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.

  • 1
    $\begingroup$ Be careful! $z\mapsto |1+z+z^2|$ is real-valued and not constant and therefore not holomorphic! $\endgroup$ Jun 19, 2017 at 16:35
  • 1
    $\begingroup$ $|z|$ is not analytic at 0 $\endgroup$
    – gt6989b
    Jun 19, 2017 at 16:35
  • $\begingroup$ Numerically integrating seems to give $\approx 0.6$ $\endgroup$
    – user399601
    Jun 19, 2017 at 16:42
  • $\begingroup$ @LutzL Actually, it's not a duplicate; the results are different, although the methods are similar. I have provided solutions to both of them. Also, none of the choices in the OP are correct, as also noted by user399601. See my solution below. $\endgroup$ Jun 21, 2017 at 17:27

3 Answers 3


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?


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.


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 $I_1=...$

  • $\begingroup$ Why do you put a square on the absolute value? It wasn't there to begin with. $\endgroup$ Jun 19, 2017 at 19:13
  • $\begingroup$ Cye Waldman..I misread it.. $\endgroup$ Jun 19, 2017 at 20:02

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