I wrote this note in 2008 to introduce complex numbers. It is posted for the Math Teachers at Play blog carnival.

What is the ${\sqrt[4]{-16}}$?

In the set of Real Numbers, ${\mathbb{R}}$, the solution is undefined. We must consider the set of Complex Numbers, ${\mathbb{C}}$. Complex numbers are numbers of the form of ${a+bi}$, where ${a}$ and ${b}$ are real numbers: ${a,b\in\mathbb{R}}$ and ${i}$ is the imaginary unit. Note ${i}$ is defined by ${i^{2}=-1}$ or equivalently ${i=\sqrt{-1}}$.

Thus, we have:

$\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & (-16)^{1/4}\\ & = & (-1\cdot16)^{1/4}\\ & = & (-1\cdot2^{4)^{1/4}}\\ & = & (-1)^{1/4}\cdot2^{4/4}\\ & = & (-1)^{1/4}\cdot2\\ & = & ((-1)^{1/2})^{1/2}\cdot2\\ & = & (\sqrt{-1})^{1/2}\cdot2\\ & = & i^{1/2}\cdot2\\ & = & 2\sqrt{i}.\end{array}$

We now have ${\sqrt[4]{-16}=2\sqrt{i}}$. So we must ask "What is the ${\sqrt{i}}$?". To evaluate ${\sqrt{i}}$ we will use Euler’s formula, ${e^{ix}=\cos\left(x\right)+i\sin\left(x\right).}$ So by substituting ${x=\frac{\pi}{2}}$, giving

$\displaystyle \begin{array}{rcl} e^{i(\frac{\pi}{2})} & = & \cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)\\ & = & 0+i(1)\\ & = & i.\end{array}$

Note we have isolated ${i}$ and now arrived at the following equality: ${i=e^{i(\frac{\pi}{2})}}$. Taking the square root of both sides gives

$\displaystyle \begin{array}{rcl} \sqrt{i} & = & \sqrt{e^{i(\frac{\pi}{2})}}\\ & = & e^{i(\frac{\pi}{2})(\frac{1}{2})}\\ & = & e^{i(\frac{\pi}{4})}.\end{array}$

Thus, we have now have ${\sqrt{i}=e^{i(\frac{\pi}{4})}}$ and following through application of Euler’s formula to ${x=\frac{\pi}{4}}$, gives

$\displaystyle \begin{array}{rcl} \sqrt{i} & = & e^{i(\frac{\pi}{4})}\\ & = & \cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\\ & = & \frac{1}{\sqrt{2}}+i(\frac{1}{\sqrt{2}})\\ & = & \frac{\sqrt{2}}{2}+i(\frac{\sqrt{2}}{2})\\ & = & \frac{\sqrt{2}+\sqrt{2}i}{2}.\end{array}$

So going back to the original problem and substituting ${\sqrt{i}=\frac{\sqrt{2}+\sqrt{2}i}{2}}$, gives

$\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & 2\sqrt{i}\\ & = & 2(\frac{\sqrt{2}+\sqrt{2}i}{2})\\ & = & \sqrt{2}+\sqrt{2}i.\end{array}$

Hence, the solution, ${\sqrt[4]{-16}=\sqrt{2}+\sqrt{2}i}$ , is a complex number!