Contour integration example question

Question

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

Answer 1

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 $f(0)=e^0/3=1/3$.

Answer 2

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:

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