The following is a rigorous proof of De Moivre's theorem by means of mathematical induction.

The theorem put simply is that:

Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$

In order to find the $n^{th}$ power of $z$, the following rule applies:

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

where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.

Base step:

We must prove that what is stated is true for $n=1$, hence

$$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$


which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step.

Inductive step:

When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence

$$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$

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

$$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$

from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used:

$$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$


$$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$


$$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$

by using these, the original equation has now become

$$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$

then, after factoring out $\theta$, the theorem becomes


which completes the proof.

Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$

  • $\begingroup$ You could assume $|z|=1$ to make the writing concise. $\endgroup$
    – user9464
    Sep 7 '17 at 15:15
  • $\begingroup$ When $n$ is not an integer, you would need to know the definition for $z\mapsto z^n$. $\endgroup$
    – user9464
    Sep 7 '17 at 15:18
  • $\begingroup$ This proof is a good practice for elementary purposes. Much simpler and elegant proof is available, though. $\endgroup$
    – High GPA
    Sep 8 '17 at 23:49
  • 1
    $\begingroup$ It's not possible to prove for $n\in\Bbb R$ under consistent definition of $z^n$, as shown in this answer. What you can do is place restrictions on $\theta$, upon which it can hold for $n\in\Bbb R$. I'd suggest proving it for $n\in\Bbb Z$ first though. $\endgroup$ Sep 9 '17 at 0:02

You're very close!

The first issue I see is a minor one. The identity $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ seems to be unnecessary. I just can't tell what use you make of it, since the preceding identity does what you need.

The second issue I see is a major one. The identity $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta+\sin n\theta\cdot\sin\theta$$ is incorrect. It should instead be $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta,$$ which allows you to draw the desired conclusion. (I suspect this may have been a typo.)

  • $\begingroup$ Thank you for the feedback, indeed the second error is a typo and I've fixed it, thank you. The identity I make use of to reduce the number of terms back down to 2 and hence layout the final answer in the form it is needed. Why would you say is it unnecessary, because I might be overlooking something? $\endgroup$ Sep 9 '17 at 2:51
  • $\begingroup$ It's just the first identity in disguise. I would probably do one more rewrite as $$r^{n+1}\bigl(\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta+i(\cos n\theta\cdot\sin\theta+\sin n\theta\cdot\cos\theta)\bigr),$$ then bring up the two trig identities. $\endgroup$ Sep 9 '17 at 14:22

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