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!
Best Answer
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 $z^3$: $$ f(z)= e \cdot \left( 1 + (z + z^2 + z^3 + \ldots) + \frac 12(z^2 + 2 z^3 + \ldots) + \frac 16(z^3 + \ldots) + \ldots \right) \\ = e \cdot \left(1 + z + \frac 32 z^2 + \frac{13}{6} z^3 + \ldots\right) $$