winding number partial fractions

Question

I need to compute the winding numbers for a particular contour integration (which I understand and is not relevant for this question). However, before doing so I need to use partial fractions to get $A$ and $B$ such that

$$\frac{1}{(z-2i)(z-2)}=\frac{A}{z-2i}+\frac{B}{z-2}.$$

I have so far cross multiplied and got,

$$1=A(z-2)+B(z-2i)=Az-2A+Bz-2Bi.$$

I am not used to partial fractions of this form, I am used to just equating real and imaginary parts to find A and B. I used Maple to get the answer:

$$\frac{1}{(z-2)(z-2i)} = \frac{\frac{1}{4}+\frac{i}{4}}{z-2}-\frac{\frac{1}{4}+\frac{i}{4}}{z-2i}$$

But I cannot work out how to get there.

Any help would be much appreciated.

Answer 1

When you have all poles $\rho$ being simple the formula is $$\sum_\rho \frac{1}{z-\rho} \mathrm{Res}_{z=\rho} f(z).$$ To find the residue differentiate the denominator.

With $f(z) = \frac{p(z)}{q(z)}$ you get $$\sum_\rho \frac{1}{z-\rho} \frac{p(\rho)}{q'(\rho)}.$$

Answer 2

You want to equate $1$ and $Az-2A+Bz-2Bi$ as polynomials: first equate the coefficients of $z$ and then the constant coefficient. As (linear) polynomials in $z$, the LHS is $0z+1$ and the RHS is $(A+B)z+(-2A-2Bi)$. Now, the coefficient of $z$ on the LHS is $0$, so you have $A+B=0$; likewise, the constant coefficient on the LHS is $1$, so $1=-2A-2Bi$. This is a system of two linear equations in two variables, so you can just solve it using your favorite method of doing those (though substitution using the first eq to substitute into the second should come out particularly cleanly).