Integrating f(z)=$z/(16-z^2)(z+i)$ over a circle $|z|=5$

i know the approach of how to solve the question as i need to trifurcate f(z) such that it becomes something like $(A/4-z) + (B/4+z) + C/(z+i) $ and then simply apply cauchy integral formula but i am unable to convert it into partial factors form if anyone can help with the partial factor part please.

Solution 1:

You are correct about the approach. Like you said, you need to find $A,B,C$ so that

$$ \frac{z}{(16-z^2)(z+i)} = \frac{A}{-z+4}+\frac{B}{z+4}+\frac{C}{z+i} $$

To do that, just add up the rational fractions on the right the same way you would add up three fractions, by using a common denominator:

$$ \frac{z}{(16-z^2)(z+i)} = \frac{A(z+4)(z+i)+B(-z+4)(z+i)+C(-z+4)(z+4)}{(-z+4)(z+4)(z+i)} $$

The denominators are already the same so we can stop worrying about them. The numerator on the left is very simple: it's just $z$. The numerator on the right looks scary, but all you need to do is expand it:

$$ \text{numerator on right} = (A-B-C)z^2+((4+i)A+(4-i)B)z+4iA+4iB+16C $$

Students are sometimes confused since you have one equation with three unknowns. But this is an identity of polynomials: all the coefficients must be the same. Therefore,

$$ \begin{align}A-B-C &= 0 \\ (4+i)A+(4-i)B &= 1 \\ 4iA+4iB+16C &=0 \end{align} $$

And now you have three simple linear equations in three unknowns. I believe you can take it from here.

Solution 2:

Since $z^2-16=(z+4)(z-4)$ and $\deg z<\deg\bigl((16-z^2)(z+i)\bigr)$ you can write$$-\frac z{(z^2-16)(z+i)}\left(=\frac z{(16-z^2)(z+i)}\right)$$as$$\frac A{z+4}+\frac B{z-4}+\frac C{z+i}.\tag1$$But $(1)$ is equal to$$\frac{(A+B+C)z^2+\bigl((4+i) B-(4-i) A\bigr)z-4 i A+4 i B-16 C}{(z^2-16)(z+i)}.$$So, solve the system$$\left\{\begin{array}{l}A+B+C=0\\(4+i) B-(4-i) A=-1\\-4 i A+4 i B-16 C=0\end{array}\right.$$in order to get $A$, $B$ and $C$.

Solution 3:

Start with $$\frac z{(16-z^2)(z+i)}=\frac A{4-z} + \frac B{4+z} + \frac C{z+i}$$ and multiply everything by $(16-z^2)(z+i)$. You will get $$z = A(4+z)(z+i) + B(4-z)(z+i) + C(4-z)(4+z).\tag{1}$$ Finally, let $z = 4,-4,-i$ in $(1)$ to get the coefficients $A$, $B$ and $C$. For example, if we let $z = 4$, we get $$4 = 8(4+i)A \implies A = \frac 1{8+2i}.$$