How to deal with the radius when applying Euler's formula to determine the roots of a number?

Question

I am trying to find $1^{1/3}$. I use Euler's formula $1^{1/3} = (re^{i\theta + 2\pi*k})^{1/3}$

Why are you able to directly this time calculate r? Is r implicitly defined as magnitude?

Answer 1

To find $1^{1/3}$, you want $z\in\mathbb{C}$ such that $$ z^3=1. $$ Now writing $z=re^{i\theta}$, one has $$ r^3e^{i3\theta}=1 $$ Taking the absolute value on both sides, and noting that $r>0$, you get $r=1$. For $\theta$, note that $$1=e^{2\pi i}$$ which implies that $$ 3\theta+2k\pi=2\pi,\quad k\in\mathbb{Z}. $$

Answer 2

in this formula, $r$and $\theta$ are both real numbers, and $r$ is nonnegative.

So you get $$ r^{1/3} \left[ \cos \frac{\theta}{3} + i \sin \frac{\theta}{3} \right]. $$

Post comment additions

Let me add a little detail. The first problem is that $a^b$ has two different meanings.

1. We define $a^b$, for $a$ real and positive, using logarithms (it's $\exp{b \ln a}$), which are in turn defined by integrals, and $\exp$, which is defined as the inverse function of the natural log.

2. We define $z^b$, for $z$ complex and $b$ real, by a somewhat different rule.

The challenge is that when you now write $z^b$ and $z$ happens to be a complex number whose imaginary part is zero...you don't know which rule to use. To keep it clear, I'm going to use $pow(a, b)$ to refer to the thing described in item 1 above. Then we can describe #2 as follows:

Write $z = r (\cos t + i \sin t)$, where $r \ge 0$ is real, and so is $t$. Then $$ z^b = pow(r, b) (\cos s + i \sin s) $$ where $s = (t + 2k\pi)b$ for any integer $k$.

This is a little odd, in the sense that there are many possible values for $s$, hence many possible powers of $z$. But what the heck: we have two square roots for nonnegative real numbers already, and just happen to be lucky that there's an easy way to consistently choose one of them (the nonnegative one) to call $\sqrt{x}$.

In the case where $z = 1$ and $b = \frac{1}{3}$, we end up with $$ 1^{1/3} = pow(1, 1/3) (\cos s + i \sin s). $$ Now $pow(1, 1/3) = 1$, so this simplifies to just $$ 1^{1/3} = \cos s + i \sin s. $$ where $s = t/3 + 2\pi k/3$, where $t$ is the "angle" for $z = 1$. That angle happens to be $0$, so $s$ is just "any multiple of $2\pi/3$.