Consider the branch of the function $f(z)=\log(z^2+1)$ defined on the domain $C-\{x+iy:x=0, |y|\ge1\}$ so that $f(0)=0$. Compute...

Question

Consider the branch of the function $f(z)=\log(z^2+1)$ defined on the domain $C-\{x+iy:x=0, |y|\ge1\}$ so that $f(0)=0$. Compute $f(1), f(-1), f(1+i), f(1-i)$

I am confused by how the domain restriction effects the computation. If i consider our constraints, I see branch cuts from the y axis from 1 inclusive to infinity and -1 inclusive to infinity

Is this correctly stated?

Then, I just plug in values for z and get

$f(1)=\log 2$

$f(-1)=\log 2$

$f(1+i)=\log(1-1+2i+1)=\log(2i+1)=\log\sqrt5+arctan(2)$

And similarly, $f(1-i)=\log\sqrt5+arctan(-2)$.

Am I missing something?

The problem then asks to compute $\int_\gamma zf(z)dz$ where $\gamma$ is the arc from -1 to 1 defined by $\gamma=\{x+iy:x^2+2y^2=1,y\ge0\}$

How does the arc change the problem? I don’t really know where to start with this part. I know Cauchy integral theorem but i think that requires a closed loop? I am having a hard time visualizing $\gamma$ and then knowing how to proceed

Thanks

Answer 1

Consider $\Gamma=[-1,1]$, the line segment from $-1$ to $1$. Then by Cauchy integral theorem we have$$ \int_\gamma zf(z) dz=\int_\Gamma zf(z)dz.$$ RHS is equal to$$ \int_{-1}^1 xf(x)dx=\int_{-1}^1 x\log(x^2+1) dx=0,$$ since the integrand is an odd function.