0

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?

Suresh
  • 663
  • 1
    It depends on how you define the complex function $f(z) = z^{2/3}$ (or which "branch" you choose). – Martin R Jan 30 '18 at 14:54

3 Answers3

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. $(-1)^{\frac 23}=((-1)^2)^{\frac 13}=1^{\frac 13}=1=((-1)^{\frac 13})^2=(-1)^2=1$

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.

zwim
  • 28,563
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.

  • 1
    But $-1 = e^{-i \pi}$ and $-1 = e^{3i \pi}$ hold as well, giving different results for $f(-1)$. – Martin R Jan 30 '18 at 14:53
  • Yes, @MartinR I am aware. But, the OP can easily relate to the identity, right.... –  Jan 30 '18 at 14:55
  • 1
    From your $(e^{i\pi})^{2/3} = e^{2i \pi/3}$ the question author (who apparently is a beginner in complex analysis and not aware of "branches") might get impression that one can calculate $(z^a)^b = z^{ab}$ generally. – Martin R Jan 30 '18 at 15:05
0

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 $1$.

David C. Ullrich
  • 89,985