# Complex Taylor Series by substitution

I need to find the first few terms or so of the Taylor series centered at $z_0 = 0$ regarding these functions:

a) $e^{z\sin z}$

b)$(1+z)^z = e^{z \ln (1+z)}$

c)$\cos (1 + z^3)$

d) $e^{e^z}$

Although it's possible to continuously take derivatives of these functions, I feel that substitution would be a lot quicker and less painful. For a), should I use the series for $e^z$, and make this substitution:

$$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$ $$e^{z\sin z} = \sum_{n=0}^\infty \frac{(z\sin z)^n}{n!}$$

For b), if my logic is correct for a), this can be solved in a similar manner, although the $\ln(1+z)$ term seems interesting, as this has its own pretty well known MacLaurin series.

For c), the substitution is $$\cos z = \sum_{n=0} ^ \infty (-1)^n \frac{z^{2n}}{(2n)!}$$ $$\cos (1+z^3) = \sum_{n=0} ^ \infty (-1)^n \frac{(1+z^3)^{2n}}{(2n)!}$$

And likewise, for d), I need to do

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

Is this an appropriate way to go about this, or is there something else I'm missing that can really help?

Let me consider the first case $$y=e^{z \sin(z)}$$ What I could do is first $$\log(y)={z \sin(z)}=z^2-\frac{z^4}{6}+\frac{z^6}{120}-\frac{z^8}{5040}+O\left(z^{10}\right)$$ and now, using $$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+O\left(t^5\right)$$ in which $t$ could replace by the expression given for $\log(y)$. This would lead, using binomial expansions (not very funny, for sure !) to something like $$y=e^{z \sin(z)}=1+z^2+\frac{z^4}{3}+\frac{z^6}{120}-\frac{11 z^8}{560}+O\left(z^{10}\right)$$
For sure, as you wrote, you could do $$e^{z\sin z} = \sum_{n=0}^\infty \frac{(z\sin z)^n}{n!}$$ and use (limiting to order $10$) $$\big({z \sin(z)}\big)^1=z^2-\frac{z^4}{6}+\frac{z^6}{120}-\frac{z^8}{5040}+O\left(z^{10}\right)$$ $$\big({z \sin(z)}\big)^2=z^4-\frac{z^6}{3}+\frac{2 z^8}{45}+O\left(z^{10}\right)$$ $$\big({z \sin(z)}\big)^3=z^6-\frac{z^8}{2}+O\left(z^{10}\right)$$ $$\big({z \sin(z)}\big)^4=z^8+O\left(z^{10}\right)$$ and get the same result faster than with what I did (on purpose) at the start of my answer.