Finding the Laurent series of a function at $z=i$

Question

The question asks me to find the Laurent series of $$f(z) = {5z+2e^{3z}\over(z-i)^6}\,\,at\,\,z=i$$I know the following $$e^z=\sum_{n=0}^\infty {z^n\over n!}$$ What I want to know, is if I can do this: $$={1\over (z-i)^6}\sum_{n=0}^\infty ({2(3z)^n\over n!}+5z)$$ $$=\sum_{n=0}^{n=6}(z-i)^{n-6}({2(3z)^n\over n!}+5z)+\sum_{n=7}^{\infty}(z-i)^{n-6}({2(3z)^n\over n!}+5z)$$ This is all I could think of possibly doing and I'm not even sure if it is even something that I can even do, and I don't really know how to continue. I also need to find the radius of convergence but I feel once I have the series it will be easy to apply whatever test necessary to acquire it.

Answer 1

Noting $$ 5z+2e^{3z}=5(z-i)+5i+2e^{3i}e^{3(z-i)}==5(z-i)+5i+2e^{3i}\sum_{k=0}^\infty\frac{1}{k!}(z-i)^k $$ one has $$ f(z) = {5z+2e^{3z}\over(z-i)^6}={5i\over(z-i)^6}+{5\over(z-i)^5}+2e^{3i}\sum_{k=0}^\infty\frac{1}{k!}(z-i)^{k-6}.$$

Answer 2

For this kind of problem I recommend changing the variable to push the pole to zero. Here I'd set $w=z-i$. The problem becomes: find the Laurent series of $$g(w)=\frac{5(w+i)+2e^{3w+3i}}{w^6}$$ at $w=0$. The numerator is $$5i+5w+2e^{3i}\left(1+3w+\frac{3^2w^2}{2}+\frac{3^3w^3}{3!}+\cdots\right)$$ etc.