I'm attempting the following problem in the context of complex analysis:
Express $h(z) = \dfrac{2z + 4i}{z^2 + 2z + 2}$ as partial fractions.
I got $\dfrac{2z + 4i}{z^2 + 2z + 2} = \dfrac{2z + 4i}{(z + 1 + i)(z + 1 – i)}$
$= \dfrac{A}{z + 1 + I} + \dfrac{B}{z + 1 – I}$
$\implies 2z + 4i = (A+B)z + (A + B) + i(B – A)$
$\therefore A + B = 2$ AND $B – A = 4$ (AND $A + B = 0$????)
$\therefore$ (If we exclude $A + B = 0$) $B = 3$ AND $A = -1$
But we also had that $A + B = 0$ above, which is where my confusion comes from.
If I then proceed with what I've done, I get $\dfrac{2z + 4i}{(z + 1 + i)(z + 1 – i)} = \dfrac{-1}{z + 1 + i} + \dfrac{3}{z + 1 – I}$
$= \dfrac{2z + 2 + 4i}{(z + 1 + i)(z + 1 – i)}$, which is obviously false.
I've only just started doing partial fraction decomposition with complex numbers, so I'm unfamiliar with the subtleties relative to partial fraction decomposition with real numbers. As I flagged above, we cannot have both $A + B = 2$ and $A + B = 0$, so I'm unsure of why my reasoning is erroneous and what the correct reasoning is?
I would greatly appreciate it if people could please take the time to clarify this by explaining why my reasoning is erroneous and what the correct reasoning/procedure is.
Best Answer
Write $Ai+B$ (where you wrote $A$) and $Ci + D$ (where you wrote $B$) solve for this in the same way. You should obtain $A = 1$; $B = 3$; $C = -1$; $D = -1$.
And the final form should be $\frac{i+3}{z+1-i} + \frac{-i-1}{z+1+i}$