Problem source

Give a parametric form for the curve in the Argand diagram determined by $\left|z-\mathrm{i}\right|=2.$ 
	Let $w=(z+\mathrm{i})/(z-\mathrm{i}).$ Find and sketch the locus, in the Argand diagram, of the point which represents the complex number $w$ when
\begin{questionparts}
\item  $\left|z-\mathrm{i}\right|=2;$ 
\item  $z$ is real; 
\item  $z$ is imaginary. 
\end{questionpart}

Solution source

There are many possible parametric forms, for example $z = i + 2e^{it}, z = 2\ cos \theta + (1 + 2\sin \theta)i$ etc. It is a circle radius $2$ about the point $i$.


\begin{questionparts}
\item \begin{align*}
w &= \frac{z+i}{z-i} \\
&= \frac{2i + 2e^{it}}{2e^{it}} \\
&= 2 + ie^{-it}
\end{align*}

This is obvious a circle radius $1$ about the point $2$.

\begin{center}
    \begin{tikzpicture}[scale=1]
        \draw[->] (-4, 0) -- (4,0) node [right] {$\textrm{Re}$};
        \draw[->] (0, -4) -- (0,4) node [above] {$\textrm{Im}$};

        \draw[ultra thick, red] (2,0) circle (1);
        \node at (2,0) {$(2,0)$};
        
    \end{tikzpicture}
\end{center}

\item If $z$ is real, then 
\begin{align*}
w &= \frac{z+i}{z-i} \\
&= \frac{(z+i)^2}{z^2+1} \\
&= \frac{z^2-1 + 2zi}{z^2+1}
\end{align*} We can quickly notice this describes a circle radius $1$ about $0$.

Alternatively, $|z+i| = |z-i| \Rightarrow |\frac{z+i}{z-i}| = 1$ so we must be talking about points on the unit circle. Since this is a Mobius transform we know it maps lines and circles to lines and circles, therefore it must map to the unit circle;

\item If $z$ is purely imaginary, say $it$ then:

\begin{align*}
w &= \frac{z+i}{z-i} \\
&= \frac{(it+i)(i-it)}{(-1+t)^2} \\
&= \frac{t^2-1}{(t-1)^2} 
\end{align*}

Which is purely real, and can take all real values.

\begin{center}
    \begin{tikzpicture}[scale=1]
        \draw[->] (-4, 0) -- (4,0) node [right] {$\textrm{Re}$};
        \draw[->] (0, -4) -- (0,4) node [above] {$\textrm{Im}$};

        \draw[ultra thick, red] (-4,0) -- (4,0);
        
    \end{tikzpicture}
\end{center}



\end{questionparts}