How can I write the Taylor series for $e^z\sin(z)$ at $z=0$ without making the procedure too complicated?
Isn't there an easier way than to compute it's derivatives and find a pattern?
[Math] Taylor series for $e^z\sin(z)$
complex-analysistaylor expansion
Best Answer
Yes, there are simpler methods based on the power series expansion of $\exp$. I indicate two equivalent methods, the first one being more elementary than the second one.
Method 1: Use Euler's formula $\sin(z) = \dfrac{e^{iz} - e^{-iz}}{2i}$.
Method 2: For all $x \in \Bbb R$,
$$ e^{x}\sin(x) = \Im(e^{(1+i)x}) = \sum_{n=0}^\infty\frac{\Im((1+i)^n)}{n!}x^n= \sum_{n=0}^\infty \frac{2^{n/2}\sin(\frac{n\pi}{4})}{n!}x^n. $$ By analytic continuation, the identity still holds for $z \in \Bbb C$.