What is the correct value?

Question

My confusion is:

$(-9)^{2/3} = ((-9)^{2})^{1/3} = ((-9)^{(1/3)})^{2} = 4.32$

But my calculator shows math error, and google says:

$(-9)^{2/3} = 2.16+3.74i$

Answer 1

When we want to take a number to an integer power, we can simply multiply repeatedly and/or take reciprocals. When we want to take a positive real number to an arbitrary power, we usually use logarithms.

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Logarithms and Exponentials

When considering raising a positive real number, $a$, to an arbitrary power, $x$, we usually compute as follows $$ a^x=e^{x\log(a)}\tag{1} $$ When $a\in\mathbb{R}^+$, $\log(a)$ is the usual natural logarithm that one gets by pressing the ln key on a scientific calculator. This can be computed in many ways; here is a series converges for all $x\gt0$: $$ \log(x)=2\sum_{k=0}^\infty\frac1{2k+1}\left(\frac{x-1}{x+1}\right)^{2k+1}\tag{2} $$ For some $x$, $(2)$ might converge very slowly, and there are ways to speed it up, but the important part here is that given a positive real number, we have a formula that gives us its real logarithm. One property that can be gotten from $(2)$ without much trouble is that $\log\left(\frac1x\right)=-\log(x)$.

The exponential function, $e^x$, has a quickly convergent series: $$ e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\tag{3} $$ Putting together $(2)$ and $(3)$ as shown in $(1)$, we can compute the $a^x$ for any positive real $a$ and real $x$.

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Complex Exponentials

Eulers Formula says that for any $x\in\mathbb{R}$, $$ e^{ix}=\cos(x)+i\sin(x)\tag{4} $$ This can be verified using $(3)$ and the series for sine and cosine: $$ \sin(x)=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}\qquad\text{and}\qquad\cos(x)=\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\tag{5} $$ One consequence of $(4)$ is that, for $n\in\mathbb{Z}$, $$ e^{2\pi in}=1\tag{6} $$ This causes difficulty when trying to compute complex logarithms because, for any $n\in\mathbb{Z}$ $$ x=e^{\log(x)+2\pi ni}\tag{7} $$ This means we could just as easily use $\log(x)+2\pi i$ or $\log(x)-6\pi i$ for the logarithm of $x$. That is, $$ e^{\log(x)+2\pi i}=e^{\log(x)-6\pi i}=x\tag{8} $$ This ambiguity in the complex logarithm leads us to consider branches of the logarithm function.

When raising to integer powers, this ambiguity does not matter. That is, for $x\in\mathbb{Z}$, $$ e^{x(\log(a)+2\pi in)}=e^{x\log(a)}e^{2\pi inx}=e^{x\log(a)}=a^x\tag{9} $$ However, for non-integer exponents, this can cause problems.

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Rational Powers

If $x$ is not rational, then there can be infinitely many values for $a^x$. However, for a rational exponent, $x=\frac pq$, there can be up to $q$ possible values for $a^x$: $$ e^{x(\log(a)+2\pi ni)}=e^{x\log(a)}e^{2\pi inp/q}\tag{10} $$ where, depending on $n$, $np/q$ can take on up to $q$ values mod $1$.

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The Particular Case

In your case, for the logarithm of $-9$, we could choose any $n\in\mathbb{Z}$ and use $$ \log(9)+(2n+1)\pi i\tag{11} $$ and get $$ (-9)^{2/3}=e^{\frac23(\log(9)+(2n+1)\pi i)}\tag{12} $$ Note that $(12)$ gives the same value for $n$ and for $n+3$. Thus, there are $3$ values represented in $(12)$: $$ \begin{align} \begin{array}{lc} n=0:&9^{2/3}\left(\cos\left(\frac{2\pi}3\right)+i\sin\left(\frac{2\pi}3\right)\right)=-2.163374 + 3.747074\,i\\ n=1:&9^{2/3}\left(\cos\left(\frac{6\pi}3\right)+i\sin\left(\frac{6\pi}3\right)\right)=4.326749\\ n=2:&9^{2/3}\left(\cos\left(\frac{10\pi}3\right)+i\sin\left(\frac{10\pi}3\right)\right)=-2.163374 - 3.747074\,i \end{array} \end{align} $$

Answer 2

It is because it is showing one of the three possible roots, one of them being $4.32$, and the other two are $2.16 + 3.74i$ and its conjugate