I'm going insane with this question from a previous exam:
How do I get the partial fraction decomposition of: $${15 \over (z-3i)(2z-3)}$$
I don't understand how to 'equate' anything here. If we have that $$15=a(2z-3)+b(z-3i)$$ then how am I meant to get $a$ and $b$ ? Equating $z$ terms I can get that $2a+b=0$ but I don't understand at all how to deal with the $i$ term.
You want to find $A,B$ such that $$ \dfrac{15/2}{(z-3i)(z-\frac 32)}=\frac{A}{z-3i} + \frac{B}{z-\frac 32} $$
So, what you need to equate is $$ B(z-3i) +A(z-\frac 32) = \frac{15}{2} $$ or,
$$ (A+B) z - (3iB+\frac 32 A) = \frac{15}{2}. $$
Since this equality must hold for all $z$, you must have $A+B=0$ and $3iB+\frac 32 A = -\frac{15}{2}$.
$$ \dfrac{15/2}{(z-3i)(z-\frac 32)}=-\dfrac{1+2i}{z-3i}+\frac{1+2i}{z-\frac 32} $$
$$ \begin{cases}A=-B \\ (3i-\frac 32)B = -\frac{15}{2}\end{cases}\Leftrightarrow \begin{cases}A = -1-2i \\ B=\frac{15}{2(3i-\frac 32)} = 1+2i\end{cases} $$