0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $i-3$.

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

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 $1$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.