# Residue of a function

• Sep 6th 2012, 07:11 AM
uhfwheh
Residue of a function
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
• Sep 6th 2012, 08:12 AM
Prove It
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

------------------------------------------------------------------------
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 \begin{align*} \textrm{Res}\, (f, c) = \lim_{ z \to c } (z - c) f(z) \end{align*}. So in this case where \displaystyle \begin{align*} f(z) = \frac{1}{\sin{z}} \end{align*} we have

\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*}
• Sep 6th 2012, 08:16 AM
uhfwheh
Re: Residue of a function
Quote:

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

\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
• Sep 6th 2012, 08:17 AM
uhfwheh
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
• Sep 6th 2012, 08:23 AM
Prove It
Re: Residue of a function
You wouldn't evaluate residue at \displaystyle \begin{align*} \frac{\pi}{2} \end{align*} because your function doesn't have a pole there...