ; x, y \in \mathbb{R}$.

Then the first of the given equations becomes

$$
\left[\operatorname{Im}\left(x+y i+\frac{1}{2} i\right)\right]^{2}=|x+y i+2|^{2}+\frac{5}{4}
$$

or

$$
\left(y+\frac{1}{2}\right)^{2}=(x+2)^{2}+y^{2}+\frac{5}{4}
$$

By simplifying, we get the set of points $(x, y)$ that lie on the parabola $y=x^{2}+4 x+5$.

The second equation $\operatorname{Im}(x+y i)-2 \operatorname{Re}(x+y i)=k$ determines the set of points $(x, y)$ that lie on the line $y=2 x+k$.

By solving the system of equations, we get the common points of the parabola and the line:

$$
\begin{aligned}
& y=x^{2}+4 x+5 \\
& y=2 x+k \\
& x^{2}+4 x+5=2 x+k \\
& x^{2}+2 x+5-k=0
\end{aligned}
$$

The obtained quadratic equation will have one real solution if its discriminant ($D=$ $\left.b^{2}-4 a c\right)$ is equal to zero, that is

$4-4(5-k)=0 \Rightarrow k=4$.