You've just finished a problem where you needed to solve the equation:

\begin{align*}x^4 = 16\end{align*}

After solving for the roots, which were \begin{align*}2,-2,2i,-2i\end{align*}

Can you accomplish this?

Read on, and by the end of this Concept, you'll understand how to plot and interpret the geometry of complex roots.

### Watch This

In the second part of this video you'll learn about the geometry of complex roots.

James Sousa: Determining the Nth Roots of a Complex Number

### Guidance

It's always good to get an intuitive feel for values by plotting them. This tendency extends into the complex numbers as well.

The five roots of the equation \begin{align*}x^5 - 32 = 0\end{align*}

The \begin{align*}n^{th}\end{align*}

#### Example A

Calculate the two roots for \begin{align*}x^2 = 1\end{align*}

**Solution:**

\begin{align*}x^2 = 1\\ r &= \sqrt{x^2+y^2}\\ r &= \sqrt{(1)^2+(0)^2}\\ r &= 1\\ \theta &= \tan^{-1} \left(\frac{0}{1}\right)=0\end{align*}

Write an expression for determining the two roots of \begin{align*}1 = 1 + 0i\end{align*}

\begin{align*}1^{\frac{1}{2}} &= [1( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{2}}\\ &= 1\left(\cos \frac{2\pi k}{2}+i \sin \frac{2\pi k}{2}\right)k=0, 1\\ x_1 &= 1\left(\cos \frac{0}{2}+i \sin \frac{0}{2}\right) \rightarrow 1(\cos 0 +i \sin 0)= 1 && for \ k=0\\ x_2 &= 1\left(\cos \frac{2\pi}{2}+i \sin \frac{2\pi}{2}\right) = -1 && for \ k=1\\ \end{align*}

These roots are plotted here:

#### Example B

Calculate the three roots for \begin{align*}x^3 = 1\end{align*} and represent them graphically.

**Solution:** In standard form, \begin{align*}1 = 1 + 0i \ r = 1\end{align*} and \begin{align*}\theta = 0\end{align*}. The polar form is \begin{align*}1 + 0i = 1 [\cos (0 + 2\pi k) + i \sin (0 + 2\pi k)]\end{align*}. The expression for determining the cube roots of \begin{align*}1 + 0i\end{align*} is:

\begin{align*}(1+0i)^{\frac{1}{3}}=1^{\frac{1}{3}} \left(\cos \frac{0+2 \pi k}{3}+i \sin \frac{0+2 \pi k}{3} \right)\end{align*}

When \begin{align*}k = 0, k = 1\end{align*} and \begin{align*}k = 2\end{align*} the three cube roots of 1 are \begin{align*}1, -\frac{1}{2}+i \frac{\sqrt{3}}{2}, -\frac{1}{2} -i \frac{\sqrt{3}}{2}\end{align*}. When these three roots are represented graphically, the three points, on the circle with a radius of 1 (the cubed root of 1 is 1), form a triangle.

#### Example C

Calculate the four roots for \begin{align*}x^4 = 1\end{align*} and represent them graphically

**Solution:**

\begin{align*}x^4 = 1\\ r &= \sqrt{x^2+y^2}\\ r &= \sqrt{(1)^2+(0)^2}\\ r &= 1\\ \theta &= \tan^{-1} \left(\frac{0}{1}\right)=0\end{align*}

Write an expression for determining the cube roots of \begin{align*}1 = 1 + 0i\end{align*}

\begin{align*}1^{\frac{1}{4}} &= [1( \cos (0+2\pi k)+i \sin (0+2\pi k)]^{\frac{1}{4}}\\ &= 1\left(\cos \frac{2\pi k}{4}+i \sin \frac{2\pi k}{4}\right)k=0, 1, 2, 3\\ x_1 &= 1\left(\cos \frac{0}{4}+i \sin \frac{0}{4}\right) \rightarrow 3(\cos 0 +i \sin 0)= 1 && for \ k=0\\ x_2 &= 1\left(\cos \frac{2\pi}{4}+i \sin \frac{2\pi}{4}\right) = 0 + i = i && for \ k=1\\ x_3 &= 1\left(\cos \frac{4\pi}{4}+i \sin \frac{4\pi}{4}\right) = -1 - 0i = -1 && for \ k=2\\ x_4 &= 1\left(\cos \frac{6\pi}{4}+i \sin \frac{6\pi}{4}\right) = 0 - i = -i && for \ k=3\\ \end{align*}

These roots are plotted here:

### Guided Practice

1. In the examples above, you saw the complex roots determined for a number of different polynomial orders, such as \begin{align*}x^2, x^3, x^4\end{align*}. What conclusion can you draw about the number of complex roots there are in relation to the order of the polynomial being solved?

2. What is the spacing in polar coordinates between the roots of the polynomial \begin{align*}x^6 = 12\end{align*}?

3. Solve for the roots of the equation \begin{align*}x^2 - 3x + 5 = 0\end{align*} and plot them.

*Solutions:*

1. You can conclude that the total number of roots is the same as the order of the polynomial under consideration. For example, \begin{align*}x^2\end{align*} will have 2 roots, while \begin{align*}x^3\end{align*} will have 3 roots, etc.

2. Since there are six total roots, and all of the roots are equally spaced around a circle in the complex plane, there are \begin{align*}\frac{360^\circ}{6} = 60^\circ\end{align*} between roots.

3. This equation can be solved using the familiar quadratic formula. Notice that previously we used quadratics that gave real solutions. However, there is no reason to limit the solution set to the real numbers, now that you know how to utilize complex numbers as well.

\begin{align*} x = \frac{-(-3) \pm \sqrt{(-3)^2-(4)(1)(5)}}{(2)(1)}\\ =\frac{3 \pm \sqrt{-11}}{2}\\ =\frac{3 \pm i\sqrt{11}}{2}\\ = \frac{3}{2} \pm \frac{i\sqrt{11}}{2}\\ \end{align*}

A plot of these roots looks like this:

where the vertical axis is the imaginary number line.

### Concept Problem Solution

You can see a plot of these roots here. Notice, as mentioned before, that the roots are placed equidistant around a circle that could be drawn in the plane with a radius of two.

### Explore More

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