# How do you find the imaginary roots of a number?

This question involves a few major concepts across several lessons - for a deeper dive and for a full gambit of practice and common tricks, make sure to check out each Relevant Mister Math Lesson in the list above - particularly the bottom three!

## DeMoivre's Theorem

First thing's first - when you take the $n$-th root of a number, real or imaginary, there are actually $n$ answers. One of the most efficient ways to find them all is to write the number in complex polar form (commonly called $\mathrm{cis}$ form, pronounced /siss/) and use DeMoivre's Theorem.When you write a number in complex form $a + bi$ (where $b$ = 0 for real numbers), recall that you can plot that number in the Real-Complex coordinate plane: This works just like any coordinate $(x , y)$ pair that we started working with Algebra 1. And once we've mastered some coordinate plan trigonometry, we can take any point we plot in the Real-Complex plane and rewrite it in polar form. Instead of describing the cooridinates of the point in terms of rectangular coordinates a and b (left/right and up/down measurement), we can plot it in terms of distance from the origin and direction angle: If we plot complex numbers in the plane in polar form, then a point that is r units from the origin in the direction that has an angle of theta, then we can say the number is $r \, \mathrm{cis}(\theta)$. Check out the lesson on polar imaginary numbers with $\mathrm{cis}$ » for a more in-depth look at plotting imaginary numbers in the plane in this manner.The final key is to this that we can take the root of a complex number in $\mathrm{cis}$ form using DeMoivre's Theorem. It works as such:$$\left[r \, \mathrm{cis}(\theta)\right]^{n} = r^n \, \mathrm{cis}(n \theta)$$To take the square root, we would raise $r \, \mathrm{cis}(\theta)$ to the (1/2) power. To take the (1/5) root we would raise $r \, \mathrm{cis}(\theta)$ to the 1/5 power. Taking roots by using exponenets is something you can learn more about here ». The trick that will allow us to find all of the roots we want is to use coterminal values of $\theta$.Let's start with positive real numbers. As an example, we will find the cube roots of 1. In the Real-Imaginary coordinate plane, using rectangluar coordinates, we can represent $1$ as $1 + 0i$. Here's where it plots to: Since this point is $1$ unit away in the direction that has an angle of $0$ radians, we can write $1$ as:$$1 = 1 \, \mathrm{cis} (0)$$However, we can also use the value $2\pi$ for theta, because $2\pi$ radians is coterminal with $0$ radians. Here's the trick: we will start with $0$ for $\theta$, and write more coterminal $\mathrm{cis}$ values so that we have as many as there are roots. Here we are told that there are $3$ cube roots of $1$, trusting when I earlier said there are always $n$ number of $n$-th roots.$$1 = 1 \, \mathrm{cis}(0)$$$$1 = 1 \, \mathrm{cis}(2\pi)$$$$1 = 1 \, \mathrm{cis}(4\pi)$$Each of these 3 forms represents the number 1 in $\mathrm{cis}$ form. Now for the payoff! Take the cube root of each of those values, using DeMoivre's Theorem:$$1^\frac{1}{3} = 1^\frac{1}{3} \left[ 1 \, \mathrm{cis}\left(0 \, \left(\frac{1}{3}\right)\right) \right] = 1 \, \mathrm{cis}(0)$$$$1^\frac{1}{3} = 1^\frac{1}{3} \left[ 1 \, \mathrm{cis}\left(2\pi \, \left(\frac{1}{3}\right)\right) \right] = 1 \, \mathrm{cis}\left(\frac{2\pi}{3}\right)$$$$1^\frac{1}{3} = 1^\frac{1}{3} \left[ 1 \, \mathrm{cis}\left(4\pi \, \left(\frac{1}{3}\right)\right) \right] = 1 \, \mathrm{cis}\left(\frac{4\pi}{3}\right)$$Finally, use the definition of $\mathrm{cis}$, along with the known values of cosine and sine at these angles, we get the answers.$$1^\frac{1}{3} = 1 \, \mathrm{cis}(0)$$$$= 1[\cos(0) + i \sin(0)] = 1[1 + 0] = \boxed{1}$$$$1^\frac{1}{3} = 1 \, \mathrm{cis}\left(\frac{2\pi}{3}\right) = 1 \left[\cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right)\right]$$$$= 1\left[-\frac{1}{2} + i \left(\sqrt\frac{3}{2}\right)\right] = \boxed{-\frac{1}{2} + i \sqrt\frac{3}{2}}$$$$1^\frac{1}{3} = 1 \, \mathrm{cis}\left(\frac{4\pi}{3}\right) = 1 \left[\cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right)\right]$$$$= 1\left[-\frac{1}{2} + i \left(-\sqrt\frac{3}{2}\right)\right] = \boxed{-\frac{1}{2} - i \sqrt\frac{3}{2}}$$Note that if we did keep going we would get$$1^\left(\frac{1}{3}\right) = 1^\left(\frac{1}{3}\right) \left[ 1 \, \mathrm{cis}\left(6\pi \left(\frac{1}{3}\right)\right) \right]$$$$= 1^\left(\frac{1}{3}\right) \, \mathrm{cis}\left(\frac{6\pi}{3}\right)$$$$= 1 [\cos(2\pi) + i \sin (2\pi)] = 1 [1 + 0] = \boxed{1}$$This is the same answer we got when we used $\theta$ = $0$. And if we went again one more, we would get the same answer as when we used $2\pi$. The same three values would be repeated because there are exactly three cube roots of a number. This is in line with what we said before - that for any number, there are $n$ number of $n$-th roots. Plotting the results on the Real-Imaginary Plane also helps us see why the answers repeat: We can see that the answers are equally spaced, and that if you start at the answer $1$ (which is probably the only cube root of $1$ that you knew before this), and you move one-third of the unit circle, you keep landing on the same three spots - the three third roots of $1$.
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