Imaginary numbers

  • Invented to solve problems where an equation has no real roots e.g. \(x^2 + 16 = 0\), the idea of declaring the existence of a quantity \(i\) such that \(i^2 = -1\) allows us to express the solution as

\[ x = \sqrt{-16} = \sqrt{16i^2} = \pm4i \]

The set represented by \(\mathbb{I}\) defines an imaginary number as

\[ i^2 = -1 \]

Powers of i

If \(i^2 = -1\) then \(i^4 = i^2i^2 = -1 * -1 = 1\)

Therefore we have the sequence

\[ \begin{array}{ccccc} \hline i & i^2 & i^3 & i^4 & i^5 & \ldots \\ \hline i & -1 & -i & 1 & i & \ldots \\ \hline \end{array} \]

Complex numbers

A complex number is just the sum of a real and an imaginary number

\[ z = a + bi \quad a,b \in \mathbb{R}, \quad i^2 = -1 \]

Operations on complex numbers

Given two complex numbers

\[ z_1 = a_1 + b_1i \\ z_2 = a_2 + b_2i \]

Addition and subtraction

\[ z_1 \pm z_2 = a_1 \pm a_2 + (b_1 \pm b_2)i \]

Product

\[ \begin{align*} z_1z_2 &= a_1a_2 + a_1b_2i + a_2b_1i + b_1b_2i^2 \quad \text{given that $i^2 = -1$} \\ &= (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i \end{align*} \]


Given the complex number

\[ z = a + bi \]

Norm (modulus or absolute value)

\[ |z| = \sqrt{a^2 + b^2} \]

Complex conjugate

The product of two complex numbers where the only difference between them is the sign of the imaginary part is

\[ (a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2 \]

This quantity \(a - bi\) is called the complex conjugate of \(z\) (denoted as \(z^*\)), it implies that

\[ zz^* = |z|^2 \]

Inverse

\[ z^{-1} = \frac{1}{z} \]

Multiplying the numerator and denominator with the conjugate of \(z\) (so that we have a real part on the denominator)

\[ z^{-1} = \frac{1}{z} \frac{z^*}{z^*} = \frac{z^*}{zz^*} = \frac{z^*}{|z|^2} \]

Square root of \[i\]

We're trying to find a complex number \(z\) such that

\[ \sqrt{i} = z \\ i = z^2 \]

Assuming that \(z\) is the complex number \(z = a + bi\)

\[ \begin{align} i &= (a + bi)^2 \nonumber \\ &= (a + bi)(a + bi) \nonumber \\ &= a^2 - b^2 + 2abi \label{square-imaginary} \end{align} \]

Therefore

\[ (a^2 - b^2) + (2ab)i = 0 + 1i \]

Equaling real and imaginary parts

\[ \begin{align*} a^2 - b^2 &= 0 \\ 2ab = 1 \end{align*} \]

Therefore \(a = \pm b\), replacing \(a = -b\) in the second equation we obtain \(-2b^2 = 1\) which is not satisfied by any real number \(b\) therefore the case \(a = -b\) is impossible, replacing \(a = b\) in the second equation we obtain \(2a^2 = 1\) so

\[ 2a^2 = 1 \\ a^2 = \frac{1}{2} \\ a = b = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} \]

Finally the value of \(\sqrt{i}\) is

\[ \sqrt{i} = (a + bi) = \pm{\frac{1}{\sqrt{2}}} (1 + i) \]

The value of \(\sqrt{-i}\) is found in the same way (by replacing \(b = -a\) in the equation \(-2ab = 1\) found from multiplying \eqref{square-imaginary} by \(-1\))

\[ \sqrt{-i} = (a + bi) = \pm{\frac{1}{\sqrt{2}}} (1 - i) \]

Matrix representation of a complex number

The matrix \(C\) for a complex number is the sum of two other matrices representing the real \(R\) and imaginary \(I\) parts:

\[ C = R + I \]

which can be written as

\[ C = a \hat{R} + b \hat{I} \quad\quad a, b \in \mathbb{R} \]

Where \(R = 1\) and \(I = i\)

The matrix representation of \(R = 1\) in 2d is the identity matrix

\[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

To find the matrix representation of \(i\) we have to analyze the definition of \(i\) which is a quantity which squares to \(-1\), given that we already know the value of \(1\) in matrix form

\[ \begin{align*} i^2 &= -1 * \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \end{align*} \]

Squaring the following matrix gives the matrix above, then the value of \(i\) expressed in matrix form is

\[ I = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \]

Finally the value of \(C\) is

\[ C = a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \]

