The book's suggestion is that $f(\theta) = Re(e^z)$ and it tells to expand $e^z$ using taylor series and show $f(\theta) = Re(e^z) = \sum_{n=0}^{+\infty} \frac{r^n}{n!}\cos(n\theta)$
but is it correct to assume $f(\theta) = Re(e^z)$
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Why do you think it is not right? Try working backwards, writing $z=re^{i\theta}$ and take the real part of $e^z$ – Ninad Munshi Apr 12 '21 at 21:11
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Because the question is determine fourier series for $f(\theta) = e^{r\cos(\theta)}\cos(r\sin(\theta))$ and not $ f(\theta) = e^{r\cos(\theta)}$ . – Gabriel Ribeiro Apr 12 '21 at 21:13
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Your comment doesn't make sense? It would be even less sensible to think of using complex without the cos part. – Ninad Munshi Apr 12 '21 at 21:19
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I know $e^z = e^{r(cos(\theta) + isen(\theta))} $ and $Re(e^z) = e^{rcos(\theta)}$. It is easy to show the fourier of this series. But what happens with that part $cos(rsin(\theta))$ ? – Gabriel Ribeiro Apr 12 '21 at 21:24
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Oh perfect you're on the right track then! Just substitute $z=re^{i\theta}$ into the Taylor series then take the real part – Ninad Munshi Apr 12 '21 at 21:26
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The real part of Taylor series for $e^{r(cos(\theta) + isin(\theta))} = \sum_{n=0}^{+\infty} \frac{r^n}{n!}\cos(n\theta)$. – Gabriel Ribeiro Apr 12 '21 at 21:41
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I GOT IT. NOW I GOT IT! THANK YOU VERY MUCH!! – Gabriel Ribeiro Apr 12 '21 at 23:01