how does the set of inverses of complex numbers become a circle

Question

Can someone show me how this equality is achieved?

$$\left\{\frac{1}{\frac1p - ir}: r\in \mathbb{R}\right\}\cup \{0\} = \left\{w\in \mathbb{C}: |w-\frac{p}{2}|=\frac{p}{2}\right\}$$

I'll really appreciate any help!

Answer 1

$$ \left| \frac{1}{1/p-ir}-\frac{p}{2} \right| =\left| \frac{p}{1-ipr}-\frac{p}{2} \right| =\frac{|p|}{2} \left| \frac{2}{1-ipr}-1 \right| \\ =\frac{|p|}{2} \left| \frac{2-1+ipr}{1-ipr} \right| =\frac{|p|}{2} \left| \frac{1+ipr}{1-ipr} \right| $$ If $p$ is real, $$ =\frac{p}{2} \cdot 1 = \frac{p}{2}. $$ This is not true if $p$ is not real. Indeed, if $pr$ is any complex number, then its distance after +90 degree rotation (times $i$) with 1, need not be the same with that of its distance after -90 degree rotation (times $-i$).

To show converse inclusion, $$ \mathfrak{Re} 1/ \big[ \frac{p}{2} (1 +e^{i\theta}) \big] =\frac{2}{p} \mathfrak{Re} \big[ 1/(1 +e^{i\theta}) \big] \\ =\frac{2}{p} \mathfrak{Re} \big[ (1 -e^{i\theta})/2 \big] =\frac{2}{p} \cdot \frac{1}{2} =\frac{1}{p}. $$ As desired.

Answer 2

\begin{eqnarray} z\in\left\{\frac{1}{\frac1p - ir}: r\in \mathbb{R}\right\}\cup \{0\} &\iff& \forall r\in \mathbb{R}: z=\frac{p}{1-ipr}\\ &\iff& \forall r\in \mathbb{R}: \frac{2}{p}\left(z-\frac{p}{2}\right)=\frac{1+ipr}{1-ipr}\\ &\iff& \Big|z-\frac{p}{2}\Big|=\frac{p}{2} \end{eqnarray} because for real $p$, the map $w=\dfrac{1+ipr}{1-ipr}$ maps imaginary axis to unit circle.