($\cos \phi + i\sin \phi)^n = \cos (n \phi) + i\sin (n \phi)$

Saw this in the De Moivre's formula proof and some other calculations involving complex numbers, but I do not understand why the equation is true.

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    $\begingroup$ Did you mean $(\cos \phi + i\sin \phi)^n = \cos (n \phi) + i\sin (n \phi)$? $\endgroup$ – Fernando Revilla Jan 15 '17 at 16:53
  • $\begingroup$ Yes, sorry I haven't considered it's specific to DeMoivre's formula. Edited the question, I saw it in some calculations involving complex numbers but still can't understand why it stands. $\endgroup$ – Jani Jan 15 '17 at 17:01

DeMoivre's formula.

$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$

De Moivre's formula can easily be derived from Euler's formula.

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

$ e^{i n \theta} =(\cos n \theta + i \sin n \theta)$

And you can use induction to prove it. Here's link with full details.

  • $\begingroup$ Haha, and Euler's formula can easily be derived from...wait, that's not easier to prove! $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 17:10
  • $\begingroup$ Thanks, I finally put it together, so the step in the DeMoivre proof works because of Euler's formula which has its own separate proof. $\endgroup$ – Jani Jan 15 '17 at 17:10
  • $\begingroup$ @Simple Art I think you need to read full article from Wikipedia. $\endgroup$ – Kanwaljit Singh Jan 15 '17 at 17:11
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    $\begingroup$ @Simple Art I accepted this answer because the only problematic line for me in the DeMoivre proof was the one in question, based on Euler's formula. ( I understand it's not trivial to prove but for my purposes I can just accept it as true ). My basic misunderstanding was I thought it might be some trigonometric identity which turned out to be not the case. $\endgroup$ – Jani Jan 16 '17 at 8:48
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    $\begingroup$ @Jani Hm, ok. I added that part into my answer for negative values. And I think when I say this, I speak for both Kanwaljit and I: "Euler's formula is awesome, so please learn it anyways :D" $\endgroup$ – Simply Beautiful Art Jan 16 '17 at 13:06

This is DeMoivre's formula:

$$(\cos\phi+i\sin\phi)^n=\cos n\phi+i\sin n\phi$$

which may be proven by induction:

First of all, this is true for $n=1$. Now, we prove that if it is true for $k$, then it is true for $k+1$:

$$\begin{align}(\cos\phi+i\sin\phi)^{k+1}&=(\cos\phi+i\sin\phi)(\cos\phi+i\sin\phi)^k\\&=(\cos\phi+i\sin\phi)(\cos k\phi+i\sin k\phi)\\&=\cos\phi\cos k\phi-\sin\phi\sin k\phi+i(\sin\phi\cos k\phi+\cos\phi\sin k\phi)\\&\stackrel{(*)}=\cos(k+1)\phi+i\sin(k+1)\phi\end{align}$$

That is, if it holds for $k=1$, it holds true for $k+1=2$, and if it holds for $k=2$, it holds true for $k+1=3$, etc.

$(*)$ we used the sum of angles formula.

Likewise, this extends to negative $n$:

$$\begin{align}(\cos\phi+i\sin\phi)^{-n}&=((\cos\phi+i\sin\phi)^n)^{-1}\\&=(\cos n\phi+i\sin n\phi)^{-1}\\&=\frac1{\cos n\phi+i\sin n\phi}\\&=\frac1{\cos n\phi+i\sin n\phi}\frac{\cos n\phi-i\sin n\phi}{\cos n\phi-i\sin n\phi}\\&=\frac{\cos n\phi-i\sin n\phi}{\cos^2n\phi+\sin^2n\phi}\\&\stackrel{(**)}=\cos n\phi-i\sin n\phi\\&\stackrel{(***)}=\cos(-n\phi)+i\sin(-n\phi)\end{align}$$

$(**)$ we used the pythagorean identity $\sin^2+\cos^2=1$

$(***)$ we used symmetry formulas $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$

  • $\begingroup$ Yes, edited the question that's the line I don't get. I understand all the other parts of the induction, but this step is still not clear. $\endgroup$ – Jani Jan 15 '17 at 17:02
  • $\begingroup$ @Jani I've updated on the induction. $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 17:09

The "identity is wrong. Consider $n=2$. Then the left hand side is $$ \cos^2 \theta + \sin^2 \theta + 2\cos \theta \sin\theta = 1 + 2\cos \theta \sin\theta = 1 + \sin(2\theta) $$ And the right hand side is $$ \cos (2\theta) + \sin(2\theta) $$ So unless $\cos (2\theta) = 1$ the equality fails.

  • $\begingroup$ You did not take the complex unit into account. $\endgroup$ – MrYouMath May 29 '18 at 15:56
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    $\begingroup$ @MrYouMath When I posted my answer, the complex $i$ was not present in the problem. An hour later, at 6:18, an edit was made to the problem inserting the $i$. So you have downvoted me for what was a correct answer to the problem as posed at the time of the answer. $\endgroup$ – Mark Fischler May 31 '18 at 15:05
  • $\begingroup$ Sorry, I just looked at the time of answer and compared it with the answer of Simply Beautiful Art did not compare it with the edited time. I cannot remove the downvote at the moment because you would need to edit your answer. I know it is frustrating when the OPs change the question. Still, it is not useful to let this answer here when it is not adapted to the edited question. $\endgroup$ – MrYouMath May 31 '18 at 16:29

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