1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

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.)

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.