Roots

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roots of complex

Balancing Stories

In the real numbers, we have been following an off-balanced story for roots.

Let $r$ be a real number.

Sometimes, there are two square roots of $r$ and sometimes there are none. It depends on the sign of $r$. The Complex Numbers have repaired this story. There are always two solutions to the equation $z^2 = r$. Even $0$ works when we bring in multiplicities.

The Complex Numbers have balanced the story for square roots.

$\blacktriangleright $ What about other roots?

If $r$ is a positive real number, then we can at least establish that there is a root of any order. Like, one.

Let $r$ be a positive real number. Let $n$ be a natural number.

In the real numbers, we can express each positive number in terms of $e$: $r = e^{\ln (r)}$

And, we can express the whole story of roots in terms of $e$.

\[ \sqrt [n]{r} = r^{\tfrac {1}{n}}= e^{\tfrac {1}{n} \ln (r)} \]

$\blacktriangleright $ Each positive real number has at least one $n^{th}$ root for any natural number $n$.

What if $n$ is not a natural number?

The expression $e^{\tfrac {1}{n} \ln (r)}$ doesn’t care if $n$ is a natural number.

$\blacktriangleright $ Each positive real number has at least one $n^{th}$ root for any positive real number $n$.

The expression $e^{\tfrac {1}{n} \ln (r)}$ doesn’t care if $n$ is positive or negative.

$\blacktriangleright $ Each positive real number has at least one $n^{th}$ root for any nonzero real number $n$.

We are stuck now. Our logarithm function does not have a domain that includes negative real numbers.

But, negative real numbers are Complex numbers. Maybe we can balance this story inside the Complex Numbers.

the Complex Story

We want to do the same reasoning of roots for Complex Numbers.

Euler’s Formula tells us that $e^{i \theta } = \cos (\theta ) + i \, \sin (\theta )$, which describes all of the points on the unit circle.

We also know that every nonzero complex number is a scalar (positive real number) times a number on the unit circle: $r e^{i \theta }$.

We also know that every positive real number can be written as an exponential: $r = e^{\ln (r)}$.

Therefore, every complex number can be written as a complex exponential: $ e^{\ln (r) + i \theta }$

\[ a + b \, i = e^{\ln (r) + i \theta } \]

\[ \text { where } \, r = |a + b \, i| = \sqrt {a^2 + b^2} \, \text { and } \, \theta = \arctan \left (\frac {b}{a} \right ) \pm 2k\pi \]

Let’s begin extending the story of roots with roots of unity.

$n^{th}$ roots of Unity

Focusing on the Complex Numbers on the unit circle, they can be written in the form $e^{i \theta }$.

This would include all of the solutions to $z^n = 1$, with $n$ a natural number (the roots of unity).

If we are looking for solutions to $(e^{i \theta })^n = 1$, then we need $n \cdot \theta = 2k\pi $, for some $k \in \mathbb {N}$.

Therefore, the $n^{th}$ roots of $1$ are equally spaced on the unit circle.

\[ 1=e^{\tfrac {0\cdot 2\pi }{n}\,i}, e^{\tfrac {1\cdot 2\pi }{n}\,i}, e^{\tfrac {2\cdot 2\pi }{n}\,i}, e^{\tfrac {3\cdot 2\pi }{n}\,i}, \cdots , e^{\tfrac {(n-1)\cdot 2\pi }{n}\,i}, e^{\tfrac {n\cdot 2\pi }{n}\,i}=1 \]

Begin with $1$, then go $\frac {1}{n^{th}}$ around the circle, then another, then another, until you get back to the beginning.

Cube roots of Unity

There are $3$ third roots of unity.

$1$ is the first.

Then move $\frac {1}{3}$ of a circle ($120^{\circ }$) to $e^{\tfrac {2 \pi }{3} i} = \cos \left (\frac {2 \pi }{3}\right ) + i \, \sin \left (\frac {2 \pi }{3}\right ) = \frac {-1}{2} + i \, \frac {\sqrt {3}}{2}$

Then move another $\frac {1}{3}$ of a circle ($240^{\circ }$) to $e^{\tfrac {4 \pi }{3} i} = \cos \left (\frac {4 \pi }{3}\right ) + i \, \sin \left (\frac {4 \pi }{3}\right ) = \frac {-1}{2} - i \, \frac {\sqrt {3}}{2}$

Then move another $\frac {1}{3}$ of a circle, which returns you to $1$.

Fourth roots of Unity

There are $4$ fourth roots of unity.

$1$ is the first.

A quarter circle gets you to $e^{\tfrac {\pi }{2}} = i$.

Another quarter circle gets you to $e^{\pi } = -1$.

Another quarter circle gets you to $e^{\tfrac {3\pi }{2}} = -i$.

Another quarter circle gets you back to $1$.

Fifth roots of Unity

There are $5$ fifth roots of unity.

$1$ is the first.

