Value of $f(-1)$ where $f(x) =x^{2/3}$.

Question

If f is a complex function defined by
$$f(x)= x^{\frac23}$$then how can we evaluate $f(-1)$?

My Approach: $$(-1)^{2/3}=1^{1/3}=1$$ but the value of $f(-1)$ is given as approximately $-0.5 + 0.87i$, so how can I evaluate it?

Answer 1

Note that using Euler's famous identity, we can write: $$-1 = e^{i\pi} $$ Hence, $$f(-1)=(-1)^{2/3} = e^{2i \pi/3}$$ $$=-\frac12+\frac{\sqrt 3}2i = - 0.5+0.866i$$ where we have used Euler's formula in the last step.

Answer 2

If we limit ourselves to real calculations then $(-1)^{\frac 23}=1$ is well defined because we are in the case where the exponent $\in\mathcal Q_{odd}$ as I defined in this post:

Exponential Equation with base -1, 0, and 1

$\mathcal x\in Q_{odd}\iff x=\frac pq$ with $\gcd(p,q)=1$ and $q$ is odd.

In this case we can interchange the order of exponentiation, this is why it is well defined. $(-1)^{\frac 23}=((-1)^2)^{\frac 13}=1^{\frac 13}=1=((-1)^{\frac 13})^2=(-1)^2=1$

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Now if you go complex, there will always be three cubic roots, consisting of a real root multiplied by $1,j,j^2$ which are the cubic roots of unity.

Reminder: $\displaystyle j=e^{\large{i\frac{2\pi}3}}=-\frac 12+i\frac{\sqrt{3}}2$

The operation is done like this: $(-1)^{\frac 23}=\exp(\frac 23\ln(-1))$

Where $\ln(-1)=i(\pi+2k\pi)$ for $k\in\mathbb Z$ is a multivalued function.

Thus $(-1)^{\frac 23}=\exp(i\frac{(4k+2)\pi}3)$ which takes only $3$ values $1,j,j^2$ as we have seen previously, when $k$ varies.

Answer 3

A real number has exactly one cube root. But a (non-zero) complex number has three complex cube roots - that other value for $f(-1)$ that you got from somewhere is one of the other cube roots of $1$.