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Question

If $$z = re^{i\theta}$$, then prove that $$\left|e^{iz}\right| = e^{-r} sin \theta$$. 


Solution

$$z=re^{i\theta}$$
$$=r(cos\theta+isin\theta)$$
$$z'=e^{iz}$$
$$=e^{ir(cos\theta+isin\theta)}$$
$$=e^{r(-sin\theta+icos\theta)}$$
$$=e^{-rsin\theta}.e^{icos\theta}$$
$$=R.e^{i\phi}$$ where $$R=e^{-rsin\theta}$$ and $$\phi=cos\theta$$
Hence
$$|z'|$$
$$=|e^{iz}|$$
$$=R$$
$$=e^{-rsin\theta}$$

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