# Review of a proof for De Moivre's theorem using mathematical induction

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

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

$$=r(\cos\theta+i\sin\theta)$$

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:

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

$$or$$

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

$$and$$

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

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

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

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 $$\{n|n\in\mathbb R,n\geq1\}$$

• You could assume $|z|=1$ to make the writing concise.
– user9464
Sep 7 '17 at 15:15
• When $n$ is not an integer, you would need to know the definition for $z\mapsto z^n$.
– user9464
Sep 7 '17 at 15:18
• This proof is a good practice for elementary purposes. Much simpler and elegant proof is available, though. Sep 8 '17 at 23:49
• 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. Sep 9 '17 at 0:02

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.)
• 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. Sep 9 '17 at 14:22