The complex plane

The powers of \(i\) give rise to the sequence \((1, i, -1, -i, 1, \ldots)\) which is quite similar to the pattern \((x, y, -x, -y, x, \ldots)\), the resemblance is no coincidence as complex number belong to a 2-dimensional plane, this complex plane allows us to visualize complex numbers using the horizontal axis for the real part and the vertical axis for the imaginary part

$$1, i, -1, -i$$

We can see that the positions of \(i^0 = 1, i^1 = i, i^2 = -1, i^3 = -i, \ldots\) suggest that the multiplication of a complex number by \(i\) is equivalent to rotating through 90 degrees

e.g.

\[ \begin{align*} z_1 &= 2 + i \\ z_2 &= (2 + i)(i) = -1 + 2i \\ z_3 &= (-1 + 2i)(i) = -2 - i \\ z_4 &= (-2 - i)(i) = 1 - 2i \\ z_5 &= (1 - 2i)(i) = 2 + i = z_1 \end{align*} \]

A complex number is rotated \(\pm 90^{\circ}\) by multiplying it by \(\pm i\)

Let's graph the roots of \(\sqrt{i} = \pm \frac{1}{\sqrt{2}} (1 + i)\)

We can see that \(\tfrac{1}{\sqrt{2}} (1 + i)\) is exactly at \(45^{\circ}\) and \(- \tfrac{1}{\sqrt{2}} (1 + i)\) is exactly at \(225^{\circ}\)

Let's multiply the complex number \(2 + i\) by \(\sqrt{i}\) (it should rotate it by \(45^{\circ}\))

\[ \begin{align*} z_1 &= 2 + i \\ z_2 &= (2 + i)(\sqrti + \sqrti i) = \sqrti + 3 \sqrti i \end{align*} \]

Multiplying \(z_2\) by \(\sqrt{i}\) again should be equal to multiplying \(z_1\) by \(i\) (because \(z_2\) is already rotated by \(45^{\circ}\))

\[ \begin{align*} z_2 &= \sqrti + 3 \sqrti i \\ z_3 &= (\sqrti + 3 \sqrti i)(\sqrti + \sqrti i) \\ &= (\frac{1}{2} - \frac{3}{2}) + (\frac{1}{2} + \frac{3}{2})i \\ &= -1 + 2i \end{align*} \]

Which is exactly what we find if we multiply \(z_i\) by \(i\), these observations suggest that we can build a complex number which can rotate another complex number by any angle

A complex number is rotated \(45^{\circ}\) by multiplying it by \(\sqrti + \sqrti i\)

A complex number is rotated \(225^{\circ}\) by multiplying it by \(-\sqrti + \sqrti i\)

Polar representation

Instead of using coordinates in the complex plane we can represent a polar number with the length of the vector from the origin to the complex coordinate and the angle between the complex vector and the positive real axis

\[ r = |z| = \sqrt{a^2 + b^2} \\ \theta = arctan(\frac{b}{a}) \]

The horizontal component of \(z\) is then \(r * cos(\theta)\) and the vectical component is \(r * sin(\theta)\), expressing the complex number using these quantities

\[ \begin{align*} z &= a + bi \\ &= r * \cos \theta + ri\; \sin \theta \\ &= r \; (\cos \theta + i \sin \theta) \end{align*} \]

Euler provided the identity

\[ \begin{equation}\label{rotor} e^{i\theta} = \cos \theta + i \sin \theta \end{equation} \]

Which allows us to represent any complex number as

\[ z = r\,e^{i\theta} \]

Given two polar numbers

\[ z = r\,e^{i\theta} \\ w = s\,e^{i\phi} \\ \]

Their product is

\[ zw = rs\, e^{i(\theta + \phi)} = rs [ \cos (\theta + \phi) + i \sin (\theta + \phi)] \]

Which effectively rotated the complex number \(z\) by \(\phi\) angles! However the quantity \(zw\) was scaled \(s\) units, to avoid scalling we can normalize \(w\) (i.e. making \(r = 1\) which is equal to \eqref{rotor})

A rotor is a complex number that rotates another complex number by an angle \(\theta\) (through multiplication) and has the form

\[ e^{i\theta} = \cos \theta + i \sin \theta \]

Rotating a complex number \(x + yi\) by an angle \(\theta\)

\[ \begin{align*} x' + y'i &= (x + yi)(\cos \theta + i \sin \theta) \\ &= (x \cos \theta - y \sin \theta) + (x \sin \theta + y \cos \theta)i \end{align*} \]

Which in matrix form is

\[ \begin{bmatrix} x' & -y' \\ y' & x' \end{bmatrix} = \begin{bmatrix} x & -y \\ y & x \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \]

Note that because of the way the complex product is defined, the multiplication between two complex numbers commutes

\[ \begin{align*} x' + y'i &= (\cos \theta + i \sin \theta)(x + yi)\\ &= (x \cos \theta - y \sin \theta) + (x \sin \theta + y \cos \theta)i \end{align*} \]