Find the Taylor series for $f(z)=e^z$ about $z_0=1+i$.

Question

Find the Taylor series for $f(z)=e^z$ about $z_0=1+i$.

I know that I want to use the geometric series for $e^z$ which goes $1+z+\frac{z^2}{2!}+\frac{z^3}{3!}...$, but this is centered around $z_0=0$. How would I go about changing this for $z_0=1+i$?

Answer 1

Since $e^z = e^{z_0} \cdot e^{z-z_0}$, we have

$$ e^z = e^{z_0} \cdot \sum_{k=0}^{\infty} \frac{(z-z_0)^k}{k!} = e^{1+i}\cdot\sum_{k=0}^{\infty} \frac{(z-1-i)^k}{k!}. $$

Answer 2

HINT:

In general, the Taylor series is defined as follows: $$ f(z_0)+\frac{f'(z_0)}{1!}(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\frac{f'''(z_0)}{3!}(z-z_0)^3+\cdots $$ So what you have written as the Taylor expansion is actually a special case of this, since all the derivatives of $e^x$ at $0$ are just $1$. If you can find a way to express the derivatives of $e^x$ at $1+i$, then all you need to do is to plug those in.