A fifth of a circle gets you to $e^{\tfrac {2\pi }{5}} = \cos (\tfrac {2\pi }{5}) + i \, \sin (\tfrac {2\pi }{5})$.

Another fifth of a circle gets you to $e^{\tfrac {4\pi }{5}} = \cos (\tfrac {4\pi }{5}) + i \, \sin (\tfrac {4\pi }{5})$.

Another fifth of a circle gets you to $e^{\tfrac {6\pi }{5}} = \cos (\tfrac {6\pi }{5}) + i \, \sin (\tfrac {6\pi }{5})$.

Another fifth of a circle gets you to $e^{\tfrac {8\pi }{5}} = \cos (\tfrac {8\pi }{5}) + i \, \sin (\tfrac {8\pi }{5})$.

Another fifth of a circle gets you back to $e^{\tfrac {10\pi }{5}} = \cos (\tfrac {10\pi }{5}) + i \, \sin (\tfrac {10\pi }{5}) = 1$.

A Peek Ahead....

Can we write those fifth roots of unity with traditional square roots rather than trigonometric function values?

We need to peek ahead a few days....

We’ll need a double angle formula: $\cos (2x) = 2(\cos (x))^2 - 1$

Let $a = \cos \left (\frac {\pi }{5}\right )$

Let $b = \cos \left (\frac {2\pi }{5}\right )$

$\frac {\pi }{5}$ is in the first quadrant and $\frac {4\pi }{5}$ is a supplementary angle in the second quadrant. This tells us that

\[ \cos \left (\frac {4\pi }{5}\right ) = -\cos \left (\frac {\pi }{5}\right ) = -a \]

The double angle formula says $\cos (2x) = 2(\cos (x))^2 - 1$. Applying this twice gives us

$b = 2a^2 - 1$
$-a 2b^2 - 1$

Subtracting gives $a + b = 2 (a^2 - b^2) = 2 (a+b)(a-b)$

$\frac {\pi }{5}$ and $\frac {2\pi }{5}$ are both in the first quadrant, therefore both $a$ and $b$ are both positive.

Therefore, $a+b \ne 0$.

Therefore, $a - b = \frac {1}{2}$ or $b = a - \frac {1}{2}$

All of that tells us that

\[ a - \frac {1}{2} = 2a^2 - 1 \]

or,

\[ 4a^2 - 2 a - 1 =0\]

We can now use the quadratic formula. We need the postive solution.

\[ a = \cos \left (\frac {\pi }{5}\right ) = \frac {\sqrt {5} + 1}{4} \]

\[ b = \cos \left (\frac {2\pi }{5}\right ) = a - \frac {1}{2} = \frac {\sqrt {5} - 1}{4} \]

We can then use $(\cos (\theta ))^2 + (\sin (\theta ))^2 = 1$ to get

\[ \sin \left (\frac {2\pi }{5}\right ) = \frac {\sqrt {10 - 2 \sqrt {5}}}{4} \]

Which finally gives us

\[ e^{\tfrac {2\pi }{5}} = \cos (\tfrac {2\pi }{5}) + i \, \sin (\tfrac {2\pi }{5}) = \frac {\sqrt {5} + 1}{4} + \frac {i \sqrt {10 - 2 \sqrt {5}}}{4} \]

Our ideas of numbers have grown as we have used them.

When we were counting, we used the whole numbers.

The whole numbers were not enough when we started spliting things apart. We filled in the gaps with rational numbers, fractions.

The rational numbers were not enough when we started measuring lengths. We filled in the gaps with irrational numbers.

Now that we are thinking of solutions to equations, we are discovering that the real numbers are missing numbers. We have filled in the gaps with the Complex Numbers.

The difference here is that filling in the gaps before maintained the 1-dimensional number line. Filling in the gaps to equations has revealed that our numbers are really 2-dimensional. It is a plane of numbers rather than a line of numbers.

This has added A LOT of geometry to our number descriptions.

We are now blurring the distinction between geometry and arithmetic.

We think of complex numbers as dots in the Cartesian plane: $(a, b)$
We think of complex numbers as vectors in the Cartesian plane: $\langle a, b \rangle $
We think of complex numbers as sums of rectangular directions: $a + b \, i$
We think of complex numbers as circular (polar) directions: $(r, \theta )$
We think of complex numbers as extensions of numbers on the unit circle: $r(\cos (\theta )+i \, \sin (\theta ))$
We think of complex numbers as complex exponentials: $r\cdot e^{i\, \theta } = e^{\ln (r)+i\, \theta }$

It turns out that the Complex Numbers have all of the solutions to all of the equations they can make. The Complex Numbers are “complete”. That means we can work confidently now knowing that we are no longer missing any numbers that we need.

We can now write a famous identity that ties all of our important constants together ($e$, $i$, $\pi $, $1$, and $0$).

Euler’s Identity

\[ e^{i \, \pi } + 1 = 0 \]

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more examples can be found by following this link
More Examples of Complex Exponentials

2025-05-17 22:31:35