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.
A+B=2
That part is right.AND B−A=...
That part doesn't follow. By equating the free terms you get $4i=A+B+i(B-A)\,$. Since $A+B=2$ this implies $4i=2+i(B-A)$ so $A-B=-4-2i\,$. Now you know $A+B$ and $A-B$, and all that remains to be done is solve for $A,B$. $\endgroup$