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I'm currently trying to solve this however I get to the point where I have,

$$\int_{0}^{2\pi} \frac{ie^{\exp(it)}}{\exp(it)+3}.dt$$

am I on the right tracks? if yes could you help with the integration

jimmy
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  • Do you want this done using the residue theorem, or something more basic? – Chappers Apr 16 '15 at 01:01
  • if possible both would be much appreciated, if not then something more basic as I currently struggle to fully understand the residue theorem – jimmy Apr 16 '15 at 01:06

2 Answers2

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Put $f(z)=e^{z}/(z+3)$, since the only singularity of $f$ is when $z=-3$, $f$ is analytic on $\gamma$ and on the inside of $\gamma$ (which I denote int$(\gamma)$) because $-3 \not\in \gamma$ and $-3 \not\in \text{int}(\gamma)$. Note that $0 \in \text{int}(\gamma)$, then by the Cauchy Integral Formula we have $$ \int_{\gamma}\frac{e^z}{z(z+3)}dz= \int_{\gamma}\frac{f(z)}{z} dz = 2\pi i \ f(0)= \frac{2\pi i }{3} $$ because $f(0)=e^0/3=1/3$.

Alonso Delfín
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  • why is it that on the denominator of $f(z)$ you've removed a factor of $z$, also what do you mean by analytic on $\gamma$ and on the inside – jimmy Apr 16 '15 at 01:24
  • f(z) is not the one you have to integrate, is actually $f(z)/z=e^z/[z(z+3)]$. I have made some edits to clarify why $f$ is analytic on $\gamma$, essentially this means that $f$ has no poles neither on the curve $\gamma$ or on the inside (because -3 is the only pole of $f$). Perhaps you would like to draw $\gamma$ and locate the singularity. Tell me if it is clearer now. @jimmy – Alonso Delfín Apr 16 '15 at 01:29
  • I feel like I'm missing something simple, I understand that $-3$ is our only singularity point, however how do we know that $-3 \not\in \gamma$ or $-3 \not\in int(\gamma)$? – jimmy Apr 16 '15 at 01:35
  • Well note that since $\gamma=e^{it}$, $0 \leq t \leq 2\pi$, then $\gamma$ is actually the contour of the unit disk, that is that $z \in \gamma $ if and only if $|z|=1$ and $z \in \text{int}(\gamma)$ if and only if $|z|<1$ and of course $|-3|>1$. Now you see it ? @jimmy – Alonso Delfín Apr 16 '15 at 01:50
  • I see!!! thank you! :) – jimmy Apr 16 '15 at 01:54
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Although its a bit more involved than Alonso's method, for study purposes here is the solution using residue theorem: Note that $$ \int_{\gamma} \frac{e^z}{z(z+3)} dz$$ has two singularities. Namely, $z = 0$ and $z = -3$ with only one of these ($z=0$) falls inside the contour $\gamma$.

The residue theorem: $$ \int_{C} f(z) dz = 2 \pi i \sum_{n=1}^k Res_{a_n} (f(z)) $$ So:

$$ \int_{\gamma} \frac{e^z}{z(z+3)} dz = 2\pi i Res_0(f(z))$$

Since the residue of $f(z)$ at $0$ is the coefficient of the $z^{-1}$ term of the Laurent series at $0$. We can write: $$\frac{e^z}{z} \cdot \frac{1}{z+3} = \frac{e^z}{z} \cdot \frac{\frac{1}{3}}{1-(-\frac{z}{3})} = \sum_{n=0}^{\infty}\frac{z^{n-1}}{n!} \cdot \sum_{n=0}^{\infty} \frac{1}{3} \left (-\frac{z}{3} \right )^n = \left ( \frac{1}{z} + 1 + \frac{z}{2} + ... \right ) \left ( \frac{1}{3} + \frac{-z}{9} + \frac{z^2}{27} + ... \right ) = \frac{1}{3z}+\frac{2}{9} + \frac{5z}{54} +...$$

So $Res_0(f(z)) = \frac{1}{3}$. Since the pole is simple, you could have also used $\lim_{a \to 0} (z-a)f(z) = \frac{1}{3}$. Finally:

$$ \int_{\gamma} \frac{e^z}{z(z+3)} dz = \frac{2\pi i }{3}$$

kmeis
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  • when you say falls inside the contour, do you mean between $0$ and $2\pi $ – jimmy Apr 16 '15 at 01:48
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    The contour $\gamma$ is a circle so by falls inside, I mean in the interior or $int(\gamma)$ as noted in Alonso's. – kmeis Apr 16 '15 at 01:50