MAT334-2018F > Quiz-6

Q6 TUT 5101

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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"

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