DLA The complex plane

Section A.3 The complex plane

In this section.

Subsection A.3.1 The complex plane

We visualize the field of complex numbers geometrically by assigning geometric meanings to the two independent parameters that make up an arbitrary complex number

$z = a + bi\text{.}$ Let's interpret the real part

$\Re(z) = a$ as an

$x$ -coordinate and

$\Im(z) = b$ as a

$y$ -coordinate in the plane.

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In this version of the plane, the

$x$ -axis is referred to as the real axis and the

$y$ -axis is referred to as the imaginary axis. Pythagoras tells us that the distance from the origin to the complex point

$z$ is equal its modulus

$\cmodulus{z}\text{.}$ The angle

$\theta$ formed by the ray from the origin through

$z$ and the positive real axis is called the argument of

$z\text{.}$

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Complex conjugation has a geometric interpretation — it is equivalent to reflection in the real axis.

Complex conjugation as a reflection in the complex plane.

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For every angle

$\theta\text{,}$ a complex number of the form

$\cos\theta + \ci\sin\theta$ has unit modulus due to the circle identity:

$\begin{equation*} \cmodulus{\cos\theta + \ci\sin\theta} = \sqrt{\cos^2\theta + \sin^2\theta} = \sqrt{1} = 1. \end{equation*}$

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Because of this, complex numbers of this form sit on the unit circle.

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The unit circle meets the axes at the four points

$\begin{align*} 1 \amp = \cos 0 + \ci\sin 0, \amp \ci \amp = \cos\frac{\pi}{2} + \ci\sin\frac{\pi}{2},\\ -1 \amp = \cos \pi + \ci\sin \pi, \amp -\ci \amp = \cos\frac{3\pi}{2} + \ci\sin\frac{3\pi}{2}. \end{align*}$

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Subsection A.3.2 Polar coordinates

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For reasons we won't go into here, it is reasonable to define complex exponentiation with natural base

$e$ (Euler's number) as

$\begin{gather} e^{\ci\theta} = \cos\theta + \ci\sin\theta \text{.}\label{equation-complex-appendix-plane-euler}\tag{$\star$} \end{gather}$

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This is known as Euler's formula.

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As in the previous subsection, these points all have modulus

$\cmodulus{e^{\ci\theta}} = 1$ and sit on the unit circle in the complex plane. We can use these points to describe every point in the complex plane in terms of a radius and angle. If

$r$ is a nonnegative real number, then for

$z = r e^{\ci\theta}$ we have

$\begin{equation*} \cmodulus{z} = \cmodulus{r}\cmodulus{e^{\ci\theta}} = r. \end{equation*}$

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Thus, instead of describing a point in the complex plane in terms of a horizontal component (the real part) and a vertical component (the imaginary part), we can describe a point in terms of a radius (the modulus) and an angle to the positive real axis (the argument).

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An expression

$z = re^{\ci\theta}$ is called the polar form of the complex number

$z\text{.}$

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While we usually take the argument

$\theta$ to be in the domain

$0\le\theta\lt 2\pi\text{,}$ it is useful in computations to allow it to take on any real value, but simplify complex polar expressions using the identity

$\begin{equation*} e^{\ci(\theta + 2m\pi)} = e^{\ci\theta}, \end{equation*}$

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valid for arbitrary integer

$m\text{.}$

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Consider what happens when we multiply complex numbers in polar coordinates: if

$z_1 = r_1 e^{\ci\theta_1}$ and

$z_2 = r_2 e^{\ci\theta_2}\text{,}$ then

$\begin{equation*} z_1 z_2 = (r_1 r_2) (e^{\ci\theta_1} e^{\ci\theta_2}). \end{equation*}$

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Notice that the radii multiply as well. We know an exponent law that would help us simplify the exponential part of the product expression above, but is it valid for complex exponentials?

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Proposition A.3.1. Complex arguments add.

Multiplication on the complex unit circle corresponds to addition of the complex arguments. That is,

$\begin{equation*} e^{\ci\theta_1} e^{\ci\theta_2} = e^{\ci(\theta_1 + \theta_2)}. \end{equation*}$

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Proof idea.

By definition (i.e. Euler's formula), we have

\begin{equation*} e^{\ci\theta_1} e^{\ci\theta_2} = (\cos\theta_1 + \ci \sin\theta_1)(\cos\theta_2 + \ci \sin\theta_2). \end{equation*}

Expanding this out using FOIL, you will find that the real part is

\begin{equation*} \cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2, \end{equation*}

and the imaginary part is

\begin{equation*} \cos\theta_1\sin\theta_2 + \sin\theta_1\cos\theta_2. \end{equation*}

So by the addition identities for $\sin$ and $\cos\text{,}$ we have

\begin{equation*} e^{\ci\theta_1} e^{\ci\theta_2} = \cos(\theta_1 + \theta_2) + \ci\sin(\theta_1 + \theta_2) = e^{\ci(\theta_1 + \theta_2)}, \end{equation*}

as desired.

