Integral of real part of $z$ around the unit circle

Question

Q 32 If C denotes the counterclockwise unit circle, the value of the contour integral is

What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?

At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.

But I tried using parameterization and got the answer to be i$\pi$. So I am guessing there is a residue of 0.5 at 0 maybe?

So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?

Answer 1

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.

First, we note that $\text{Re}(z) = \frac12(z+\overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $\overline{z}=\frac1z$. So, on the circle $|z|=1$ we care about, $$\text{Re}(z) = \frac12\left(z+\frac1z\right)$$ That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $\frac az$ around a circle - we needed that integral to prove the theorem in the first place)

Answer 2

$\displaystyle{1 \over 2 \pi i} \int_0^{2 \pi} \cos t \ ie^{it} dt = {1 \over 2 \pi} \int_0^{2 \pi} {1 \over 2 } (e^{it} + e^{-it}) \ e^{it} dt = {1 \over 2}$.