# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Quiz-6 => Topic started by: Victor Ivrii on November 17, 2018, 04:12:01 PM

Title: Q6 TUT 5101
Post 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:
$$f(z)= \frac{e^z-1}{e^{2z}-1}.$$
Title: Re: Q6 TUT 5101
Post by: Yatong Yu on November 17, 2018, 04:19:12 PM
let $f(z)=e^z-1$  $g(z)=e^{2z}-1$

$e^{2z}-1=0$ so $e^{2z}=1$  $e^z=\pm1$

when  $e^z=+1$
$f(z)=e^z-1=0$ $f'(z)=e^z=1\neq0$    so order=1
$g(z)=e^{2z}-1=0$ $g'(z)=2e^{2z}=2\neq0$    so order=1
1-1=0   removable

when  $e^z=-1$
$f(z)=e^z-1=-2\neq0$    so order=0
$g(z)=e^{2z}-1=0$ $g'(z)=2e^{2z}=2\neq0$    so order=1
1-0=1   simple pole
Title: Re: Q6 TUT 5101
Post by: Zihan Wan on November 17, 2018, 04:19:43 PM
solution is attached
Title: Re: Q6 TUT 5101
Post by: Qing Zong on November 17, 2018, 05:03:51 PM
The value at the point should be 1/2
Title: Re: Q6 TUT 5101
Post by: Qing Zong on November 17, 2018, 05:09:38 PM
This is extra step for 1/2
Title: Re: Q6 TUT 5101
Post by: Meerna Habeeb on November 17, 2018, 08:50:29 PM
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.
Title: Re: Q6 TUT 5101
Post by: Victor Ivrii on November 28, 2018, 05:03:29 AM
Meerna is right, but you need to write "pole of order 1" or "zero of order 1" without skipping "pole" or "zero"