Find the values of all the cube roots of $\cos \pi + i \sin \pi$

Question

How should I use de Moivre’s theorem to solve the above question?

Some hints would be helpful. Thank you

Answer 1

We can say that: $w^3=r^3(\cos(3\theta)+i\sin(3\theta))=\cos\pi+i\sin\pi$ Then, using De Moivre formula, you obtain: $r=1$ and $$3\theta_k=\pi+2k\pi, k=0,1,2$$

The solutions are: $$w_0=\cos(\pi/3)+i\sin(\pi/3) \: \vee \: w_1=\cos(\pi)+i\sin(\pi)) \: \vee \: w_2=\cos(5\pi/3)+i\sin(5\pi/3))$$

Answer 2

DeMoivre's theorem is a logical deduction from Euler's equation. In this case $-1=e^{i\pi}$ and the question is which rotations can you triple to get that which is $\pi$.

The obvious one is $\dfrac{\pi}{3}$ so one of the roots is $\exp(\frac{i\pi}{3})$

But if you remember the periodicity of angles, consider this situation. If an obtuse angle was tripled it could lead to an angle greater than $2\pi$. So you have to consider this possibility.

Let's say that for an angle $\theta$ defining a complex number as $e^{i\theta}$ , $3\theta \gt 2\pi$. Then maybe, instead of that angle directly landing on $\pi$ it literally landed on $3\pi$, which is the corollary of this angle because they're separated by the period $2\pi$.

Then another answer would be $\dfrac{3\pi}{3}=\pi$. Meaning that one of the cube roots of $-1$ is itself which is trivial.

Then, tripling an angle $\gt \pi$ could make you do two full turns, so let's consider that too. In this case the corollary for $\pi$ is two extra laps which is $5\pi$. Hence another angle is $\dfrac{5\pi}{3}$ leading to another answer: $\exp(\dfrac{5\pi}{3})$

If you tried this with 3 turns you'd get an answer $+2\pi$ which is a repetition of the first answer.

So that's DeMoivre's theorem. For fourth roots consider upto three turns instead for the reason mentioned above. For $\sqrt[n]{z}$ consider upto $n-1$ turns.

How is this written down rigorously?

$\sqrt[n]{re^{i\theta}}=\sqrt[n]{r} \cdot (\cos (\dfrac{\theta+2\pi k}{n})+i\sin (\dfrac{\theta+2\pi k}{n}))$ where $k \in \{0,1,2 \ldots n-1\}$

Then you just consider every possibility for $k$. Peace.