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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.

ThunderHex
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    $15 = (2a+b)z + (-3a-3ib)$ must hold for all values of $z$. That gives you two equations for $a$ and $b$. – Martin R Nov 08 '21 at 09:09
  • "Equating z terms I can get..." $;-;$ Then equate non-$z$ terms next. – dxiv Nov 08 '21 at 09:12
  • @dxiv I still don't understand = how does $-15 = 3a + 3bi$ help at all? – ThunderHex Nov 08 '21 at 09:17
  • @ThunderHex At that point you have two linear equations in two unknowns, which should be easy to solve for $a,b \in \mathbb C$. – dxiv Nov 08 '21 at 09:20
  • @dxiv I'm totally lost here. I don't understand how to deal with $3bi$ – ThunderHex Nov 08 '21 at 09:24
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    @ThunderHex By elimination, for example, the first equation is $b=-2a$, then the second equation is $-15=3a-6ai$. What's the difficulty with solving that for $a$? – dxiv Nov 08 '21 at 09:29
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    @dxiv Managed to get it, thanks. We were never taught complex numbers so I'm very rusty – ThunderHex Nov 08 '21 at 09:36

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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 $3iB+\frac 32 A = -\frac{15}{2}$.

$$ \dfrac{15/2}{(z-3i)(z-\frac 32)}=-\dfrac{1+2i}{z-3i}+\frac{1+2i}{z-\frac 32} $$

$$ \begin{cases}A=-B \\ (3i-\frac 32)B = -\frac{15}{2}\end{cases}\Leftrightarrow \begin{cases}A = -1-2i \\ B=\frac{15}{2(3i-\frac 32)} = 1+2i\end{cases} $$

PierreCarre
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