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We can directly obtain several more important facts about polar expressions from this proposition.

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Corollary A.3.2. Multiplication of polar expressions.

If $z_1 = r_1 e^{\ci\theta_1}$ and $z_2 = r_2 e^{\ci\theta_2}\text{,}$ then

$\begin{equation*} z_1 z_2 = (r_1 r_2) e^{\ci(\theta_1 + \theta_2)}. \end{equation*}$

Geometrically, if $z$ is a point in the complex plane, then multiplying $z$ by the polar expression $r e^{\ci\theta}$ scales $z$ toward/away from the origin by scale factor $r\text{,}$ and rotates by $\theta$ radians about the origin.

Multiplying by a polar expression in the complex plane.

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Corollary A.3.3. Inverse of a polar expression.

If $z = r e^{\ci\theta}$ then $\inv{z} = \inv{r} e^{-\ci\theta}\text{.}$

In particular, $\inv{(e^{\ci\theta})} = e^{-\ci\theta}\text{.}$

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Proof idea.

Just multiply $r e^{\ci\theta}$ and $\inv{r} e^{-\ci\theta}$ together and use Corollary A.3.2 to simplify. You will find that the simplified product is equal to $1\text{.}$

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Corollary A.3.4. DeMoivre's formula.

For every integer $n\text{,}$ we have $(e^{\ci\theta})^n = e^{\ci n \theta}\text{.}$

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Proof idea.

A proper proof would use the method of proof by induction, but we will be a little less formal.

If $n$ is a postive integer, then exponent $n$ means multiplying $n$ copies of the complex exponential:

\begin{equation*} (e^{\ci\theta})^n = e^{\ci\theta} \cdot e^{\ci\theta} \cdot \dotsm \cdot e^{\ci\theta}. \end{equation*}

Using Proposition A.3.1, we can combine the expression on the right above into one complex exponential where $n$ copies of the argument $\theta$ are added together.

If $n$ is a negative integer, we can combine the positive case above with Corollary A.3.3.

And if $n$ is zero, then both sides of the formula are equal to $1$ (adhering to the convention $z^0 = 1$ for $z \ne 0$ on the left).

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Subsection A.3.3 Roots of unity

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When we allow only real solutions, the equation

$x^n = 1$ (equivalently,

$x^n-1=0$ ) has two solutions when

$n$ is even and one solution when

$n$ is odd. However, when we allow complex solutions as well, The Fundamental Theorem of Algebra (Complex Version) tells us there should always be exactly

$n$ solutions (including repeated roots). Each solution is called an

$\nth$ root of unity.

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Suppose complex number

$z$ is an

$\nth$ root of unity, so that

$z^n = 1$ is true. Then Statement 9 of Proposition A.2.14 tells us that

$\cmodulus{z}^n = 1\text{.}$ But a modulus is always a real number, and cannot be negative (Statement 2 of Proposition A.2.14). So there is only one way

$\cmodulus{z}^n = 1$ can be true, and that is if

$\cmodulus{z} = 1\text{.}$

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Proposition A.3.5. Modulus of roots of unity.

Every complex root of unity has modulus $1\text{,}$ and hence lies on the unit circle in the complex plane.

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In light of the proposition, in polar coordinates we can represent a root of unity purely in terms of its argument:

$z = e^{\ci\theta}$

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Example A.3.6. Cubic roots of unity.

Let's determine all three complex solutions to $x^3 = 1\text{.}$ As above, let's set $x = e^{\ci\theta}$ and try to determine the values of $\theta$ that lead to solutions. From DeMoivre's formula, we have $x^3 = e^{\ci(3\theta)} \text{,}$ which transforms our equation into

$\begin{equation*} e^{\ci(3\theta)} = 1. \end{equation*}$

Clearly we obtain a solution by setting $\theta = 0 \text{,}$ which corresponds to our known real solution $x = 1 \text{.}$ Reasoning geometrically in the complex plane, the other solutions correspond to values of $\theta$ so that $3\theta$ returns back to the positive real axis, i.e. when $3\theta$ is a multiple of $2\pi \text{.}$ So our full slate of solutions could be expressed as

$\begin{equation*} e^{\ci(0\pi/3)}, \quad e^{\ci(2\pi/3)}, \quad e^{\ci(4\pi/3)}, \end{equation*}$

and we can visualize these solutions as regularly placed around the unit circle in the complex plane.

Cubic roots of unity on the complex unit circle.

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The pattern of the previous example continues for higher-order roots of unity.

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Proposition A.3.7. Polar form of roots of unity.

The complete set of $\nth$ roots of unity are

$\begin{equation*} e^{0\pi\ci/n}, \quad e^{2\pi\ci/n}, \quad e^{4\pi\ci/n}, \quad \dotsc, \quad e^{2k\pi\ci/n}, \quad \dotsc, e^{2(n-1)\pi\ci/n}. \end{equation*}$