Find the first four non-zero terms of the Maclaurin series of $f(z)=e^\frac{1}{1-z}$ Consider the following complex function:
$$f(z)=e^\frac{1}{1-z}$$
Find the first four non-zero terms of the Maclaurin series for $f(z)$, so around $z_0=0$.
I've asked my lecturer what's the best way to solve this and she simply said "just differentiate the function 4 times and find its value at $z=0$". I'm refusing to accept this and I'm sure there's a smarter way to approach this.
Unfortunately I'm clueless about this one and have no idea how to even start. I tried thinking of a function $g(z)$, which I know its Maclaurin series already, such that $g(z)f(z)=1$ and then comparing coefficients but no luck of finding one.
Any help would be grealy appreciated!
 A: One can compose the Taylor series for the exponential function with the geometric series. The “trick” is to write
$$
 \frac{1}{1-z} = 1 + \frac{z}{1-z}
$$
where the second term on the right has a Taylor series without constant term, so that it can be substituted into the exponential series:
$$
f(z)=e^{1/(1-z)} = e \cdot e^{z/(1-z)} = e \cdot \sum_{k=0}^\infty \frac{1}{k!} \left( \frac{z}{1-z}\right)^k \\
= e \cdot \sum_{k=0}^\infty \frac{1}{k!} z^k(1 + z + z^2 + z^3 + \ldots )^k
$$
Now collect all terms with powers from $z^0$ up to $z^3$:
$$
f(z)= e \cdot \left( 1 + (z + z^2 + z^3 + \ldots) + \frac 12(z^2 + 2 z^3 + \ldots) + \frac 16(z^3 + \ldots) + \ldots \right) \\
= e \cdot \left(1 + z + \frac 32 z^2 + \frac{13}{6} z^3 + \ldots\right)
$$
A: From the Maclaurin series of $$\dfrac{1}{1-z}=1+z+z^{2}+z^{3}+O\left( z^{4}\right) $$ and $$\exp(z)=  1+z+\dfrac{1}{2!}z^{2}+\dfrac{1}{3!}z^{3}+O(
z^{4}) ,$$  we can obtain the expansion of $\exp \left( \frac{1}{1-z}\right)$ as explained in the following steps:
\begin{eqnarray*}
\exp \left( \frac{1}{1-z}\right)  &=&\exp \left( 1+z+z^{2}+z^{3}+O\left(
z^{4}\right) \right)\\&=& \exp \left( 1 \right)\exp \left( z \right)\exp \left( z^2 \right)\exp \left( z \right)\exp \left( z^3 \right) \exp \left( O(z^4) \right)  \\
&=&\underset{\exp \left( 1\right) }{\underbrace{e}}\,\underset{\exp \left(
z\right) }{\underbrace{\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O(
z^{4})\right)  }}\times  \\
&&\qquad\underset{\exp \left( z^{2}\right) }{\times \underbrace{\left(
1+z^{2}+O( z^{4}) \right) }}\, \underset{\exp \left( z^{3}\right) }{
\underbrace{\left( 1+z^{3}+O( z^{4}) \right ) }} \\
&=&e\left( 1+z+\frac{1}{2}z^{2}+\frac{1}{6}z^{3}+O( z^{4})
\right) \underset{\exp \left( z^{2}\right) \exp \left( z^{3}\right) }{
\underbrace{\left( 1+z^{2}+z^{3}+O( z^{4}) \right ) }} \\
&=&e\left( 1+z+( 1+\frac{1}{2}) z^{2}+( 1+1+\frac{1}{6})
z^{3}+O( z^{4}) \right) \\
&=&e+ez+\frac{3}{2}ez^{2}+\frac{13}{6}ez^{3}+O( z^{4}) .
\end{eqnarray*}
