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?

  • $\begingroup$ 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

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

  • $\begingroup$ 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

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