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I am trying to find:

$$ \overline{\frac{-i \Gamma}{2\pi}\log\left(\frac{a^2}{\overline z}-b\right)} $$

Notice the conjugate sign above the $z$ and that $z$ is complex.

$\Gamma , a,b$ are real constants

I am aware of that $\overline{\log(\overline {Z})}=\log(Z)$

I am just confused on how to write this conjugate. Can I just change the minus sign to a plus and remove the conjugate sign above the $z$?

Thanks

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  • $\begingroup$ yes sorry, I didn't know how to write pi in the format. $\endgroup$ – juper Apr 25 '17 at 21:53
  • $\begingroup$ What is $\Gamma$? $\endgroup$ – Joe Apr 25 '17 at 21:56
  • $\begingroup$ it is the strength of a vortex - so just a constant. Sorry, I should have explained this. $\endgroup$ – juper Apr 25 '17 at 21:56
  • $\begingroup$ Are $a,b$ reals? $\endgroup$ – dxiv Apr 25 '17 at 21:56
  • $\begingroup$ yes sorry, I shall clarify this all in the question. apologies all, $\endgroup$ – juper Apr 25 '17 at 22:00
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It should be as easy as saying

$$\overline{\frac{-i \Gamma}{2\pi}\log\left(\frac{a^2}{\overline z}-b\right)}=\frac{i \Gamma}{2\pi}\log\left(\frac{a^2}{ z}-b\right)$$

In every example I have ever seen, the conjugate of the whole is the same as systematically taking the conjugate of each of the complex parts. The are many examples in Complex conjugate.

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  • $\begingroup$ yes that's what I thought. Thanks for confirming! $\endgroup$ – juper Apr 26 '17 at 13:32
  • $\begingroup$ Yeah, it's not an obvious thing and I think people over-think it. Just replace all $i$'s by $-i$. $\endgroup$ – Cye Waldman Apr 26 '17 at 13:58

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