solving equations with complex numbers, with one and multiple equations

Question

The question is as follows, I am being asked a question in which I have to solve for $a \in \mathbb{C}$. Find all $a\in \mathbb{C}$ such that $a^3+1=0$, $a^6-1=0$ and $a^4-1=0$

Answer 1

Let's first take a look at $a^3+1=0$, which rewrites into \begin{equation} a^3=-1 \end{equation} Then, by using Euler's identity (I assume you are familiar with this), we can rewrite the latter into \begin{equation} a \in \mathbb{C} \implies a = re^{\theta i} \implies a^3 = r^3e^{3\theta i} \end{equation} and \begin{equation} -1 = e^{\pi i} \end{equation} Thus \begin{equation} e^{\pi i} = r^3e^{3\theta i } \end{equation} Which implies both \begin{equation} r^3 = 1 \quad \textrm{and} \quad \pi i = 3\theta i \end{equation} And thus \begin{equation} r=1 \quad \textrm{and} \quad \theta = \pi/3+\frac{2}{3}\pi k \end{equation} And thus your solution is \begin{equation} a = e^{i(\pi/3+\frac{2}{3}\pi k)} \end{equation}

If you are having difficulties understanding this problem, and why this is the solution, you could try to substituting our solution into the equation.

Also, $-1+0i$ is one of our solutions, try for yourself that you could guess that solution without any knowledge of complex numbers, that should not be too hard. For the other problems, try using the same method.

If you do not understand nor know the exponential form of a (complex) number, please try to learn that in advance before solving such problems