Write as the sum of a series The question asks to write 
$\dfrac{1}{1-i-z}$ as the sum of a series such that $\left|-z-i\right|<1$
but I genuinely have no idea how to do it, or even where to start.
 A: hint: $$1+r+r^2+r^3+\cdots=\frac{1}{1-r}$$
where $|r|<1$
A: Just notice that your function is of the form $\displaystyle -\frac{1}{z-z_0}$. The key idea then is that
$$
 \frac{1}{z-z_0} = \sum_{n = 0}^\infty (z- z_0)^n
$$
when $|z - z_0| < 1$
A: $$
\frac 1 {1-i-z}
$$
You want the series to converge if $\left| -z-i \right|<1.$ That is the same as $\left|z+i\right|<1,$ and so the same as $\left|z-(-i)\right|<1.$ We therefore want to write this as a sum of powers of $z-(-i),$ i.e. of $z+i.$ The function is
$$
\frac 1 {1 - (z+i)}.
$$
Recall that
$$
\frac 1 {1-r} = 1+r+r^2+r^3+r^4+\cdots. \tag 1
$$
Thus we need
$$
1 + (z+i) + (z+i)^2 + (z+i)^3 + (z+i)^4 + \cdots.
$$
The answer would be different if, for example, we had wanted it to converge if $|z-5|<\text{something}.$ In that case we would want a sum of powers of $z-5.$ We would have
$$
|1-i-z| = |z- 1+i| = |(z-5)-(i-4)|
$$
and then we would apply line $(1)$ above to write
\begin{align}
& \frac 1 {1-i-z} = \frac 1 {(z-5) - (i-4)} = \frac 1 {4-i} \cdot \frac 1 {1 - \left(\dfrac{z-5}{i-4}\right)} \\[12pt]
= {} & \frac 1 {4-i} \left( 1 + \frac{z-5}{i-4} + \left( \frac{z-5}{i-4} \right)^2 + \left( \frac{z-5}{i-4} \right)^3 + \left( \frac{z-5}{i-4} \right)^4 + \cdots \right).
\end{align}
This converges if $\left|\dfrac {z-5}{i-4} \right| < 1,$ i.e. if $\left|z-5\right| < |i -4|,$ i.e. if $|z-5|< \sqrt{17}.$
