This is my first post so I apologise for the notation not being very good.(adsbygoogle = window.adsbygoogle || []).push({});

I have been given some notes to check over by my PhD supervisor, and the notes he has given me state that:

[tex]i \omega \log \left(\frac{\sqrt{r^2-\omega ^2}-i \omega }{r}\right)[/tex] = [tex]\tan ^{-1}\left(\frac{\omega }{\sqrt{r^2-\omega ^2}}\right)[/tex]

But I can't seem to reproduce this answer using ArcTan(x) = (i/2)ln(1-ix/1+ix). Can someone please either show me how to obtain this ArcTan result from the exponential equation I have given or tell me that it's wrong.

Thank you.

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# Does iωLog[(-iω + √((r^2-ω^2)/r)] = ArcTan[ω/√(r^2-ω^2)]?

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