Partial Fraction Decomposition with Complex Number $\frac{1}{z^2 - 2i}$.

How do I decompose the fraction

$$\dfrac{1}{z^2 - 2i}$$

into partial fractions? I understand how to do partial fraction decomposition with real numbers, but I am unsure of how to do it with complex numbers. I attempted to find examples online, but all examples are with real numbers -- not complex.

I would greatly appreciate it if people could please take the time to demonstrate this.

Note that $z^2-2i=(z+\sqrt{2i})(z-\sqrt{2i})$ and $\sqrt{2i}=\sqrt{2}e^{i\pi/4}=1+i$. To simplify, let $b=1+i$, then $$\frac{1}{z^2-2i}=\frac{1}{(z+b)(z-b)}$$ From here it actually doesn't matter if you regard $b$ as real or complex, the process to find the partial fractions is the same as long as the terms are linear in $z$. So we let $$\frac{1}{(z+b)(z-b)}=\frac{A}{z+b}+\frac{B}{z-b}$$ for some $A,B\in \mathbb C$. Adding the two fractions on the right hand side we get that $$A(z-b)+B(z+b)=1$$ and so $$A+B=0$$ $$-bA+bB=1$$ which has solution $$A=-\frac{1}{2b}$$ $$B=\frac 1{2b}$$ Plugging in the original $b=1+i$ we have that $$\frac{1}{2b}=\frac12\frac 1{(1+i)}\frac{(1-i)}{(1-i)}=\frac 14(1-i)$$ Therefore $$\frac{1}{z^2-2i}=-\frac{\frac 14(1-i)}{z+1+i}+\frac{\frac 14(1-i)}{z-1-i}.$$ As you can see the process for computing the partial fraction coefficients with complex rationals is equivalent to that of real numbers.

the roots of $z^2 - 2i$ are $1+i$ and $-(1+i)$

$$ (1+i)^2 = 1^2 + 2 \cdot 1 \cdot i + i^2 = 1 + 2i - 1 = 2i $$