Subject: Mathematics — Complex Numbers
Question: Let $S$ be the set of all complex numbers $z$ satisfying $|z - 2 + i| \geq \sqrt{5}.$
If the complex number $z_{0}$ is such that
$\frac{1}{\left| z_{0} - 1 \right|}$ is the maximum of the set
$\left\{ \frac{1}{|z - 1|}:z \in S \right\},$
then the principal argument of
$\frac{4 - z_{0} - \overline{z_{0}}}{z_{0} - \overline{z_{0}} + 2i}$
is
Answer: $- \frac{\pi}{2}$
Solution: $\frac{1}{|z - 1|},\quad z \in S,$
attains its maximum when $|z - 1|$ attains its minimum.
Now the curve
$$|z - 2 + i| \geq \sqrt{5}$$
looks like the following figure.
[IMAGE] $|z - 2 + i| \geq \sqrt{5}$.
Consider
$$\frac{4 - z_{0} - \overline{z_{0}}}{z_{0} - \overline{z_{0}} + 2i}\frac{4 - z_{0} - \overline{z_{0}}}{z_{0} - \overline{z_{0}} + 2i} = \frac{4 - 2Rez_{0}}{2iImz_{0} + 2i}\frac{4 - z_{0} - \overline{z_{0}}}{z_{0} - \overline{z_{0}} + 2i} = \frac{2\left( 2 - Rez_{0} \right)}{2i\left( Imz_{0} + 1 \right)}$$
$$\frac{4 - z_{0} - \overline{z_{0}}}{z_{0} - \overline{z_{0}} + 2i} = - i\lambda,\quad\text{where }\lambda > 0.$$
Since
$$2 - Rez_{0} > 0\quad\text{and}\quad Imz_{0} > 0,$$
therefore, the principal argument of
$$\frac{4 - z_{0} - \overline{z_{0}}}{z_{0} - \overline{z_{0}} + 2i}$$
is $- \frac{\pi}{2}$.
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