Evaluate$\int_{\gamma(0;1)} \frac{\cos z}{z}dz$. Write in parametric form and deduce that$$\int^{2\pi}_0 cos(\cos\theta)\cosh(\sin\theta)d\theta=2\pi$$

By Cauchy's integral formula, $\int_{\gamma(0;1)} \frac{\cos z}{z}dz=2\pi i(\cos0) = 2\pi i $ , but could anyone help with parametrization and deducing the above integral?


The circle is a parametrization over $\theta \in [0,2 \pi]$. Now let $z=e^{i \theta} \implies dz = i z d\theta$. Also note that

$$\cos{z} = \cos{(\cos{\theta}+i \sin{\theta})} = \cos{(\cos{\theta})} \cos{(i \sin{\theta})} - \sin{(\cos{\theta})} \sin{(i \sin{\theta})}$$

Use the fact that $\cos{i x} = \cosh{x}$ and $\sin{i x} = i \sinh{x}$ to get

$$\cos{z} = \cos{(\cos{\theta})} \cosh{(\sin{\theta})} - i \sin{(\cos{\theta})} \sinh{( \sin{\theta})}$$


$$\oint_{\gamma(0,1)} dz \frac{\cos{z}}{z} = i \int_0^{2 \pi} d\theta \left [ \cos{(\cos{\theta})} \cosh{(\sin{\theta})} - i \sin{(\cos{\theta})} \sinh{( \sin{\theta})} \right ] = i 2 \pi$$

Equating real and imaginary parts, the sought-after result follows.


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