Q6 TUT 5101

[Victor Ivrii]

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:
$$
f(z)= \frac{e^z-1}{e^{2z}-1}.
$$

[Yatong Yu]

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

[Zihan Wan]

solution is attached

[Qing Zong]

The value at the point should be 1/2

[Qing Zong]

This is extra step for 1/2

[Meerna Habeeb]

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.

[Victor Ivrii]

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