# Quiz 3 Summer 2011

Find the residue of the function
$f(z)=\frac{e^z}{z\sin z}$
at z=0.

Solution:

$\frac{e^z}{z\sin z}=\frac{1+z+\ldots}{z^2-z^4/6+\ldots}=1/z^2+a_{-1}/z+\ldots$

where

$(1/z^2+a_{-1}/z+\ldots)(z^2-z^4/6+\ldots)=1+z+\ldots$

or

$Res(f;z=0)=a_{-1}=1$

Question 2:

Find all the poles of the function
$f(z)=\frac{e^z\sin^2(z)}{(z^2+z)^3}$
and for each one state its order.

Solution:

Poles at z=0, z=-1. At z=0 numerator vanishes to order 2, while denominator vanishes to order 3, hence the pole is of order 1. At z=-1 the numerator doesn't vanish, while denominator vanishes to order 3. Hence the pole is of order 3.

Question 3:

Find
$\lim\limits_{z\to 0} \frac{\sin(z^3) -z^3}{z\cos(z^4) - z}$

Hint: Taylor series work better than L'Hopital's rule when applied to quotients of analytic functions.

Solution:

$\lim\limits_{z\to 0} \frac{\sin(z^3) -z^3}{z\cos(z^4) - z}=\lim\limits_{z\to 0} \frac{(z^3-z^9/6+\ldots) -z^3}{z(1-z^8/2+\ldots) - z}=\lim\limits_{z\to 0} \frac{-z^9/6+\ldots}{-z^9/2+\ldots}=1/3$