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Is it so that partial fractions with complex roots can work sometime, and sometime not?

I have tried to check a result by WA here, and tried to solve it manually:

\begin{equation} X(z)=\frac{104z+30}{z^2+4z+5}=\frac{A}{(z-2+i)}+\frac{B}{(z+2-i)} \end{equation}

But I get that A does not exist: $(A\cdot 0)=104i-238$ which is false. So B equally does not exist.

Wolfram alpha does not give any solution either.

Is there something important to note when one has complex denominators?

Thanks

Luthier415Hz
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1 Answers1

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Since $z^2+4z+5=(z+2+i)(z+2-i)$, you should search for numbers $A$ and $B$ such that$$\frac{104z+30}{z^2+4z+5}=\frac A{z+2+i}+\frac B{z+2-i}.$$But, since$$\frac A{z+2+i}+\frac B{z+2-i}=\frac{(A+B)z+(2-i) A+(2+i) B}{z^2+4z+5},$$you simply solve the system$$\left\{\begin{array}{l}A+B=104\\(2-i)A+(2+i)B=30.\end{array}\right.$$You will get that $A=52-89i$ and that $B=52+89i$.

José Carlos Santos
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