The equation $|z - z_0| = r$ denotes the circle of radius $r$.
Express $z = x + \iota y$ and simplify the expression to see what geometric shape it represents.
Solution: We recall that the equation
\[
\vert z - z_0 \vert = r,
\]
denotes the circle with center $z_0$ and radius $r$.
The set is
\begin{align*}
\left\{ z \in \mathbb{C} : \vert z - 1 - \iota \vert = 1 \right\} = \{ z \in \mathbb{C} : \vert z - (1 + \iota ) = 1
\vert \} ,
\end{align*}
which is a circle with center $1 + \iota $ and radius $1$.
The set is
\begin{align*}
1 < \vert 2z - 6 \vert < 2 & \implies \frac{1}{2} < \vert z - 3 \vert < 1 \\ & \implies \vert z-3 \vert> \frac{1}{2}
\text{ and } \vert z - 3 \vert < 1. \end{align*}
The set is
\begin{align*}
\vert z - 1 \vert ^2 + \vert z + 1 \vert ^2 < 8. \end{align*} Take $z=x + \iota y$. Then \begin{align*} \vert z - 1
\vert ^2 + \vert z + 1 \vert ^2 < 8 & \implies \vert (x-1) +\iota y \vert ^2 + \vert (x + 1) + \iota y \vert ^2 < 8 \\
& \implies (x-1)^2 + y^2 + (x+1)^2 + y^2 < 8 \\ & \implies 2x^2 + 2y^2 + 2 < 8 \\ & \implies x^2 + y^2 < 3.
\end{align*}
The above represents an open disc centered at $(0,0)$ and of radius $\sqrt{3} $.
The set is
\begin{align*}
\vert z - 1 \vert + \vert z + 1 \vert \leq 2.
\end{align*}
Take $z = x + \iota y$, then from the previous part, we have
\begin{align*}
\sqrt{(x-1)^2 + y^2} + \sqrt{(x + 1)^2 + y^2} \leq 2.
\end{align*}
The above is an ellipse with interior whose foci are $(1,0)$ and $(-1,0)$. This one can also be seen geometrically that
the set is the locus of all point whose sum of distance from $1$ and $-1$ is constant $2$, which is an ellipse.
The set is
\begin{align*}
\vert z - 1 \vert < \vert z \vert . \end{align*} Take $z=x + \iota y$, then \begin{align*} \vert (x - 1) + \iota y \vert
< \vert x + \iota y \vert & \implies (x-1)^2 + y^2 < x^2 + y^2 \\ & \implies -2x + 1 < 0 \\ & \implies x>
-\frac{1}{2}.
\end{align*}
