- #1

terhorst

- 11

- 0

I never took complex analysis in undergrad and always regretted it, so I'm working through the book Visual Complex Analysis on my own. Really enjoying it so far.

Actually you can view the problem http://books.google.de/books?id=ogz5FjmiqlQC&printsec=frontcover&hl=en#PPA117,M1", it's #21 at the top of the page.

Show that the Fourier series for [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] is [tex]\sum^{\infty}_{n=0} \frac{\cos n \theta}{n!}[/tex].

The author suggests substituting [tex]z=re^{i \theta}[/tex] into the power series for [tex]e^z[/tex] and then isolating the real and imaginary parts.

I don't really understand why the substitution. So then you end up with [tex]\sum \frac{r^n [\cos(n \theta) + i \sin (n \theta)]}{n !}[/tex]. But I don't see how this gets me any closer to equating [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] and [tex]\sum \frac{r^n \cos(n \theta)}{n!}[/tex].

A nudge in the right direction would be appreciated. Thanks for looking.

## Homework Statement

Actually you can view the problem http://books.google.de/books?id=ogz5FjmiqlQC&printsec=frontcover&hl=en#PPA117,M1", it's #21 at the top of the page.

Show that the Fourier series for [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] is [tex]\sum^{\infty}_{n=0} \frac{\cos n \theta}{n!}[/tex].

## Homework Equations

The author suggests substituting [tex]z=re^{i \theta}[/tex] into the power series for [tex]e^z[/tex] and then isolating the real and imaginary parts.

## The Attempt at a Solution

I don't really understand why the substitution. So then you end up with [tex]\sum \frac{r^n [\cos(n \theta) + i \sin (n \theta)]}{n !}[/tex]. But I don't see how this gets me any closer to equating [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] and [tex]\sum \frac{r^n \cos(n \theta)}{n!}[/tex].

A nudge in the right direction would be appreciated. Thanks for looking.

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