Question

What is the residue of the function $\frac{1-e^{2z}}{z^4}$ at its pole?

• A. $\frac{4}{3}$
• B. $-\frac{4}{3}$
• C. $-\frac{2}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $-\frac{4}{3}$

$f\left(z\right)=\frac{1-e^{2z}}{z^4}$

Diagram

$=-2\left(\frac{1}{z^3}\right)-2\left(\frac{1}{z^2}\right)-\frac{4}{3}\left(\frac{1}{z}\right)-\frac{2}{3}-\frac{2}{3}\frac{32}{120}\left(z\right)+....$
Which is Laurent expansion of $f\left(z\right)$ about $z=0$ and it is pole of order $3$.
So residual of $f\left(z\right)$ at $z=0$ = coefficient of $\left(\frac{1}{z}\right)=-\frac{4}{3}$