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I can't seem to understand how to solve this. I mean, if we weren't dealing with complex numbers, then I suppose it is clearly 1, but I don't know how to approach this. Apparently the answer is $\cos(2\sqrt{2} k \pi) + i\sin (2 \sqrt{2} k \pi)$, but I don't know how to go through with this. Do I begin with setting it to $e^{\sqrt{2}ln(1)}$? Even then, $ln(1) = 0$, and $e^\sqrt{2}$ is just that... I'm not sure how to go about this.

In short, how do I go from $1^{\sqrt{2}}$ to $\cos(2\sqrt{2} k \pi) + i\sin (2 \sqrt{2} k \pi)$?

David
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2 Answers2

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Hint: Recall that $e^{x+iy}$ where $x,y\in\mathbb{R}$ is equal to $e^x(\cos(y)+i\sin(y))$. What complex values $z$ give $e^z=1$?

Expanding the hint: We start with $e^{\sqrt{2}\log(1)}$. Solving the above equation, we see that $e^{z}$ is one at precisely $2\pi ik$ for $k\in\mathbb{Z}$, so $\log(1)=\{2\pi i k|k\in\mathbb{Z}\}$. Plugging in, we have the set $e^{2\sqrt{2}\pi i k}$ for $k\in\mathbb{Z}$. Expanding according to the above, we have $e^{2\sqrt{2}\pi i k}=\cos(2\sqrt{2}\pi k)+i\sin(2\sqrt{2}\pi k)$ and we arrive at the answer.

The punchline here is that $\log(z)$ is still the inverse of $e^z$, but $e^z$ is no longer 1-1 and therefore $\log(z)$ can only be a local inverse, and there's some choice involved in which branch to pick.

KReiser
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  • Either the number is purely imaginary, or x = 1. Right? – David Sep 12 '13 at 01:56
  • $x\neq 1$. Solve the equation! What value of $x,y$ can you plug in to get 1? – KReiser Sep 12 '13 at 02:08
  • Whoops, you're right, I didn't see that. x clearly has to be 0, and then we're left with figuring out when cos(y) = 1, and sin(y) = 0. This happens at $0, \pi, 2\pi, etc.$ – David Sep 12 '13 at 02:28
  • Though... the answer doesn't make sense to me. $cos(2\sqrt{2}k\pi)$ does not equate to 1. – David Sep 12 '13 at 02:36
  • I just came back to this one, and I'm wondering: Why are you solving $e^z = 0$ and not $e^z = 1$? – David Sep 12 '13 at 07:07
  • Sorry, typo. I'm solving $e^{z}=1$. $e^z=0$ has no solutions. – KReiser Sep 12 '13 at 07:16
  • Okay, so I understand that when solving $\cos(y) + i\sin(y)$ we need $\cos(y)$ to be 1, and $\sin(y)$ to be 0. We know x = 0 for the sake of $e^x = 1$, but why do you get $2 \pi i k$ and not $2 \pi k$ for $\cos(y) + i\sin(y)$ (I assume that is what you're solving)? And then finally, why does that particularly equate to log(1)? Apologies for what seem like trivial questions, I just really want to understand exactly what's going on. – David Sep 12 '13 at 07:25
  • That's fine, questions is how we learn! For the part dealing with $\cos(y)+i\sin(y)=1$, I get $y=2\pi k$, which means the complex number I get which solves $e^{x+iy}=1$ is $x+iy=2\pi i k$. For the second question, recall what the definition of $\log$ is: if $a=e^b$, then $b=\log a$. – KReiser Sep 12 '13 at 07:29
  • Aha! I see now. Yes, it must be $2 \pi i k$ because it is the imaginary part of the number. Alright, then we know that $e^{x + iy} = 1$, hence, $x + iy = 2 \pi i k = \log(1)$. $\log(1)$ is obtained by the fact that $log(e^{x+iy}) = log(1)$. From this, we know that $log(1) = 2 \pi i k$, and because $1^{\sqrt{2}} = e^{\sqrt{2}\log(1)}$, we have $e^{\sqrt{2}2 \pi i k} = \cos(2\sqrt{2} \pi k) + i \sin(2 \sqrt{2} \pi k)$. I believe that is right. – David Sep 12 '13 at 07:40
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$e^0 = 1$. However, remember that $e^{2\pi ki}$ for $k \in \mathbb{Z}$ is also $1$. Therefore, you can rewrite $1^{\sqrt{2}}$ as $e^{{(2\pi ki)}^{\sqrt{2}}}$, and simplify using exponent laws to $e^{2\sqrt{2}\pi ki}$. From there, use Euler's identity.

u8y7541
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