Problem source

The variable non-zero complex number $z$ is such that 
\[
\left|z-\mathrm{i}\right|=1.
\]
Find the modulus of $z$ when its argument is $\theta.$ Find also the modulus and argument of $1/z$ in terms of $\theta$ and show in an Argand diagram the loci of points which represent $z$ and $1/z$. 
Find the locus $C$ in the Argand diagram such that $w\in C$ if, and only if, the real part of $(1/w)$ is $-1$.

Solution source

\begin{center}
    \begin{tikzpicture}[scale=1.5]
        \draw[->] (-3,0) -- (3,0) node[right] {Re};
        \draw[->] (0,-3) -- (0,3) node[above] {Im};

        \draw (0,1) circle (1);

        \draw (0,0) -- ({cos(20)}, {1 + sin(20)}) -- (0,1);

        % \filldraw[domain = 5:175, samples=180, variable = \x]  
            % plot ({1/(sqrt(2)*abs(sin(\x)))*cos(-\x)}, {1/(sqrt(2)*abs(sin(\x)))*sin(-\x)});

        \draw (-3,-0.5) -- (3,-0.5);
    \end{tikzpicture}
\end{center}



$z$ is a point on the circle shown:

Therefore using the cosine rule $|z|^2 = 1^2 + 1^2 - 2\cdot 1 \cdot 1 \cdot \cos (2 \theta) = 2 -2\cos 2\theta = 2\sin^2 \theta \Rightarrow |z| = \sqrt{2}|\sin \theta|$

$\frac{1}{z}$ has modulus $\frac{1}{\sqrt{2}|\sin \theta|}$ and argument $-\theta$.

$|\frac{1}{z} - i| = 1 \Rightarrow |1-iz| = |z| \Rightarrow |-i-z| = |z|$ ie we're looking for the points on the perpendicular bisector of $0$ and $-i$.


$\textrm{Re}\left (\frac{1}{w}\right) = -1 \Rightarrow -1 = \textrm{Re} \left (\frac{1}{a+ib} \right) = \frac{a-ib}{a^2+b^2} = \frac{a}{a^2+b^2} \Rightarrow a^2+b^2 = -a \Rightarrow (a+\tfrac12)^2+b^2 = \tfrac14$ so we are looking at a circle radius $\tfrac12$ centre $-\frac12$