Partial fractions of $1/(z^2+2)$

Question

How does one split $1/(z^2+2)$ into partial fractions?

Normally I would factorise, but I cannot spot the solutions of $z^2+2=0$.

Answer 1

Factor $z^2 + 2 = (z + \sqrt{2}i)(z - \sqrt{2}i)$. Note $(z + \sqrt{2}i) - (z - \sqrt{2}i) = 2\sqrt{2}i$. So we have

$$\frac{1}{z^2 + 2} = \frac{1}{2\sqrt{2}i} \frac{(z + \sqrt{2}i) - (z - \sqrt{2}i)}{(z + \sqrt{2}i)(z - \sqrt{2}i)} = \frac{1}{2\sqrt{2}i}\left(\frac{1}{z - \sqrt{2}i} - \frac{1}{z + \sqrt{2}i}\right)$$

Answer 2

Tip:

$z^2+2=0$ at $z_0=\sqrt{2}i$ and $z_1=-\sqrt{2}i$

So $z^2+2=(z-\sqrt{2}i)(z+\sqrt{2}i)$. (Check this)

Write down $\frac{1}{z^2+2}=\frac{1}{(z-\sqrt{2}i)(z+\sqrt{2}i)}$

Now, can you find $A,B \in \mathbb C$ such that $\frac{1}{(z-\sqrt{2}i)(z+\sqrt{2}i)}=\frac{A}{z+\sqrt{2}i}+\frac{B}{z-\sqrt{2}i}$?

Answer 3

Start with $z^2+2=(z-i\sqrt{2})(z+i\sqrt{2})$ So we are looking for $$\frac{1}{z^2+1}=\frac{a}{z-i\sqrt{2}}+\frac{b}{z+i\sqrt{2}}$$ Because the LHS is real $a$ and $b$ are complex conjugate. Multiplying by $z-i\sqrt{2}$ we get $$\frac{1}{z+i\sqrt{2}}=a+\bar{a}\frac{z-i\sqrt{2}}{z+i\sqrt{2}}$$ and this is valid for all $z$ especially for $z=i\sqrt{2}$ which cancels the second term of the RHS and yields $a=\frac{1}{2i\sqrt{2}}$