Complex Taylor series for sin(z)?

**cassius** · November 26th 2011, 09:39 AM

I know the general formula for a Taylor series about z_0 (z subscript 0) in the complex plane, here http://upload.wikimedia.org/wikipedi...279ed2da10.png, and the Taylor formula for sinx in the real plane, http://upload.wikimedia.org/wikipedi...75a8fd748a.png
i also know the expansion for e^z about z_0 in the complex plane

now i need to work out the Taylor formula formula for sin(z) about the point z_0, not zero. i can see it should resemble the real formula, but i don't know how to approach it.

would you use the definition of complex sine = (e^iz - e^(-iz))/2i and go on from there ??

**chisigma** · November 26th 2011, 09:54 AM

Originally Posted by cassius

I know the general formula for a Taylor series about z_0 (z subscript 0) in the complex plane, here http://upload.wikimedia.org/wikipedi...279ed2da10.png, and the Taylor formula for sinx in the real plane, http://upload.wikimedia.org/wikipedi...75a8fd748a.png
i also know the expansion for e^z about z_0 in the complex plane

now i need to work out the Taylor formula formula for sin(z) about the point z_0, not zero. i can see it should resemble the real formula, but i don't know how to approach it.

would you use the definition of complex sine = (e^iz - e^(-iz))/2i and go on from there ??

Your approach is fully correct... take into account that the taylor series expansion of $e^{i z}$ and $e^{-i z}$ around a point $z_{0}$ are...

$e^{i z}= e^{i z_{0}}\ \sum_{n=0}^{\infty} i^{n} \frac{(z - z_{0})^{n}}{n!}$ (1)

$e^{-i z}= e^{-i z_{0}}\ \sum_{n=0}^{\infty} (-i)^{n} \frac{(z - z_{0})^{n}}{n!}$ (2)

Kind regards

$\chi$ $\sigma$

**cassius** · November 27th 2011, 08:28 AM

Dear Chisigma,

you kindly answered my question about the taylor expansion of complex sine, http://www.mathhelpforum.com/math-he...-z-192734.html
i made some progress on the problem, but the result i got is a rather long expression, i attach the image here http://savepic.su/813764.gif

is there any way it can be simplified?
thank you so much for your help

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