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I have the fraction $\frac{z+4}{(z+1+2i)(z+1-2i)}$. I want to partial decompose this fraction, but I am not seeing how to do it. I know that the answer is a=$\frac{1}{2}$+$\frac{3i}{4}$ and b=$\frac{1}{2}$-$\frac{3i}{4}$, where $$\frac{z+4}{(z+1+2i)(z+1-2i)}= \frac{a}{z+1+2i}+\frac{b}{z+1-2i},\quad a,b \in \mathbb C. $$

I tried to write $a=c+di$ and $b=e+fi$ but I didn't got the answer. I also tried $a(z+1-2i) + b(z+1+2i)$ but this is not right because it gave me $a+b=1$ and $a+b=4$.

Thank you for your help in advance!

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    $\begingroup$ Simply finding $a$ and $b$ with $a(z+1-2i)+b(z+1+2i)=z+4$ works. Your $a+b=4$ is an error - nobody can tell you what you did wrong because you don't say how you got $a+b=4$. $\endgroup$ – David C. Ullrich Jun 21 at 11:09
  • $\begingroup$ Hi, David! Yes, I am sorry I didn't explain. What I did was a(z+1−2i) + b(z+1+2i)=z+4 $\Leftrightarrow$ az + a - 2a$i$ + bz + b + 2b$i$ = z+4 $\Leftrightarrow$ $\begin{cases} a + b = 1 \\ a + b = 4 \\ -2a + 2b = 0 \end{cases}$ $\endgroup$ – Learn_Math Jun 21 at 11:19
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    $\begingroup$ I can't tell whether it's a typo, but there's at least one mistake in the algebra in that comment ($2zi$ should be something else...) More important. how do you get those three equations on the right side of the $\iff$???????? $\endgroup$ – David C. Ullrich Jun 21 at 11:22
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    $\begingroup$ Ah. There's no such thing as "the coefficient of $i$". You meant "looking at the imaginary part...". Looking at the real and imaginary parts separately would be valid, but it's much trickier than you think! Because you can't tell, for example, what the real and imaginary parts of $az+i$ are just by looking at it, because $a$ and $z$ are complex numbers. $\endgroup$ – David C. Ullrich Jun 21 at 11:46
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    $\begingroup$ Or to put it in terms of the language you used: The "coefficient of $i$" in the expression $az+i$ is not $1$, because $a$ and $z$ both have $i$s hidden in them... $\endgroup$ – David C. Ullrich Jun 21 at 11:47
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Starting with $a(z+1-2i) + b(z+1+2i)=z+4$ should give the right answer; in particular it doesn't lead to $1=4$. Your comment about "the coefficient of $i$" makes it pretty clear what went wrong: Rewrite as $$(a(z+1)+b(z+1))+i(-2a+2b)=z+4;\tag{*}$$now looking at "the coefficient of $i$" gives $$-2a+2b=0,$$etc. That's wrong, because for example the imaginary part of $i(-2a+2b)$ is not $-2a+2b$, and the imaginary part of $z+4$ is not $0$, because $a$, $b$ and $z$ are complex numbers.

Instead simply do the usual thing. Collect terms: $$(a+b)z+(a(1-2i)+b(1+2i))=z+4.$$Now the constant term in (*) shows that $$(1-2i)a+(1+2i)b=0,$$the coefficient of $z$ shows that $$a+b=1,$$and you can simply solve those two equations for $a$ and $b$.

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Let $k=1+2i$, then we want to find $a$ and $b$ such that

$$\frac{z+4}{(z+k)(z+\bar{k})}=\frac{a}{z+k}+\frac{b}{z+\bar{k}}$$

Or $$z+4=(z+\bar{k})a+(z+k)b$$

Then setting $z=-\bar{k}$ and $z=-k$ we obtain $b=\frac{-\bar{k}+4}{k-\bar{k}}=\frac{-(1-2i)+4}{(1+2i)-(1-2i)}=\frac{3+2i}{4i}=\frac{1}{2}-\frac{3}{4}i$ and $a=\frac{1}{2}+\frac{3}{4}i$ respectively.

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