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I want to prove this geometrically.

For function $g : \mathbf{C} \rightarrow \mathbf{R}$ and g is some continuous function.

The value of $\int_0^{2\pi} g(re^{i(\theta + \phi)}) \, d\phi$ is independent of $\theta$?

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    What would be a geometric proof of this? That the measure is translation invariant?2012-06-12

2 Answers 2

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The mapping $ \varphi \mapsto re^{i(\theta+\varphi)} $ maps the line to the circle of radius $r$ centered at $0$, and wraps once around the circle every time $\varphi$ increases by $2\pi$. You go around the whole circle once. It doesn't matter at which point on the circle you start. The parameter $\theta$ only identifies the starting point.

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Here is a diagram of the path of integration:

$\hspace{4cm}$enter image description here

$\displaystyle\int_0^{2\pi}g\left(re^{i(\theta+\phi)}\right)\,\mathrm{d}\phi$ is the integral of $g$ over the green arc, from $re^{i\theta}$ to $re^{i2\pi}$, and then over the red arc, from $re^{i2\pi}$ to $re^{i(\theta+2\pi)}$.

$\displaystyle\int_0^{2\pi}g\left(re^{i\phi}\right)\,\mathrm{d}\phi$ is the integral of $g$ over the red arc, from $re^{i0}$ to $re^{i\theta}$, and then over the green arc, from $re^{i\theta}$ to $re^{i2\pi}$.

Both integrate $g$ over the circle of radius $r$ centered at the origin.

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    @AD: hopefully, this passes as geometric :-)2012-06-12