# Day 34:

Complex numbers

### Complex Numbers

The set $$\mathbb{C} = \{a+bi : a,b\in\R\}$$, where $$i$$ is a number such that $$i^2=-1$$, is called the complex numbers.

Let $$z\in\mathbb{C}$$ and assume $$z=a+bi$$ for some $$a,b\in\R$$.

• The conjugate of $$z$$, denoted $$\overline{z}$$, is the complex number $$a-bi$$.
• The modulus of $$z$$, denoted $$|z|$$, is the number $$\sqrt{a^{2}+b^{2}}$$.
• The real part of $$z$$, denoted $$\Re(z)$$, is the number $$a$$.
• The imaginary part of $$z$$, denoted $$\Im(z)$$, is the number $$b$$. (Note that the imaginary part of $$z$$ is a real number!)

Some facts about complex numbers:

1. $$z\overline{z} = (a+bi)(a-bi) = a^2-abi+abi-(bi)^2 = a^2+b^2 = |z|^2$$.
2. $$z=\Re(z) + i\Im(z)$$
3. $$\left|\dfrac{z}{|z|} \right|^2 = \left|\dfrac{a+bi}{|z|}\right|^2 = \left|\dfrac{a}{|z|} + \dfrac{b}{|z|}i\right|^2 = \dfrac{a^2}{|z|^2} + \dfrac{b^{2}}{|z|^{2}} = \dfrac{a^2+b^2}{|z|^2} = 1$$

### Complex Numbers

Theorem. If $$z\in\mathbb{C}$$ and $$|z|=1$$, then there is a number $$\theta\in[0,2\pi)$$ such that $$z=\cos(\theta)+i\sin(\theta).$$

Corollary. If $$z\in\mathbb{C}$$, then there is a number $$\theta\in[0,2\pi)$$ and a nonnegative number $$r\geq 0$$ such that $$z=r\left(\cos(\theta)+i\sin(\theta)\right).$$

Proof. Take $$r=|z|$$.

### Complex Numbers in Polar Form

If $$\theta\in\R$$, then we define the complex number

$e^{i\theta} = \cos(\theta)+i\sin(\theta).$

By the previous theorem, if $$z\in\mathbb{C}$$, then there exists a number $$\theta\in[0,2\pi)$$ and $$r\geq 0$$ such that

$z=re^{i\theta}.$

The expression $$re^{i\theta}$$ is called the polar form of $$z$$. The number $$\theta$$ is called the argument of $$z$$, and is sometimes written $$\arg(z)$$.

Warning! If $$\varphi=\theta+2\pi$$, then

$e^{i\varphi} = \cos(\varphi)+i\sin(\varphi) = \cos(\theta+2\pi) + i\sin(\theta+2\pi)$

So, $$e^{i\theta}$$ and $$e^{i\varphi}$$ are the same number!

$= \cos(\theta)+i\sin(\theta) = e^{i\theta}.$

### Complex Numbers in Polar Form

Exercise. Write $$1+i$$ in polar form.

First, we find the modulus:

$|1+i| = \sqrt{1^2+1^2} = \sqrt{2}.$

Then, divide by the modulus and we get

$\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i.$

Now, we want $$\theta\in[0,2\pi)$$ such that

$\cos(\theta)=\frac{1}{\sqrt{2}}\quad\text{and}\quad \sin(\theta) = \frac{1}{\sqrt{2}}$

We see that $$\theta=\frac{\pi}{4}$$, and thus

$e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i.$

Multiplying both sides by the modulus we have

$\sqrt{2}e^{i\frac{\pi}{4}} = 1+i.$

### Roots of unity

Given a positive whole number $$n$$, an $$n$$th root of unity is a complex number which is a solution to the equation $$z^n = 1$$.

Example.

• $$(-1)^2=1$$, and hence $$-1$$ is a $$2$$nd root of unity.
• $$i^4=i^2i^2=(-1)(-1)=1$$, and hence $$i$$ is a $$4$$th root of unity.
• $$1,-1$$, and $$-i$$ are all $$4$$th roots of unity.

