Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz6 => Topic started by: Victor Ivrii on November 17, 2018, 04:12:01 PM

Locate each of the isolated singularities of the given function $f(z)$ and tell whether it is a removable singularity, a pole, or an essential singularity.
If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole:
$$
f(z)= \frac{e^z1}{e^{2z}1}.
$$

let $f(z)=e^z1$ $g(z)=e^{2z}1$
$e^{2z}1=0$ so $e^{2z}=1$ $e^z=\pm1$
when $e^z=+1$
$f(z)=e^z1=0$ $f'(z)=e^z=1\neq0$ so order=1
$g(z)=e^{2z}1=0$ $g'(z)=2e^{2z}=2\neq0$ so order=1
11=0 removable
when $e^z=1$
$f(z)=e^z1=2\neq0$ so order=0
$g(z)=e^{2z}1=0$ $g'(z)=2e^{2z}=2\neq0$ so order=1
10=1 simple pole

solution is attached

The value at the point should be 1/2

This is extra step for 1/2

When $e^{2z}1=0 \to e^{z}=1$ then $z=2n\pi i$ and $(2n+1)\pi i$
When $z=2n\pi i,$
$f(2n\pi i)=e^{z}1=0$
$f'(2n\pi i)=e^{z}\ne 0$
Therefore $f(z)$ at $2n\pi i$ is of order 1
$g(2n\pi i)=e^{2z}1=0$
$g'(2n\pi i)=2e^{2z}\ne 0$
Therefore $g(z)$ at $2n\pi i $ is of order 1
We have removable singularity at $2n\pi i$ with value of $\frac{1}{2}
$
$f(2n\pi i)=\lim _{z\to 2n\pi i}\frac{e^{z}1}{e^{2z}1}=\frac{1}{2}$
When $z=(2n+1)\pi i,$
$f((2n+1)\pi i)=e^{z}1\ne 0$
Therefore $f(z)$ at $((2n+1)\pi i)$ is of order zero
$g((2n+1)\pi i)=e^{2z}1=0$
$g'((2n+1)\pi i)=2e^{2z}\ne 0$
Therefore $g(z)$ at ($(2n+1)\pi i) $ is of order 1
We have poles at $((2n+1)\pi i)$ of order 1 because order of $
g(z)f(z)=1$ so simple poles.

Meerna is right, but you need to write "pole of order 1" or "zero of order 1" without skipping "pole" or "zero"