Q: If $\tan(z) = \sum_{k=0}^\infty a_kz^k$ based at $z=0$, show that $a_k = 0$ for every even index $k$. Furthermore, calculate $a_1, a_3$.
So finding $a_1,a_3$ is relatively straightforward just by taking a few derivatives and using the fact that the power series expansion must match the functions Taylor Series, but I am having trouble proving the first part. Intuitively it makes sense. I could write $\tan(z) = \frac{sin(z)}{cos(z)}$ and then just do polynomial long-division on the Taylor series for $sin(z),cos(z)$ to see you only get odd terms in the Taylor series. Other than that, I tried using the alternative formula for the coefficients in the power series. Namely:
$$a_k = \frac{1}{2\pi i} \int_{|w| = r} \frac{\tan(w)}{w^{k+1}} dw$$
Which using the path parameterization $|w|=r \rightarrow re^{it}$ for $t \in [0, 2\pi]$ becomes:
$$a_k = \frac{1}{2\pi r^k} \int_0^{2\pi} \frac{\tan(re^{it})}{e^{ikt}} dt$$
But I was not able to accomplish much from there.
Best Answer
Hint: $\tan(-z) = -\tan(z)$...