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Messages - Meerna Habeeb

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Quiz-6 / Re: Q6 TUT 5101
« 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.

Quiz-6 / Re: Q6 TUT 0301
« on: November 17, 2018, 04:18:21 PM »
Set $ w=z-\pi $

So that $ z=w+\pi $

Using the identity $\sin (w+\pi )=-\sin w$

$f(w+\pi )=\frac{sin (w+\pi )}{((w+\pi )-\pi )^{2}}=\frac{-\sin

$-\sin w=\sum_{n=0}^{\infty }{(-1)^{n+1} \frac{w^{2n+1}}{(2n+1)!}}$

$\frac{-\sin w}{w^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}

$\frac{-\sin w}{w^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}


Substitute back $w=z-\pi $

$\frac{\sin z}{(z-\pi )^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}
\frac{(z-\pi )^{2n-1}}{(2n+1)!}}$

$=-\frac{1}{(z-\pi )}+\frac{(z-\pi )}{3!}-\frac{(z-\pi
)^{3}}{5!}+\frac{(z-\pi )^{5}}{7!}-\frac{(z-\pi )^{7}}{9!}+\ldots $

Residue of $\frac{\sin z}{(z-\pi )^{2}}$ at $z=\pi$ is -1

Which is the coefficient of $-\frac{1}{(z-\pi )}$

Or you can use $Res(f,z_{0})=\lim _{z\to z_{0}}(z-z_{0}) f(z)$

$\lim _{z\to \pi}(z-\pi ) \frac{\sin z}{(z-\pi )^{2}}=-1$

Quiz-5 / Re: Q5 TUT 0301
« on: November 02, 2018, 08:46:46 PM »
See answer attached

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