Re: Residue of a function

Quote:

Originally Posted by

**uhfwheh** I need to find the residue of f(z) = 1/(sin(z)) at z=Pi

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I had a similar exercise for z=Pi/2. I found the expansion of f(z) to be 1/z + z/6 + ...

then calculating the residue for a simple pole (?) i found that lim as z-->Pi/2 of (z - Pi/2)*(1/z + z/6 + ...) = 0

I find I get the same residue for z=Pi, but I think this is wrong

Since this is a simple pole, the residue can be calculated by $\displaystyle \displaystyle \begin{align*} \textrm{Res}\, (f, c) = \lim_{ z \to c } (z - c) f(z) \end{align*}$. So in this case where $\displaystyle \displaystyle \begin{align*} f(z) = \frac{1}{\sin{z}} \end{align*}$ we have

$\displaystyle \displaystyle \begin{align*} \textrm{Res}\,(f,\pi) &= \lim_{z \to \pi} \frac{z - \pi}{\sin{z}} \\ &= \lim_{z \to \pi}\frac{1}{\cos{z}} \textrm{ by L'Hospital's Rule} \\ &= \frac{1}{\cos{\pi}} \\ &= \frac{1}{-1} \\ &= -1 \end{align*}$

Re: Residue of a function

Quote:

Originally Posted by

**Prove It** Since this is a simple pole, the residue can be calculated by $\displaystyle \displaystyle \begin{align*} \textrm{Res}\, (f, c) = \lim_{ z \to c } (z - c) f(z) \end{align*}$. So in this case where $\displaystyle \displaystyle \begin{align*} f(z) = \frac{1}{\sin{z}} \end{align*}$ we have

$\displaystyle \displaystyle \begin{align*} \textrm{Res}\,(f,\pi) &= \lim_{z \to \pi} \frac{z - \pi}{\sin{z}} \\ &= \lim_{z \to \pi}\frac{1}{\cos{z}} \textrm{ by L'Hospital's Rule} \\ &= \frac{1}{\cos{\pi}} \\ &= \frac{1}{-1} \\ &= -1 \end{align*}$

thank you!!!

but wouldn't that also be the case if we want to find the residue at Pi/2? I'm slightly confused

Re: Residue of a function

Quote:

Originally Posted by

**uhfwheh** thank you!!!

but wouldn't that also be the case if we want to find the residue at Pi/2? I'm slightly confused

ah, never mind, you can't since cos(pi/2)=0

Re: Residue of a function

You wouldn't evaluate residue at $\displaystyle \displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ because your function doesn't have a pole there...