Complex integral

Question

Let $C$ denote the unit circle centered at the origin in Complex Plane

What is the value of

$$ \frac{1}{2\pi i}\int_C |1+z+z^2 |dz,$$ where the integral is taken anti-clockwise along $C$?

1. 0
2. 1
3. 2
4. 3

What I have answered is 0 because it seems like $f(z)$ is analytic at 0 hence by Cauchy's Theorem.

Answer 1

HINT Note that if $z = x+yi$ then $$ \left|1+z+z^2\right| = \left|\left(x^2-y^2+x+1\right) +y(2x+1)i\right| $$ Now parameterize $C$ as $x = \cos t, y = \sin t$ and $t \in [0,2\pi]$, then $$ \int_C |1+z+z^2 |dz = \int_{t=0}^{t=2\pi} \left|\left(x^2-y^2+x+1\right) +y(2x+1)i\right| $$ where $x,y$ are functions of $t$ as in the parameterization. Can you finish this?

Answer 2

We want the integral

$$\frac{1}{2\pi i}\int_C |1+z+z^2|dz=\frac{1}{2\pi}\int_0^{2\pi}|1+z+z^2|z~d\theta$$

since $z=e^{i\theta}$.

Now consider the absolute value portion,

$$ \begin{align} |1+z+z^2| &=\sqrt{(1+z+z^2)(1+z+z^2)^*}\\ &=\sqrt{(1+z+z^2)(1+z^{-1}+z^{-2})}\\ &=\sqrt{(1+z+z^2)(1+z^{-1}+z^{-2})\frac{z^2}{z^2}}\\ &=\sqrt{\left(\frac{1+z+z^2}{z} \right)^2}\\ &=\sqrt{\left(\frac{1}{z}+1+z \right)^2}\\ &=\sqrt{(1+2\cos\theta)^2}\\ &=|1+2\cos\theta| \end{align} $$

We can return to solve the integral

$$ \begin{align}\frac{1}{2\pi i}\int_C |1+z+z^2|dz &=\frac{1}{2\pi}\int_0^{2\pi}|1+2\cos\theta|(\cos\theta+i\sin\theta)~d\theta\\ &=\frac{1}{2\pi}\int_0^{2\pi}|1+2\cos\theta|\cos\theta~d\theta\\ &=\frac{1}{3}+\frac{\sqrt{3}}{2\pi}\approx0.60900 \end{align}$$

This is in agreement with our own numerical solution as well as that of others noted previously.

Answer 3

Put $z=e^{it}, 0\le t\le 2π$ and $dz=i.e^{it}dt$.

So $I_1=\int_0^{2π}|1+e^{it}+e^{i2t}|^2ie^{it}dt=i\int_0^{2π}(1+e^{it}+e^{i2t})(1+e^{-it}+e^{-i2t})e^{it}dt$

$=i\int_0^{2π}g(t)dt$, where $g(t)$ is the integrand and $g(2π+t)=g(t)$. So $I_1=...$