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The complex conjugate is distributive over addition, subtraction, multiplication and division:

$$ \overline{z+w} = \bar z + \bar w, $$ $$ \overline{z*w} = \bar z * \bar w, $$ etc.

Is it also distributive over exponentiation, i.e., is ($\overline{z^w} = \bar z ^ \bar w$ ?)

I was able to prove this for real $w$:

$$ z^w = (a*e^{i\theta})^w = a^w*e^{iw\theta} = a^w(\cos w\theta + i\sin w\theta) $$

Thus,

$$ \overline {z^w} = \overline {a^w(\cos w\theta + i\sin w\theta)} = a^w(\cos w\theta - i\sin w\theta) = a^w*e^{iw(-\theta)} = a^w*e^{(-iw\theta)} = a^w*(e^{-i\theta})^w. $$

Then, as $e^{\overline {i\phi}} = \overline {e^{i\phi}}$ for any real $\phi$,

$$ \overline {z^w} = a^w*{\overline {(e^{i\theta})}}^w = \overline {a^w * {e^{i\theta}}^w} = \overline {(a*e^{i\theta})}^w. $$

Therefore

$$ \overline {z^w} = (\bar z) ^ w $$

which for real $w$ equals $(\bar z) ^ {\bar w}$. However, I could not prove the theorem for complex $w$. Does the property still hold? If it does, how can it be proved?

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  • $\begingroup$ Definition: $z^w = \exp(w\ln z)$. Now can you do it? $\endgroup$ – GEdgar May 10 '16 at 15:01
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    $\begingroup$ If you want a grounded answer, you need to provide a rigorous definition of complex exponentiation. $\endgroup$ – Yves Daoust May 10 '16 at 17:46
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The answer is in the affirmative provided $\mathrm{Arg}(z) \neq -\pi$. For $z,w \in \mathbb{C}$ we write $z^w = e^{w\mathrm{Log}(z)}$. Also, recall that:

  • $\mathrm{Arg}(\overline{z})=-\mathrm{Arg}(z)$, for $\mathrm{Arg}(z)\neq-\pi$ and
  • $e^{\overline{z}}=\overline{e^z}$.

Now, \begin{equation} \overline{z^w}=\overline{e^{w\mathrm{Log}(z)}}=e^{\overline{w(\mathrm{ln}|z|+i\mathrm{Arg}(z))}}=e^{\overline{w}(\mathrm{ln}|z|-i\mathrm{Arg}(z))}=e^{\overline{w}(\mathrm{ln}|\overline{z}|+i\mathrm{Arg}(\overline{z}))}=\overline{z}^{\overline{w}} \end{equation}

However, if $\mathrm{Arg}(z)= -\pi$, the above argument fails. Consider $-2^i=e^{\pi}e^{i\mathrm{ln}2}$. Then $\overline{-2^i}=e^{\pi}e^{-i\mathrm{ln}2}$ but $-2^{-i}=e^{-\pi}e^{-i\mathrm{ln}2}$

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    $\begingroup$ Yes, but. To my mind, the problem is that just as there is no consistent way of defining the logarithm (so that it’s defined on the whole punctured plane), there’s also no consistent way of defining exponentiation when the exponent is not an integer. Both may be considered to be multi-valued functions, but that opens a can of worms, too. $\endgroup$ – Lubin May 10 '16 at 17:20
  • $\begingroup$ There are two "safe-havens" for defining complex exponentiation: the base can be a positive real number and the exponent can be any complex number, or the base can be any nonzero complex number and the exponent can be an integer. Stepping off these "shining paths" quickly leads to multi-valued results or worse. $\endgroup$ – hardmath May 14 '16 at 13:51
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In general, $z^w$ is a multivalued function, unless you specify a particular branch. By definition,

$$ z^w = \exp(w \log(z))$$ so $$ \overline{z^w} = \overline{\exp(w \log(z)} = \exp(\overline{w}\; \overline{\log(z)})$$

Now $\overline{\log(z)}$ will be one of the branches of $\log(\overline{z})$, i.e. $$\exp(\overline{\log(z)}) = \overline{\exp(\log(z))} = \overline{z}$$ so that $\overline{z^w}$ is one of the branches of $\overline{z}^\overline{w}$, but it will not necessarily be your favourite branch, whichever that is. And it is impossible to choose a branch consistently that will make $\overline{z^w} = \overline{z}^{\overline{w}}$ always true. For example, consider $ (-1)^{1/2}$. Depending on which branch you choose, it is either $i$ or $-i$. But in either case $$ (\overline{-1})^{\overline{1/2}} = (-1)^{1/2} \ne \overline{(-1)^{1/2}}$$

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