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I am kind of stuck on finding residue at the poles of a function.

The function is $$f(z) = \frac{e^{\pi z}}{4z^2 + 1}$$

I have determined the poles to be $\pm i/2$

Is this a single or double pole? How do I calculate the residue? Do I have to compute the integral to find the Laurent series?

Any help greatly appreciated.

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    $\begingroup$ You have two simple poles. Do you know what the residue of $$\frac{g(z)}{z-z_0}$$ in $z_0$ is if $g$ is holomorphic in $z_0$? $\endgroup$ – Daniel Fischer Sep 26 '13 at 17:28
  • $\begingroup$ I seem to have figured out the residue is 0 in both cases (wolfram agrees). $\endgroup$ – dingari Sep 26 '13 at 17:42
  • $\begingroup$ That cannot be. In a simple pole, the residue can never be $0$. The sum of the residues may be $0$ (haven't computed them). $\endgroup$ – Daniel Fischer Sep 26 '13 at 17:44
  • $\begingroup$ I used the limit described here en.wikipedia.org/wiki/… $\endgroup$ – dingari Sep 26 '13 at 18:04
  • $\begingroup$ Okay, that gives you $$\frac{e^{\pm\pi i/2}}{8(\pm\frac i2)} = \frac{\pm i}{\pm 4i} = \frac14,$$ like it should. $\endgroup$ – Daniel Fischer Sep 26 '13 at 18:09
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The denominator of $f$ factors as $4(z - \frac{i}{2})(z+\frac{i}{2})$. The exponential function has no zeros, so $f$ has two simple poles, in $\frac{i}{2}$ and in $-\frac{i}{2}$.

If $g$ is a function that is holomorphic in $z_0$, then

$$h(z) = \frac{g(z)}{z-z_0}$$

has either a simple pole or a removable singularity in $z_0$, and the residue of $h$ in $z_0$ is $g(z_0)$. The residue is $0$ if and only if $g(z_0) = 0$, if and only if $h$ has a removable singularity in $z_0$.

For the residue of $f$ in $\frac i2$, we choose

$$g(z) = \frac{e^{\pi z}}{4(z+\frac i2)}$$

and find

$$\operatorname{res}(f;\frac i2) = \frac{e^{\pi i/2}}{4i} = \frac14.$$

For the residue of $f$ in $-\frac i2$, we choose

$$g(z) = \frac{e^{\pi z}}{4(z-\frac i2)}$$

and find

$$\operatorname{res}(f;-\frac i2) = \frac{e^{-\pi i/2}}{-4i} = \frac14.$$

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  • $\begingroup$ Thanks for the answer. Do you know why WolframAlpha says zero? wolframalpha.com/input/?i=res[e^%28pi+z%29%2F%284z^2+%2B+1%29%2C+pi%2F2] Could you explain why you choose your g(z) function this way? $\endgroup$ – dingari Sep 26 '13 at 18:20
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    $\begingroup$ Your URL doesn't work for me (Wolfram can't interpret the input), but this gives $\frac14$. Looking at the URL, it seems you asked it for the residue in $\frac{\pi}{2}$, which of course is $0$, since it's not a pole. I chose $g$ as $(z - z_0)\cdot f(z)$, to have $f$ in the form $$f(z) = \frac{g(z)}{z-z_0}.$$ $\endgroup$ – Daniel Fischer Sep 26 '13 at 18:25
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    $\begingroup$ There you have a double pole. That requires a different method. For a double pole things are still simple, $$\frac{g(z)}{(z-z_0)^2} = \frac{g(z_0) + (z-z_0)g'(z_0) + (z-z_0)^2h(z)}{(z-z_0)^2} = \frac{g(z_0)}{(z-z_0)^2} + \frac{g'(z_0)}{(z-z_0)} + h(z),$$ so the residue is $g'(z_0)$. Here, $z_0 = \pi$ and $g(z) = e^{iz}$, so $g'(z_0) = ie^{iz_0} = ie^{i\pi} = -i$. $\endgroup$ – Daniel Fischer Sep 26 '13 at 19:25
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    $\begingroup$ $$\frac{ze^{iz}}{(z-\pi)^2} \leadsto (1+i\pi)e^{i\pi} = -(1+i\pi)$$ $\endgroup$ – Daniel Fischer Sep 26 '13 at 19:34
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    $\begingroup$ $h(z) = (g(z) - g(z_0) - (z-z_0)g'(z_0))/(z-z_0)^2$. If $g$ has the power series $g(z) = \sum\limits_{k=0}^\infty a_k (z-z_0)^k$, then $h(z) = \sum\limits_{m=0}^\infty a_{m+2}(z-z_0)^m$. The important thing is only that such a function $h$ exists and is holomorphic at $z_0$. $\endgroup$ – Daniel Fischer Sep 26 '13 at 19:41

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