If $$n$$ is a positive whole number, then the number $$e^{\frac{2\pi i}{n}}$$ is an $$n$$th root of unity, since

$\left(e^{\frac{2\pi i}{n}}\right)^{n} = e^{\frac{2\pi i}{n}\cdot n} = e^{2\pi i} = cos(2\pi)+i\sin(2\pi) = 1.$

There are $$n$$ complex $$n$$th roots of unity, they are

$e^{\frac{2\pi i}{n}},e^{\frac{4\pi i}{n}},e^{\frac{6\pi i}{n}},\ldots,e^{\frac{2n\pi i}{n}} = 1$

### Roots of unity

Example. The $$4$$th roots of unity are

$$e^{\frac{2\pi i}{4}} = e^{\frac{\pi}{2}i} = \cos(\tfrac{\pi}{2}) + i\sin(\tfrac{\pi}{2}) = 0 + 1i = i,$$

$$e^{\frac{4\pi i}{4}} = e^{\pi i} = \cos(\pi) + i\sin(\pi) = -1 + 0i = -1,$$

$$e^{\frac{6\pi i}{4}} = e^{\frac{3\pi}{2}i} = \cos(\tfrac{3\pi}{2}) + i\sin(\tfrac{3\pi}{2}) = 0 + (-1)i = -i,$$

$$e^{\frac{8\pi i}{4}} = e^{2\pi i} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i = 1.$$

Note that $$i+(-1)+(-i)+1 = 0$$.

### Roots of unity

Example.  The $$8$$th roots of unity are

$e^{\frac{2\pi i}{8}}, e^{\frac{4\pi i}{8}}, e^{\frac{6\pi i}{8}}, e^{\frac{8\pi i}{8}}, e^{\frac{10\pi i}{8}}, e^{\frac{12\pi i}{8}}, e^{\frac{14\pi i}{8}}, e^{\frac{16\pi i}{8}}$

In rectangular form these are

$\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i, i, -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i, -1, -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i, -i, \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i, 1$

### Roots of unity

Theorem. Let $$n\geq 2$$ be a whole number. The $$n$$th roots of unity sum to zero, that is,

Proof. Set $\sigma=\sum_{k=1}^{n}e^{\frac{2k\pi i}{n}}\quad\text{and}\quad \omega = e^{\frac{2\pi i}{n}}.$

Since $$n\geq 2$$ we see that $$\omega\neq 1$$.

Now,

$\omega\sigma = e^{\frac{2\pi i}{n}}\left(\sum_{k=1}^{n}e^{\frac{2k\pi i}{n}}\right) = \sum_{k=1}^{n}e^{\frac{2\pi i}{n}}e^{\frac{2k\pi i}{n}} = \sum_{k=1}^{n}e^{\frac{2(k+1)\pi i}{n}}$

$= e^\frac{4\pi i}{n} + e^\frac{6\pi i}{n} + e^\frac{8\pi i}{n} + \cdots + e^\frac{2n\pi i}{n} + e^\frac{2(n+1)\pi i}{n}$

$\sum_{k=1}^{n}e^{\frac{2k\pi i}{n}} = 0.$

$e^{\frac{2(n+1)\pi i}{n}} = e^{\left(\frac{2n\pi i}{n} + \frac{2\pi i}{n}\right)} = e^{\frac{2n\pi i}{n}} e^{\frac{2\pi i}{n}}= e^{2\pi i} e^{\frac{2\pi i}{n}} = e^{\frac{2\pi i}{n}}$

$= e^\frac{2\pi i}{n} + e^\frac{4\pi i}{n} + e^\frac{6\pi i}{n} + e^\frac{8\pi i}{n} + \cdots + e^\frac{2n\pi i}{n} = \sigma$

### Complex Vectors

$$\mathbb{C}^{2} = \left\{\left[\begin{matrix}a\\ b\end{matrix}\right] : a\in\mathbb{C} \text{ and } b\in\mathbb{C}\right\}$$

$$\mathbb{C}^{3} = \left\{\left[\begin{matrix}a\\ b\\c\end{matrix}\right] : a\in\mathbb{C} \text{ and } b\in\mathbb{C} \text{ and } c\in\mathbb{C}\right\}$$

$$\mathbb{C}^{n} = \left\{\left[\begin{matrix}a_{1}\\ a_{2}\\\vdots\\a_{n}\end{matrix}\right] : a_{1}\in\mathbb{C} \text{ and } a_{2}\in\mathbb{C}\text{ and }\ldots \text{ and } a_{n}\in\mathbb{C}\right\}$$

Assume we have a matrix $$A$$ with complex entries:

$A=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]$

### The adjoint

The adjoint of $$A$$, denoted $$A^{\ast}$$, is the matrix

$A^{\ast}=\left[\begin{matrix} \overline{a_{11}} & \overline{a_{21}} & \cdots & \overline{a_{n1}}\\ \overline{a_{12}} & \overline{a_{22}} & & \overline{a_{n2}}\\ \vdots & & \ddots & \vdots \\ \overline{a_{1m}} & \overline{a_{2m}} & \cdots & \overline{a_{nm}}\end{matrix}\right]$

Note that we obtain the adjoint by taking the transpose of $$A$$ and conjugating each entry.

By John Jasper

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