1
$\begingroup$

Thanks for any help in advance.

I'm trying to justify the answer to :

$$\oint_V \frac{e^{3z}}{z-\ln2} \,dz$$

over the square of vertices $\pm$$1$$\pm$$i$.

By the Cauchy's integral theorem, as there is a simple pole at $z = \ln 2$, so

$$\oint_V \frac{f(z)}{z-z_{0}} \,dz$$

becomes $2\pi i\cdot e^{3ln2} = 16\pi i$

However, if you expand $e^{3z}$, we have a fraction of

$$ \frac{1 + (3z) + \frac{(3z)^{2}}{2} + ...}{z-\ln2} $$

which gives us a residue of $1$ at $z = \ln 2$.

Then, by the residue theorem $$\oint_V f(s) \,ds$$ $ = 2\pi \cdot $ (sum of residues) $= 2\pi i \cdot (1)$

which gives us an answer of $ 2\pi i$

What is the flaw in my reasoning? Have i applied the residue theorem wrongly?

$\endgroup$
  • $\begingroup$ Correct me if I'm wrong but the pole is outside the contour? $\endgroup$ – asdf Apr 30 '18 at 10:38
  • $\begingroup$ @asdf $0<\ln2<1$. The pole is inside the contour. $\endgroup$ – Julián Aguirre Apr 30 '18 at 10:42
  • $\begingroup$ the pole's at ln2, which is roughly 0.693, inside the square $\endgroup$ – chickenpie Apr 30 '18 at 10:43
3
$\begingroup$

You are expanding $e^{3z}$ around $z=0$. You should expand it around $z=\ln2$.

$\endgroup$
  • $\begingroup$ what's problematic about expanding $e^{3x}$ about z = 0? $\endgroup$ – chickenpie Apr 30 '18 at 10:45
  • $\begingroup$ To find the residue at a pole $z_0$, you need the Laurent expansion in powers of $z-z_0$. The residue is the coefficient of $1/(z-z_0)$. $\endgroup$ – Julián Aguirre Apr 30 '18 at 10:51

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.