# Failure of De Moivre's Theorem

I know that De Moivre's Theorem does not necessarily work for non-integer powers.

The classic counter-example is by considering $\left (\cos \theta + i \sin \theta \right )^n=\cos n\theta + i \sin n \theta$ when $n=\frac{1}{2}$ and setting $\theta=0$ versus $\theta=2\pi$, which yield 1 and -1 respectively.

My question is, if this is the case, why is it that we can use the following to solve say $z^5=1$?

$z^5=\cos (\theta+2k\pi) + i \sin (\theta+2k\pi)$ where $k\in\mathbb{z}$

And then raising both sides to the power of $\frac{1}{5}$ etc?

• When you said “integer powers” in the first sentence, did you mean “non-integer powers”? And I'm not sure what your counterexample is in the second sentence, since $\pm1$ are both square roots of $1$, so it seems that the theorem is working just fine for that case. – MJD Aug 21 '14 at 15:24
• In your example $z^5=1$ you are not substituting the 1/5 in for $n$ you are simply raising an expression to the 1/5th power. – Paul Sundheim Aug 21 '14 at 15:28
• MJD, when we compute $1^\frac{1}{2}$, do we not take take the principle square root ie: the positive one? Also, I have amended my typo. Paul, but when I raise both sides to the 1/5th power, I can invoke DMT to make the argument of the RHS become $\frac{\theta+2k\pi}{5}$ – Trogdor Aug 21 '14 at 15:36
• Here we are looking at the complex plane rather than the real line. On the real line there is one fifth root for every real number. In the complex plane there are five. Often one wants to work with all the roots - de Moivre's theorem allows us to identify them all. Occasionally there are other constraints on the possible solutions which pick one possibility out as special. – Mark Bennet Aug 21 '14 at 15:48

It does work for $n=\frac{1}{2}$. By putting $\theta =0$ and $n=\frac{1}{2}$ you're working out $1^{1/2} = \pm 1$.

• Perhaps I am mistaken, but I've always been under the impression that say $4^\frac{1}{2}=2$ not -2, and likewise for 1, or have I once again been deceived by teachers? – Trogdor Aug 21 '14 at 16:03
• @Trogdor There is a subtle difference between $\sqrt{4}$ and $4^{1/2}$. We use $\sqrt{4}$ to denote the unique positive real solution to $z^2=4$. In that case $\sqrt{4} = +2$. The symbol $4^{1/2}$ means any solution to $z^2 = 4$, i.e. $4^{1/2} = \pm 2$. (Although some people don't maintain this difference.) At school they don't teach you this because $\sqrt[3]{8}$ is the positive real root of $z^3 = 8$ and that sentence won't make sense until you know about complex numbers. You're teachers don't mislead you, you have $-b \pm \sqrt{b^2-4ac}$ in the quadratic formula, They just shield you. – Fly by Night Aug 21 '14 at 16:07

To solve $z^5=1$, take $z=\cos(t)+i\sin(t)$ (since |z|=1).

So you have (by de Moivre's theorem for a positive integer power) $$z^5 = \cos(5t) + i \sin(5t) = 1 \\ \Rightarrow \cos(5t) + i \ sin(5t) = \cos(2 \pi k) + i \sin (2 \pi k)$$

On equating real and imaginary parts of this equation, you get two trigonometric equations, $\cos(5t)=\cos(2\pi k)$ and $\sin(5t)=\sin(2 \pi k)$.

The common solution is $5t = 2 \pi k + 2 \pi n$, so $t = \dfrac{2 m \pi}{5}$ is the solution. So there is no need to raise anything to fractional powers anywhere.

To convince yourself that there are only five roots, set m={0,1,2,3,4,5}; when m=5 you get the same root as m=0 and so on.

• This in no way addresses the OP question. – Fly by Night Aug 21 '14 at 16:00
• @FlybyNight Well, I wanted to show the OP that there is no "failure" of de Moivre's theorem at all. He / she seems to be comfortable with using it to take integer powers, and that's enough to find the roots in this case. – user_of_math Aug 21 '14 at 16:08
• @flybynight I'm truly mystified as to what the OP question actually is, and in such cases OP often finds it helpful to get a clear and straightforward explanation of the situation in general. – MJD Aug 22 '14 at 1:14