Would the solution to the following integral:
$$\int_{|z|=1} e^{2z} \frac1z \,dz$$
be 2$\pi$i when using Cauchy's Integral Formula?
Also, how would I use the value from this integral to evaluate the following:
$$\int_0^{2𝜋} e^{2\cosθ}\cos(2\sinθ) \,dθ$$
Best Answer
Yes, the solution would be $2\pi i$. Letting $z=e^{i\theta}$, we have $\text{d}z=ie^{i\theta}\text{d}\theta$, and your integral becomes
$$i\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta-\int_{0}^{2\pi}e^{2\cos\theta}\sin\left(2\sin\theta\right)\text{d}\theta.$$
Notice that the right integral equals $0$ because it is odd with a period of $2\pi$. Thus,
$$\int_{\left\lvert z\right\rvert=1}\frac{e^{2z}}{z}\text{d}z=i\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta=2\pi i,$$
giving $$\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta=2\pi.$$