# Give an example of a function $f$ which has a simple pole at $z=i-3$ with residue $8(i-3)$ and an essential singularity at $z=i$ with residue $6$

Give an example of a function $f$ which has a simple pole at $z=i-3$ with residue $8(i-3)$ and an essential singularity at $z=i$ with residue $6$.

I know that the function $-24/(z+3-i)$ would have a pole at $z=i-3$ with residue $8(i-3)$ but I cannot seem to make this function also have an essential singularity. Would making this function trigonometric help.

Hint. Try something of the form $$f(z)=e^{1/(z-i)^2}+\frac{A}{z-i}+\frac{B}{z-(i-3)}$$ where $A$ and $B$ are complex numbers to be found. The essential singularity is given by $e^{1/(z-i)^2}$ which has residue $0$ at $i$ and at $i-3$.

• I see that the last part of that function is where the pole would be however where is the essential singularity? – user544158 Mar 21 '18 at 14:12
• In $\exp(1/(z-i)^2)$ that has residue $0$ at $i$ – Robert Z Mar 21 '18 at 14:28
• Recall that $e^{1/w}=\sum_{k\geq 0}(1/w)^k/k!$. – Robert Z Mar 21 '18 at 14:32
• Is it clear now? Are you able to find $A$ and $B$? – Robert Z Mar 21 '18 at 14:50
• I found that if B is equal to 8(i-3), this will give me the pole i-3 with residue 8(i-3). However I am struggling to find the residue 6 for the singularity at i. – user544158 Mar 21 '18 at 14:53

Hint:

1. $e^{1/(z-i)}=1+\frac1{z-i}+\frac1{2(z-i)^2}+\dots$ has an essential singularity at $z=i$ with residue $1$.

2. $\frac1{z+3-i}$ has a simple pole at $z=i-3$ with residue $1$.

• Would you please be able to explain why hint number 1 has residue 1? – user544158 Mar 21 '18 at 15:00
• Remember that the residue at $z=i$ is the coefficient of the $\frac1{z-i}$ term in the Laurent expansion about $z=i$ – robjohn Mar 21 '18 at 16:09