When is a power real?

Question

We know that the power $a^b$ is, if $b$ is not an integer, the product of a power and a root (example : $2^{2.5}=2^2\sqrt{2}$).

But how do we know, if $a$ is negative, if $a^b$ is complex ?

For example, $(-1)^{2.5}$ is complex ($i$), but $(-1)^\frac{7}{3}$ is real. For simple exponents, it seems simple, but how to know when the exponent is more "complicated" ?

Answer 1

What about the following: a number is real if the imaginary component is zero.

Let's have a look at a = abs(a) * (cos(180°) + i * sin(180°)) = abs(a) * (e^180°)
a^b = abs(a)^b * (e^(180°*b)) = abs(a) * (cos(180°*b) + i * sin(180°*b))
This is real if the imaginary part is zero:
sin(180°*b) = 0
This is true if 180°*b is a whole multiple of 180°, which means that b is a whole number.

This means that for a negative number a: a^b is real if b is a whole number.

Answer 2

Well $(-1)^{\frac73}=\sqrt[3]{(-1)^7}=\sqrt[3]{-1}$, which has $3$ solutions, some real and some not.

In order to solve $\sqrt[3]{-1}$, draw the unit circle, and then:

• Draw the 1st solution, which is obviously $-1$, or more generally: $\cos(\pi)+\sin(\pi)i$
• Rotate it $\frac{2\pi}{3}$ radians in order to get the 2nd solution: $\cos(\pi+\frac{2\pi}{3})+\sin(\pi+\frac{2\pi}{3})i$
• Rotate it $\frac{2\pi}{3}$ radians in order to get the 3rd solution: $\cos(\pi+\frac{4\pi}{3})+\sin(\pi+\frac{4\pi}{3})i$

As you can see:

• The 1st solution is real, because $\sin(\pi)=0$
• The 2nd solution is not real, because $\sin(\pi+\frac{2\pi}{3})\neq0$
• The 3rd solution is not real, because $\sin(\pi+\frac{4\pi}{3})\neq0$

Here is a graphic illustration: