Polar Coordinate usind De Moivre’s Theorem

Question

I need the solutions for the following problem: Find all solutions over $\mathbb{C}$ to the equation $x^3=i^2$.

I tried using De Moivre’s Theorem can't get around it.

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Note: The question originally asked for "the solutions to $i^{2/3}$".

Answer 1

Note that you will get three answers (this is what Did is getting at in his comment - theoretically, $i^{2/3}$ doesn't actually exist! Instead, we are finding solutions to the equation $x^3=i^2$). In general, if $z$ is a complex number and $a, b$ are coprime integers (so, that the fraction $a/b$ is reduced, so $2/3$ rather than $4/6$) then $x^b=z^a$ will have $b$-many solutions.

Polar form: Start by writing $i$ in polar form, so write $i=re^{i\theta}$ and try to find $r$ and $\theta$. You may find it useful to draw your number on the complex plane. It corresponds to the point $(0, 1)$.

The modulus $r$ is equal to the distance of your number from the origin, so here $r=1$. Alternatively, you can use the formula $r=\sqrt{0^2+1^2}=1$.

The argument $\theta$ is equal to the angle (in radians!) made between the real line (the $x$-axis) and the line connecting your point, $(0, 1)$, to the origin, $(0, 0)$. The angle here is $\pi/2$.

Hence, $i=e^{\frac{\pi}{2}i}$.

Adding $2\pi$ doesn't matter: Now, you know that $e^{2\pi i}=1$ (why?). This implies that $e^{2k\pi i}=1$ for all integers $k\in\mathbb{Z}$ (why?). This in turn implies that $i=e^{\frac{\pi}{2}i+2k\pi i}$ (why?).

Obtaining the answers: This all implies that $i^{2/3}=(e^{\frac{\pi}{2}i+2k\pi i})^{2/3}$. Plug in $k=0, 1, 2$ to get the three answers.