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So I worked along the lines of the following:

$$ \left( \cos \left( \theta \right) + i \sin \left( \theta \right) \right)^{\alpha} = \left( e^{i \theta} \right)^{\alpha} = e^{i (\theta \alpha)} = \cos \left( \theta \alpha \right) + i \sin \left( \theta \alpha \right) $$

with $i$ the imaginary number, $\theta$ real and $\alpha$ real and $\sin , \cos$ the normal trig functions

However if we take $\alpha = \frac{1}{2}$ a fellow member saw that

$$ \left( \cos \theta + i \sin \theta \right)^{1/2} = \left\{ \begin{array}{l} \cos \left( \frac{\theta}{2} \right) + i \sin \left( \frac{\theta}{2} \right) \\ \cos \left( \frac{\theta}{2} + \pi \right) + i \sin \left( \frac{\theta}{2} + \pi \right) \end{array}\right. $$

which, from my perspective comes down to the fact that $a^2 = b$ can be solved as $a = \sqrt{b} \lor a = -\sqrt{b}$. I feel as though either I'm being very silly (as per usual) or there's something deep going on here that I'm missing

Did
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DanZimm
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    $z^\alpha$ is not so uniquely defined when $\alpha$ is not an integer ... – Hagen von Eitzen Jun 09 '13 at 19:46
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    Your hunch is correct. What happens if you square both answers? – Tpofofn Jun 09 '13 at 19:48
  • I didn't find anything. Seriously. – Kaster Jun 09 '13 at 19:48
  • @Kaster I apologize, I'll take your name out of the post – DanZimm Jun 09 '13 at 19:53
  • @Tpofofn right, you get the same thing if you square both answer. I 'm trying to understand why Euler's formula fails here.\ – DanZimm Jun 09 '13 at 19:54
  • @HagenvonEitzen hrm, do you know of any pdf's online that talk about this sort of thing? Or maybe just a website? – DanZimm Jun 09 '13 at 19:55
  • What exactly do you call "Euler formula" here? – Did Jun 09 '13 at 19:55
  • What failing "Euler's formula" are you talking about, @DanZimm ? – DonAntonio Jun 09 '13 at 19:55
  • I'm guessing @DanZimm means De Moivre's formula here, which btw works only for integer powers. – Kaster Jun 09 '13 at 19:57
  • @DonAntonio Did how come I can't use Euler's formula $e^{i \theta} = \cos \theta + i \sin \theta$ here? Clearly something goes awry in what I did so I presumed I improperly used Euler's formula. As Hagen pointed out it seems as though $(e^{i\theta})^{\alpha}$ doesn't uniquely equal $e^{i \theta \alpha}$ – DanZimm Jun 09 '13 at 19:58
  • @DanZimm , I think that you actually mean de Moivre's formula, as Kaster says. Well, that formula works for natural numbers. You may want to google it. – DonAntonio Jun 09 '13 at 19:59
  • @DonAntonio I'm asking why my derivation isn't valid, I'm not using De Moivre's formula (not trying to) only trying to use Euler's formula of $\cos \theta + i \sin \theta = e^{i \theta}$ – DanZimm Jun 09 '13 at 20:00
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    @DanZimm have you heard about periodic nature of exponential function in complex analysis? That is $e^{i \theta} = e^{i (\theta + 2 \pi)}$, so yes, it's not unique. – Kaster Jun 09 '13 at 20:01
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    Also, I think it's time to leave Euler's formula alone, since everything is fine with it :) It works as it should, but has nothing to do with the stuff mentioned in your question. – Kaster Jun 09 '13 at 20:02
  • @Kaster changed title – DanZimm Jun 09 '13 at 20:04
  • @Kaster yes I'm aware of this, I suppose uniquely wasnt the proper word to use here however I figure you understand what I meant – DanZimm Jun 09 '13 at 20:05
  • @HagenvonEitzen if you explain a bit further and put it in an answer you get the accept – DanZimm Jun 09 '13 at 20:05

3 Answers3

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The issue I'm not understanding is why you can only raise to an integer power

To stay focussed on the question asked by the OP, and at the risk of repeating what @Hagen explained from the start, there is no such thing as a function $z\mapsto z^\alpha$ on the complex plane except when $\alpha$ is an integer.

Hence you are in trouble from the second line of your post on, when you consider the nonexistent quantity $$ \left( \cos \left( \theta \right) + \mathrm i \sin \left( \theta \right) \right)^{\alpha}. $$ As an example, assume that $\alpha=\sqrt2$ and $\theta=\frac\pi2$, then $$ \cos \left( \theta \right) + \mathrm i \sin \left( \theta \right) =\mathrm i=\mathrm e^{\mathrm i(2k\pi+\pi/2)}, $$ for every integer $k$, hence legitimate looking candidates to be the $\sqrt2$th power of $\mathrm i$ are $$ \mathrm e^{\mathrm i\sqrt2(2k\pi+\pi/2)}=\mathrm e^{\mathrm i\pi/\sqrt2}\cdot\mathrm e^{\mathrm i2\sqrt2 k\pi}. $$ Since $\sqrt2$ is irrational, the set of fractional parts of the numbers $\sqrt2 k$ for every integer $k$ is dense in $[0,1]$, thus, for every $\varphi$ in $[0,2\pi]$ there exists some $k$ such that $2\pi\sqrt2 k$ is as close as one wants to $\varphi$. In particular, $\mathrm e^{\mathrm i2\sqrt2 k\pi}$ can be made as close as one wants to $\mathrm e^{\mathrm i\varphi}$, for every possible angle $\varphi$. Thus, you could declare that the $\sqrt2$th power of $\mathrm i$ is as close as one wants from every point on the unit circle.

Let me only hope that this state of affairs makes you worry about the mere existence of any quantity defined as the $\sqrt2$th power of $\mathrm i$...

Did
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Tpofofn's hint is huge:

$$\bullet\;\;\;\left(\cos\frac\theta2+i\sin\frac\theta2\right)^2=\cos^2\frac\theta2-\sin^2\frac\theta2+2i\cos\frac\theta2\sin\frac\theta2=\cos\theta+i\sin\theta$$

$$\bullet\;\;\left(\cos\left(\frac\theta2+\pi\right)+i\sin\left(\frac\theta2+\pi\right)\right)^2=\left(-\cos\frac\theta2-i\sin\frac\theta2\right)^2=\ldots$$

You can complete the exercise above and see that both leftmost expressions above are square roots of the same complex number $\,\cos\theta+i\sin\theta\;\ldots$

DonAntonio
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What you are seeing is fundamentally driven by the fact that $\sin$ and $\cos$ are periodic. You obviously know that

$$\sin(\theta) = \sin(\theta + 2\pi) = \sin(\theta +2\pi n)$$

likewise for $\cos$. We conclude therefore that the same applies to

$$e^{i(\theta+2\pi n)}.$$ When we raise Euler's formula to an integer power, this makes no difference (i.e. we get the same answer for all values of $n$), however when raised to a fractional power, different values of $n$ yield different values, which is what you observed in your post.

Tpofofn
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