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I had this exam question today and I could not figure it out. Consider the function: $$f(z)=\frac{1}{z^2+1}$$ For $\rho>1$ show that $|f(z)|\leq \frac{1}{\rho^2-1}$ for $|z|=\rho$. I thought about taking the contour integral along the circle $C:|z|=\rho$ and then using the inequality $$\bigg|\int\limits_{C}f(z)dz\bigg| \leq \max\limits_C(|f(z)|)\cdot l(C)$$ where $l(C)$ is the arclength of the contour which is $2\pi \rho$. The contour integral can be calculated using residue theorem but the integral equals 0 so this does not help me. Any suggestions?

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    $\begingroup$ No need to use complex analysis - just triangle inequality. $\endgroup$ – user27126 Jan 22 '13 at 21:54
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You can use the reverse triangle inequality. Notice that $$ |z^2+1|=|z^2-(-1)|\geq \bigl||z^2|-|-1|\bigr|=\bigl||z^2|-1|\bigr|\geq |z^2|-1=|z|^2-1. $$ If $|z|>1$, then $|z|^2-1>0$, so this implies $$ \frac{1}{|z^2+1|}\leq\frac{1}{|z|^2-1}. $$ So in your case $$ |f(z)|=\frac{1}{|z^2+1|}\leq\frac{1}{|z|^2-1}=\frac{1}{\rho^2-1}. $$

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  • $\begingroup$ Ah of course, thanks. I was looking for an answer using complex analysis so I didn't think of using the triangle equality. $\endgroup$ – Slugger Jan 22 '13 at 22:28

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