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I am trying to show that $$ \left\lvert\int_\gamma \frac{\cos(z)}z \,dz\right\rvert \le 2\pi e $$ if $\gamma$ is a path that traces the unit circle once.

The book recommends that I show that $\lvert \cos(z) \rvert \le e$ if $\lvert z \rvert = 1$ to help prove this. I know that if $\lvert f(z) \rvert \le M$ for all $z \in \gamma (I)$ then $$ \left\lvert \int_\gamma f(z) \,dz \right\rvert \le M\ell (\gamma), $$ where $\ell (\gamma)$ is the length of the path, which in this case is $2\pi$. So I can see why I would need to show $\lvert cos(z) \rvert \le e$ if $\lvert z \rvert = 1$ to prove the inequality, but I am not sure where to go from here to show that.

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3 Answers 3

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With the series expansion of $\cos$ we get for $|z|=1$:

$$| \cos z| \le \sum_{n=0}^{\infty}\frac{1}{(2n)!} \le \sum_{n=0}^{\infty}\frac{1}{n!}=e.$$

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The exponential function satisfies the inequality $\vert e^z \vert \leq e^{\vert z \vert}$. We have $$\cos(z) = \frac{1}{2}(e^{iz} + e^{-iz}),$$ thus \begin{equation*} \begin{aligned} \vert \cos(z) \vert &\leq \frac{1}{2}( \vert e^{iz} \vert + \vert e^{-iz} \vert) \\ &\leq \frac{1}{2}(e^{\vert z \vert} + e^{\vert z \vert}) \\ &= e^{\vert z \vert}. \end{aligned} \end{equation*} So when $\vert z \vert = 1$ we have $\vert \cos(z) \vert \leq e$.

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$|\cos z|=|\frac {e^{iz}+e^{-z}} 2| \leq \frac {e+e} 2=e$ because $|e^{z}| \leq e^{|z|}$.

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