Integral with complex expoentials

Question

For any natural number $n$, how would one calculate the integral

$$ \int_{0}^{2 \pi} |1 - ae^{i\theta}|^n \ d \theta $$

where $a$ is a complex number such that $|a| = 1$. I real just need $n$ to be even, but I'm not sure how much this changes anything. I also don't know how necessary $a =1$ is in the problem either. I can see this function is the distance from 1 to a circle of radius $a$ but not sure how to compute this integral.

Answer 1

Hint:

If $|a|=1$, $a=e^{i\phi}$ and this phase factor can just be dropped (you are integrating over a whole period). Hence WLOG $a=1$.

Then for even $n$,

$$|1-e^{i\theta}|^{2m}=((1-\cos\theta)^2+\sin^2\theta)^m=2^m(1-\cos\theta)^m=4^m\sin^{2m}\frac\theta2.$$

Use https://en.wikipedia.org/wiki/List_of_definite_integrals#Definite_integrals_involving_trigonometric_functions (fifth row).

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For $|a|\ne 1$, the computation remains possible but requires the expansion of $(a^2+1-2a\cos\theta)^m$ and you end up with a linear combination of integrals of even powers of the cosine (the odd powers cancel out).

Answer 2

$|1-e^{i\theta}|^{2}=(1-\cos (\theta))^{2}+\sin ^{2} (\theta)=2-2\cos (\theta)=4\sin^{2}(\frac {\theta} 2)$ If $n=2m$ and $a=1$ the given integral becomes $4^{2m}\int_0^{2\pi} (\sin(\frac {\theta} 2))^{m}d\theta =2(4^{2m})\int_0^{\pi} \sin ^{m}(\theta) d\theta$. This standard integral can be evaluated using integration by parts.

The given integral does not depend on the exact value of $a$ as long as $|a|=1$. To show this write $a=e^{it}$ with $t$ real and make the substitution $\theta'=\theta+t$. Note that for a periodic function with period $2\pi$ the value of the integral over any intervals of length $2\pi$ is same.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{2\pi}\verts{1 - a\expo{\ic\theta}}^{\, n} \,\dd\theta\,\right\vert_{{\large a\ \in\ \mathbb{C}} \atop {\large\verts{a}\ =\ 1}}} = \oint_{\verts{z}\ =\ 1}\pars{\verts{1 - z}^{2}}^{\, n/2} \,{\dd z \over \ic z} \\[5mm] = & \oint_{\verts{z}\ =\ 1} \bracks{\pars{1 - z}\pars{1 - \overline{z}}}^{\, n/2} \,{\dd z \over \ic z} = \oint_{\verts{z}\ =\ 1} \pars{1 - z - \overline{z} + z\overline{z}}^{\, n/2} \,{\dd z \over \ic z} \\[5mm] = &\ \oint_{\verts{z}\ =\ 1} \pars{1 - z - {z\overline{z} \over z} + z\overline{z}}^{\, n/2} \,{\dd z \over \ic z} = \oint_{\verts{z}\ =\ 1}\pars{2 - z - {1 \over z}}^{\, n/2} \,{\dd z \over \ic z} \\[5mm] = &\ -\ic\oint_{\verts{z}\ =\ 1}{\pars{-1 + 2z - z^{2}}^{\, n/2} \over z^{n/2 + 1} }\,\dd z = -\ic\oint_{\verts{z}\ =\ 1}{\bracks{-\pars{z -1}^{\, 2}}^{\, n/2} \over z^{n/2 + 1} }\,\dd z \end{align}