Proving De Moivre's Theorem for Negative Numbers?

[NatFex]

Or basically anything that isn't a positive integer.

So I can prove quite easily by induction that for any integer n>0, De Moivre's Theorem (below) holds.

If $\DeclareMathOperator\cis{cis} z = r\cis\theta, z^n= r^n\cis(n\theta)$

My proof below:

Spoiler

Let P(n) be the statement $z^n = r^n\cis(n\theta)$

Prove true for n = 1:
$z^1 = z = r\cis\theta = r^1\cis(1\theta)$

Assume true for n = k:
$z^k = r^k\cis(k\theta)$

Prove true for n = k + 1:
$z^{k+1} = z^k \times z\\ = r^k\cis(k\theta) * r\cis\theta\\ = r^{k+1}\cis(k\theta)\cis\theta\\ = r^{k+1}(\cos(k\theta) + i\sin(k\theta))((\cos\theta + i\sin\theta)\\ = r^{k+1}(\cos(k\theta)\cos\theta + \cos(k\theta)i\sin\theta + i\sin(k\theta)\cos\theta - \sin(k\theta)\sin\theta)$
then by compound angle identities
$= r^{k+1}(\cos(k\theta + \theta) + i\sin(k\theta + \theta))\\ = r^{k+1}\cis((k+1)\theta)$

However I struggle to do this with negatives. I tried the simple case of

$\frac{1}{r\cis\theta}$ and from there, try to get to $\frac{1}{r}\cis(-\theta)$, but I've only managed to do that by replacing the numerator of 1 with $r\cis0$, which is equivalent, and then from there using De Moivre's Theorem to get an argument of $-\theta$, which is... not very useful considering De Moivre's Theorem is what I'm trying to prove in the first place.

I just can't manipulate it into a form that would allow me to use compound angle identities like I could with positive numbers. Any ideas?

 

[member 587159]

You can use $$z^{-n} = (z^n)^{-1}$$

 

[Samy_A]

What is the relation between $\frac{1}{\cis \theta}$ and $\cis (-\theta)$?

 

[NatFex]

You can use $$z^n = (z^n)^-1$$

Yeah, but I can't prove that that last step (raising the complex number to a power of -1) will make the angle/argument inside $\cis$ negative

What is the relation between $\frac{1}{\cis \theta}$ and $\cis (-\theta)$?

I know that $\frac{1}{r\cis\theta} = (r\cis\theta)^{-1}$, but -1 is outside the boundaries for n that I have proved in my first post :(

I know it works, I would just like to prove it.

 

[member 587159]

What is $$1/z$$ equal to?

 

[Samy_A]

Yeah, but I can't prove that that last step (raising the complex number to a power of -1) will make the angle/argument inside $\cis$ negative

So what?
Where in your proof for positive $n$ did it matter whether the angle is positive or negative?

I know that $\frac{1}{r\cis\theta} = (r\cis\theta)^{-1}$, but -1 is outside the boundaries for n that I have proved in my first post :(

You didn't answer the question: what is the relation between $\frac{1}{\cis \theta}$ and $\cis (-\theta)$?

 

[NatFex]

What is $$1/z$$ equal to?

$z^{-1}$, but my proof in the OP only works for positive integers. I'm trying to use -1 as a base case to write a proof for negative numbers, but for some reason I cannot work out how to apply a similar strategy (expand then use compound identities)

So what?
Where in your proof for positive $n$ did it matter whether the angle is positive or negative?

It doesn't matter if θ is positive or negative, what matters is its multiplier, which is n.

You didn't answer the question: what is the relation between $\frac{1}{\cis \theta}$ and $\cis (-\theta)$?

Well... they are equal, but thus far my understanding of it is only intuitive, until I can prove it, which is what I'm asking for help with

 

[Samy_A]

$z^{-1}$, but my proof in the OP only works for positive integers. I'm trying to use -1 as a base case to write a proof for negative numbers, but for some reason I cannot work out how to apply a similar strategy (expand then use compound identities)
It doesn't matter if θ is positive or negative, what matters is its multiplier, which is n.
Well... they are equal, but thus far my understanding of it is only intuitive, until I can prove it, which is what I'm asking for help with

Ok, so if $z=a+ib$, where $a, b$ are real, how would you compute $\frac{1}{z}=\frac{1}{a+ib}$?
Hint: multiply numerator and denominator by $a-ib$.

Once you have proven that $\frac{1}{\cis \theta}=\cis (-\theta)$, you can easily transform the formula with a negative $n$ into a formula with the positive $-n$. And then you can use what you already have proven.

 

[NatFex]

Ok, so if $z=a+ib$, where $a, b$ are real, how would you compute $\frac{1}{z}=\frac{1}{a+ib}$?
Hint: multiply numerator and denominator by $a-ib$.

Once you have proven that $\frac{1}{\cis \theta}=\cis (-\theta)$, you can easily transform the formula with a negative $n$ into a formula with the positive $-n$. And then you can use what you already have proven.

Brilliant, just what I was looking for

I completely forgot about looking at the complex numbers in rectangular form instead, cheers!