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So i have seen things like $ \int ... d(\cos x)$ and now i have an expression with $\int d\operatorname{Im}(z)$ so i though my integral would be quite straight forward, knowing the Imaginary part of $z$.

$$ z = \rho \, e^{i \theta} $$

This is my Integral in full: \begin{equation} \int_{-\infty}^{+\infty} d\operatorname{Im}(z) \int_{-\infty}^{+\infty} d\operatorname{Re}(z) \; e^{-|z|^2} \; (z^*)^n \; z^m \label{imre-eqn} \tag{1} \end{equation}

This is the next step, which i cant get to: \begin{equation} \int_{0}^{+\infty} d\rho \; \rho^{n+m+1} \; e^{-\rho^2} \; \int_{0}^{2\pi} d\theta \; e^{i\theta (m-n)} \label{exp-eqn} \tag{2} \end{equation}

The problem is basically that i don't know where to search to find how to get from (\ref{imre-eqn}) to (\ref{exp-eqn}). any tipp? or what can i put into google to find out ?

Or, can somebody give me a more simple example of changing the integral variable from Im and Re to $\to$ what looks like spherical coordinates

that would be fantastic thankyou!

Thanks! xyz

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2 Answers 2

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You can use the identities $\Im(z)=-\frac{i}{2} (z-\bar{z})$ and $\Re(z)=\frac{1}{2}(z+\bar{z})$. You will see that $\Im(z)=\rho\sin(\theta)$ and $\Re(z)=\rho\cos(\theta)$. It should be easy from here to transform your coordinates to $(\rho,\theta)$, I hope this helps.

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  • $\begingroup$ I have put d(ρ cos( θ)) into the integral d- part but don't know how to resolve it.. Why is the solution just d θ and dρ? Ι am German and don't know how to google it in English :( how do I convert to spherical Koordinates.. Thanks for your answere x $\endgroup$
    – Ella
    Dec 17, 2021 at 23:57
  • $\begingroup$ There is a typo here. $\Im(z)=\rho\sin\theta$. Welcome Mr. Calculus; some MathJax advice - use backslashes to get special functions! If you're unsure how to render something, chances are that placing a backslash together with a sensible abbreviation will work. E.g., \Im, \rho, \sin\theta all render properly $\endgroup$
    – FShrike
    Dec 18, 2021 at 0:13
  • $\begingroup$ Op is a struggling with the transformation itself - worth fleshing out the answer a bit more $\endgroup$
    – FShrike
    Dec 18, 2021 at 0:14
  • $\begingroup$ I'm getting $\rho^{n+m\color{red}{+}1}$ and $e^{\color{red}{+}\rho^2}$ in my derivation, but I could be wrong. As for backslashes I just meant that Mr Calculus is writing im = rho sin theta when they should write \Im(z)=\rho\sin\theta @Ella $\endgroup$
    – FShrike
    Dec 18, 2021 at 0:28
  • $\begingroup$ Right. How do I get to the $ /rho ^ {n+n+1} $ how do I split the d{/rho *cos / theta "}? Just into two integrals? d$/rho} d{cos/theta"} $\endgroup$
    – Ella
    Dec 18, 2021 at 0:41
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all i needed was someone to remind me that Re and Im can be drawn just like x and y in a koordinate system! And so the dIm and dRe can be translated to polar coordinates!

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jan 11 at 15:32

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