# On complex circle $|z|=1$, evaluate $\int_C \frac{z + i}{z} \, dz$

Q: Where $C$ is the closed contour around the complex plane circle $|z|=1$ in the positive orientation, calculate:

$$\int_C \frac{z + i}{z} \, dz$$

Is this simply zero? If an antiderivative exists, the integral of a closed contour is always zero, right? That seems too easy.

The antiderivative would be $F(z) = z + i \log z$

• It equals $\int_C 1\, dz + i\int_C dz/z.$ What do you think?
– zhw.
Mar 9, 2017 at 17:39

Using the parametrization $$\gamma:[0,2\pi] \to \mathbb{C}, \quad \gamma(t)=e^{it}$$ of the unit circle, we get \begin{eqnarray} \int_C\dfrac{z+i}{z}\,dz&=&\int_C(1+iz^{-1})\,dz=\int_0^{2\pi}(1+ie^{-it})ie^{it}\,dt=\int_0^{2\pi}(-1+ie^{it})\,dt\\ &=&\left[-t+e^{it}\right]_0^{2\pi}=-2\pi \end{eqnarray}
You cannot define a $\log$ function on the whole complex plane. The integral is not zero, indeed rewrite $$\frac{z+i}{z} = 1 + i\frac{1}{z}$$ to see that the function has a simple pole at the origin with residue $i$. It follows by the residue theorem that $$\int_C\frac{z+i}{z}dz = -2\pi\ .$$
Consider the function $f(z)=z+i$ . Then for Cauchy integral formula one has: $$\int_{C}\frac{f(z)}{z-0}dz=2\pi if(0).$$ Hence , $$\int_{C}\frac{z+i}{z}dz=2\pi i(0+i)=-2\pi$$