# Trigonometric manipulation of complex number, how does this step occur?

I was reading the section about DeMoivre, and my book showed how to derive his formulas. The next part is supposed to be about finding roots of complex and real numbers. Roughly, it says:

"Let $z$ be a complex number. Suppose we wish to find the $n^{th}$ root of $z$ . Then there exists a complex number $w$ such that $w^n=z$ . Let's write $z$ in trigonometric form:

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

One $n^{th}$ root of $z$ is $w=r^{ \frac 1n} (\cos \frac \theta n +i \sin \frac \theta n)$"

This is where I have the problem.

It looks like they wrote $w^n=r( \cos \theta +i \sin \theta)$ $\to$ $w=\sqrt [n] {r( \cos \theta +i \sin \theta)}$ , but how did they get from $w=\sqrt [n] {( \cos \theta +i \sin \theta)}$ to $(\cos \frac \theta n +i \sin \frac \theta n)$? It looks like they used DeMoivre's formula which states that $z^n=r^n(\cos n\theta +i \sin n\theta)$ , but in the box with the theorem my book makes it clear: If $z=r( \cos \theta +i \sin \theta)$ , then for any $integer$ $n$ $z^n=r^n(\cos n\theta +i \sin n\theta)$ . What's going on here?

$\mathbf {P.S.}$ If you want to know what book this is it's an otherwise very good but old precalculus book that I have: "Precalculus Mathematics for Calculus 4th" by James Stewart, Lothar Redlin, and Saleem Watson.

• proofwiki.org/wiki/De_Moivre%27s_Formula Commented Jul 6, 2013 at 19:52
• Euler: $\cos \theta + i\sin\theta = e^{i\theta}$. Commented Jul 6, 2013 at 19:59

Your are correct when you say that they didn't use DeMoivre's formula because it has only been proven in your book for integers (though it actually does hold for all real powers which can be seen using Euler's formula).

The important thing to notice is that while the formula puts a restriction on $n$ it doesn't put on on $\theta$ which can be any real number. So when they write, $$\sqrt[n]{ \cos \theta + i \sin \theta} = \cos (\theta/n )+ i \sin (\theta/n)$$

They are really saying "Applying DeMoivre's formula to the right hand side of this expression yields the expression in the radical."

The argument is as follows, suppose that $w=\cos (\theta/n )+ i \sin (\theta/n)$, then taking the $n$'th power of $w$ gives,

$$w^n = \left(\cos (\theta/n )+ i \sin (\theta/n) \right)^n = \cos (n(\theta/n) )+ i \sin (n(\theta/n)) = \cos (\theta )+ i \sin (\theta)$$

From this we conclude that $w$ is a $n$'th root of $z=\cos (\theta )+ i \sin (\theta)$.

• Oh ok thank you very nice
– Ovi
Commented Jul 6, 2013 at 20:03