Dirac delta from polar coordinates to cartesian coordinates I have:
$$k_x = k \cos\theta\\k_y=k\sin\theta$$
I would like to rewrite in terms of $k_x$ and $k_y$:
$$\exp(in\theta)\,\frac{\delta(k-\alpha)}{k}$$
I start from:
$$\exp(in\theta)=\left(\cos\theta+i\sin\theta\right)^n=\left(k_x+ik_y\right)^n\,k^{-n}$$
but I don't if it helps and if yes how to go on.
 A: Your approach for eliminating $\theta$ seems preferable to the one suggested in jwg's answer. Combining it with jwg's $k=\sqrt{k_x^2+k_y^2}$ yields
$$
\exp(\mathrm in\theta)\frac{\delta(k-\alpha)}k=(k_x+\mathrm ik_y)^n\left(k_x^2+k_y^2\right)^{-(n+1)/2}\delta\left(\sqrt{k_x^2+k_y^2}-\alpha\right)\;.
$$
Depending on what you want to use this for, you may need to use the formula for composition of $\delta$,
$$
\delta(g(x))=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}\;,
$$
where the sum extends over all roots $x_i$ of $g(x)$, which are assumed to be simple.
A: There are two other formulae you need, which express $k$ and $\theta$ in terms of $k_x$ and $k_y$:
$$k = \sqrt{k_x^2 + k_y^2}$$
and 
$$ \theta = \tan^{-1}\left(\frac{k_y}{k_x}\right) $$
Note that the second one always gives a value for $\theta$ between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$. To get the actual value you should eg add $\pi$ whenever $k_y$ is negative (and renormalize if necessary). Programming languages and computer algebra systems usually have a two-argument atan2 that does this for you. 
