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Define $argument$ of $z$ to be

$[arg z] := (\theta \in \Re : z=|z|e^{i\theta})$

How do we then find $[\sqrt2^i]$ and $[i^{\sqrt2}]$ ?

$[\sqrt2^i]$ = $\frac{ln2}{2}$ from other sources. But can anyone derive this please?

Shouldn't the argument be a set of infinite values? i.e. not just a constant?

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$$[\sqrt{2}^i]=[e^{i\ln(\sqrt{2})}]=\ln(\sqrt{2})+2k\pi=\frac{1}{2}\ln(2)+2k\pi$$

$$[i^{\sqrt{2}}]=[e^{\sqrt{2}\ln(i)}]=[e^{\sqrt{2}\ln(e^{i\pi/2})}]=[e^{i\sqrt{2}\pi/2}]=\frac{\pi\sqrt{2}}{2}+2k\pi$$

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    $\begingroup$ @rebc No, you need to use your definition $[re^{i\theta}]=\theta+2k\pi$ only with $r>0$ and $\theta$ real. In this case we first need to bring it to this form $[e^{i\ln(i)}]=[e^{i\ln(e^{i\pi/2})}]=[e^{-\frac{\pi}{2}+i2k\pi}]=[e^{-\frac{\pi}{2}}e^{i2k\pi}]=2k\pi$. $\endgroup$ – Tom Feb 12 '15 at 20:15
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The argument $\theta$ of $z\in\mathbb{C}$ is usually defined as $(\theta \in\color{blue}{ [-\pi,\pi)}\mid z=|z|e^{i\theta})$. Hence, it is unique.

However, it is true that if $z=|z|e^{i\theta}$ then $\forall k\in\mathbb{Z},z=|z|e^{i\theta+2ik\pi}$.

For instance, for $\arg(\sqrt{2}^i) :$

$\sqrt{2}^i=e^{i\ln{\sqrt{2}}}=e^{i\frac{\ln(2)}2}$ and $\ln(2)/2\in[-\pi,\pi)$

Hence $\arg(\sqrt{2}^i)=\frac{\ln(2)}2$

Using the same way, i'll let you do $[i^{\sqrt2}]$ :-)

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  • $\begingroup$ thanks. according to your derivation, $[i^i] = e^{iln(i)} = ln(i)$, but is $ln(i) = 0$? $\endgroup$ – rebc Feb 12 '15 at 20:08
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    $\begingroup$ @rebc $\ln(1)=0$, not $\ln(i)$. Remember, $i=e^{i\frac{\pi}2}$ $\endgroup$ – Hippalectryon Feb 12 '15 at 20:09
  • $\begingroup$ So that means $[i^i] = i\frac{\pi}{2} $ ? But Wolfram says it's $=0$? $\endgroup$ – rebc Feb 12 '15 at 20:14
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    $\begingroup$ @rebc $i^i=e^{i(i\pi/2)}=e^{-\pi/2}$, hence its argument is $-\pi/2$. By the way you can check wolfram, it does say that $\ln(i)=i\pi/2$. $\endgroup$ – Hippalectryon Feb 12 '15 at 20:17

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