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


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 \(z^n = 1\).



  • \((-1)^2=1\), 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 \(\omega\neq 1\).


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

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

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

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

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

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.

Linear Algebra Day 34

By John Jasper

Linear Algebra Day 34

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