Trouble with partial fractions and complex numbers

Question

$$f(z) = \frac{3}{(z+1)(z-i)} = \frac{A}{z+1} + \frac{B}{z-i}$$ $$z(A + B) + B - Ai = 3$$ $$A + B = 0$$ $$B-Ai=3$$

Somehow, I end up with $B = - \frac{3}{(1+i)}$

Is that the way to go?

Answer 1

Technically, it will not be the simplest way, just vecause if you have many factors, you'll a system of linear equations which be hard/long to solve.

Especially well suited to the case of simple poles for the fraction, you can remove the denominators, then give $z$ the poles as successive values.

Here is how it runs in the present case: multiplying both sides by $(z+1)(z-i)$, you get $$3=A(z-i)+B(z+1)$$ Then setting $z=i$ and $z=-1$ successively, you obtain: $$\begin{cases} 3=A\cdot 0+B(1+i),&\text{whence}\enspace B=\dfrac{3(1-i)}2,\\[1ex] 3=A(-1-i)+B\cdot 0,&\text{whence}\enspace A=\dfrac{3(-1+i)}2. \end{cases}$$

Answer 2

One practical way to do partial fractions which are of the type $\frac{1}{(x-a)(x-b)}$ is to simply let the answer be $\frac{1}{x-a}-\frac{1}{x-b}$ and adjust the constant term after calculation. In our case we have:

$$\frac{1}{z+1}-\frac{1}{z-i}=\frac{z-i-z-1}{(z+1)(z-i)}\\=-\frac{1+i}{(z+1)(z-i)}$$ from which one directly infers that we must have:

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

or that

$$\frac{3}{(z+1)(z-i)}=-\frac{3}{1+i}\Big(\frac{1}{z+1}-\frac{1}{z-i}\Big)$$

This technique is usually much quicker than solving by forming linear equations.