Evaluate the complex line integral $\int_C \frac{1+z}{z}dz$

Question

Evaluate the complex line integral $\int_C \frac{1+z}{z}dz$ where the curve is the right semicircle centered at $0$ with a radius of $4$, going upward.

My attempt:

First, I need to define the curve as $z$ which is a function of $t$, so $z(t)$. I know that the equation of a circle is $x^2+y^2=1$, so:

$$z(t)=\frac{1}{4}(x^2(t)+y^2(t))=\frac{1}{4}(1)$$

But I'm stuck after this.

Answer 1

Let $z=re^{i\theta}$. Then $dz=ire^{i\theta}d\theta$. For the curve, $r=4,-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$. Then \begin{align} \int_C\frac{1+z}{z}dz&=\int_{-\pi/2}^{\pi/2}\frac{1+4e^{i\theta}}{4e^{i\theta}}4ie^{i\theta}d\theta\\ &=i\int_{-\pi/2}^{\pi/2}\left(1+4e^{i\theta}\right)d\theta\\ &=i\pi+4e^{i\pi/2}-4e^{-i\pi/2}\\ &=i\pi+8i\sin(\pi/2)\\ &=(8+\pi)i \end{align}

Answer 2

Hint: You shouldn't use the equations $x^2+y^2$ since the radius is $4$ and not $1$. Also, we must pay heed to the fact that we're integrating over a semi-circle, not everything.

That semi-circle can be parametrized by $z(t) = 4e^{it}$, with $-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$, so: $$\int_C\frac{1+z}{z}\,{\rm d}z = \int_{-\pi/2}^{\pi/2}\frac{1+4e^{it}}{4e^{it}}\cdot 4ie^{it}\,{\rm d}t.$$Can you procced?

Answer 3

The function $\log z + z,$ where $\log$ denotes the principal value logarithm, is an antiderivative for $(1+z)/z$ in an open connected set containing the contour. Hence to compute the integral, just evaluate $(\log z + z)\big|_{-4i}^{4i}.$