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Show that if the Laurent series $\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ represents an even function, then $a_{2n+1}=0$ for $n=0,\pm 1,\pm 2,\ldots$, and if it represents an odd function, then $a_{2n}=0$ for $n=0,\pm 1,\pm 2,\ldots$.

where

$a_n=\frac{1}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}}dz$

I know the fact that if $f(-z)=f(z)$ then $f$ is even. But I have difficulty applying this to show what i need to have.

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What happens if you replace $z$ with $-z$ and compare coefficients with the obvious odd/even counterparts? –  J. M. May 10 '12 at 12:50
    
If we are going to replace $z$ by $-z$ we have $\sum_{n=-\infty}^\infty a_n (-z-z_0)^n$ where $a_n=\frac{1}{2\pi i}\int_C \frac{f(-z)}{(-z-z_0)^{n+1}}$. So, $-\sum_{n=-\infty}^\infty a_n (z+z_0)^n$ where $a_n=\frac{1}{2\pi i}\int_C \frac{f(z)}{(z+z_0)^{n+1}}$. I dont know what to do after... –  ron May 12 '12 at 7:22
2  
This isn't true as it is currently stated. Take $f(z)=-1/z$ around $z=-1$; the expansion is $$\frac{1}{1-(z+1)}=\sum_{n=0}^\infty (z+1)^n.$$ The function $f$ is odd in the complex plane but the coefficients above are $1$, not $0$, for even $n$. The same sort of counterexample can be done for even functions. The claim is true around $z=0$ however. –  anon May 12 '12 at 15:01
    
To be honest I think Zarrax's answer is exactly what you're looking for ; your question needs to consider only the case $z_0 = 0$, and in this case he says everything that is necessary to solve the problem. You should really consider everything that's on the comments and in his answers and offer him the bounty if you still wish to give it. –  Patrick Da Silva May 12 '12 at 17:38
    
yes. Thank you. –  ron May 12 '12 at 18:12
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2 Answers

up vote 3 down vote accepted
+50

This is only true if $z_0 = 0$. Suppose $f(z)$ is even. Then $f(z) - f(-z) = 0$. But $$f(z) - f(-z) = \sum_{n = -\infty}^{\infty} a_nz^n - \sum_{n = -\infty}^{\infty} a_n(-z)^n$$ $$= \sum_{n = -\infty}^{\infty} a_nz^n - \sum_{n = -\infty}^{\infty} (-1)^na_nz^n$$ $$= \sum_{n = -\infty}^{\infty} (1 - (-1)^n)a_nz^n$$ The above Laurent series gives the zero function by evenness of $f(z)$. But by uniqueness of Laurent expansions this implies all coefficients are zero. So $(1 - (-1)^n)a_n = 0$ for all $n$. When $n$ is odd this gives that $2a_n = 0$ for all odd $n$, or equivalently that $a_n = 0$ for all odd $n$.

A similar argument works when $f(z)$ is odd.

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@rockabhai: When $n$ is even, we get $$\big(1-(-1)^n\big)a_n=(1-1)a_n=0a_n=0,$$ which is true no matter what $a_n$ is. –  anon May 12 '12 at 16:02
    
The statement $0=0$ is true independently of whatever $a_n$ is. So for any even $n$, the coefficient $a_n$ can be anything. –  anon May 12 '12 at 16:07
    
What makes you think $a_n$ depends on $z$? –  anon May 12 '12 at 16:19
    
$a_n$ is not dependent on $z$. –  Zarrax May 12 '12 at 17:02
    
@rockabhai: Would you also say $g(x)=\int_0^x zdz$ depends on $z$? It makes no sense; $z$ is what's called a dummy variable. If it helps, use a $w$ in the formula, $$a_n:=\frac{1}{2\pi i}\oint\frac{f(w)}{(w-z_0)^{n+1}}dw.$$ This coefficient depends on the function $f$, the integer $n$ and the choice of point $z_0$. It does not "depend on $w$." As we've said, here we must restrict ourselves to $z_0=0$ (the claim isn't true otherwise), and here there is a shortcut to finding the coefficient of $f(-z)$: just substitute $z\mapsto-z$ in the Larent expansion of $f(z)$. This is what Zarrax does. –  anon May 12 '12 at 17:43
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You need to have $z_0=0$ here. Then \begin{align} 2\pi i a_n &= \int_{|z|=r} \frac{f(z)}{z^{n+1}} \,dz \\ &= \int_0^{2\pi} \frac{f(re^{i\theta})}{r^{n+1}e^{i(n+1)\theta}} ire^{i\theta} \, d\theta \\ &= \frac{i}{r^n} \int_0^{2\pi}f(re^{i\theta})e^{-in\theta} \,d\theta \end{align} where $r$ lies between the inner and outer radii of the annulus of convergence and we have made the substitution $z=re^{i\theta}$. So \begin{align} 2\pi r^na_n &= \int_0^\pi f(re^{i\theta})e^{-in\theta} \,d\theta + \int_\pi^{2\pi} f(re^{i\theta})e^{-in\theta} \,d\theta \\ &= \int_0^\pi f(re^{i\theta})e^{-in\theta} \,d\theta + \int_0^\pi f(re^{i(\theta+\pi)})e^{-in(\theta+\pi)} \,d\theta \\ &= \int_0^\pi f(re^{i\theta})e^{-in\theta} \,d\theta + (-1)^n\int_0^\pi f(-re^{i\theta})e^{-in\theta} \,d\theta \\ &= \int_0^\pi\Big[ f(re^{i\theta}) + (-1)^n f(-re^{i\theta}) \Big]e^{-in\theta} \,d\theta \end{align} and now you can see exactly the effect of the symmetry of $f$ and parity of $n$.

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