Why does the graph of $x^n$ only have an imaginary component if $n$ is not an integer?

Question

I was playing around graphing equations and noticed that only non-integer exponents of x yield imaginary graphs. Try it on Wolfram Alpha. Why is that?

Answer 1

An imaginary component is a component containing $i=\sqrt{-1}=(-1^{\frac{1}{2}})$. Since $(x^{b})^{c}=x^{bc}$, any exponent in the form $x^{\frac{p}{2q}}$, where x is real and $\frac{p}{2q}$ is a reduced fraction so $p$ is not even, can be rewritten as $(x^{\frac{1}{2}})^{\frac{p}{q}}=((\sqrt x)^{p})^{\frac{1}{q}}$. When $x$ is negative this can be seen to be $(i\sqrt{|x|})^{\frac{p}{q}}=(i)^{\frac{p}{q}}(|x|)^{\frac{p}{2q}}$. Since $|x|$ is positive and real $(|x|)^{\frac{p}{2q}}$ is real and since $\frac{p}{q}$ is non-even $(i)^{\frac{p}{q}}$ has an imaginary component. So the graph of $x^{\frac{p}{2q}}$ will have an imaginary component when $x$ is negative.

As Fly by Night pointed out in a comment other fractional powers can indeed have imaginary parts because there are $n$ roots of unity for $x^{\frac{1}{n}}$. Many graphing programs will ignore these if you have an odd root, but a quick check shows that wolfram alpha is not among them and will include the information about the other roots of unity.

An integer power will always have any fractional powers exactly canceled out so that any roots of unity are taken back to unity, that is $1$, by exponentiation. Thus a graph of an integer power of real $x$ should not have an imaginary component.

Answer 2

$\sqrt[3]x$ has no imaginary component, because negative numbers have real cube roots. In the example you've given, we can rewrite it as:

$$x^{4.1}=x^4\sqrt[10]x$$

Notice that the even root will have an imaginary part when $x$ is negative.

Answer 3

This is natural. For an integer value, say $n$, there is a unique value for $x^n$. Simply multiply $x$ with itself $n$ times. Even if $n$ is negative: just divide one by the product of $x$.

\begin{array}{ccc} x^3 = x \times x \times x \\ \\ x^{-2} = \frac{1}{x \times x} \end{array}

When you come to non-integer powers, you have to ask yourself what these expressions mean. For example, $x^{1/2}$ stands for a number whose square is $x$. Hence $4^{1/2} = \pm 2$ since $(\pm 2)^2 = 4$. Even for such a simple power as $1/2$, we have two possible $y$ values for each $x$ value when we plot $y=x^{1/2}$.

Note that the symbol $\sqrt{x}$ is different to $x^{1/2}$. The symbol $x^{1/2}$ stands for a number whose square is $x$. The symbol $\sqrt{x}$ stands for the positive value of $x^{1/2}$. Hence $2^{1/2} = \pm \sqrt{2}.$

Things get even more tricky when we consider expressions like $x^{1/4}$. What does $x^{1/4}$ mean? It's a number whose fourth power is $x$. For example $1^{1/4} = \pm 1, \pm i$ where $i$ is a complex number and has the property that $i^2=-1$. Trying to plot the graph $y=x^{1/4}$ causes a problem. For all non-zero values of $x$, there are four (possibly complex) values for $y$. That is where the branches come from.

In this case, the symbol $\sqrt[4]{x}$ stands for the positive (real) value of $x^{1/4}$. Hence $2^{1/4} = \pm\sqrt[4]{2}, \pm i \sqrt[4]{2}.$