Introducing Complex Numbers

Imaginary Unit Number and Complex plane

Define the imaginary unit number (Euler's notation) as \(i^2 = -1\). Then a complex number is an expression of the form \(z=x+iy\) where

\(\text{Re}(z):=x\) is the real part of \(z\)
\(\text{Im}(z):=y\) is the imaginary part of \(z\).

If \(y=0\), then \(z\) is real, if \(x=0\), then \(z\) is pure imaginary.

Therefore, we can define the set of complex numbers as

\[\mathbb C = \{x+iy:x,y\in\mathbb R\}\]

Geometrically, each complex number can be placed on a 2-D coordinate system \((x, y)\), such system is defined as a complex plane.

Note that \(z\) can also be represented by the polar coordinate

\[x = r\cos\theta, y = r\sin\theta\]

so that the complex number \(z\) can also reside in the polar coordinate

\[z = x+iy = r(\cos\theta+i\sin\theta)\]

And we can define the modulus of \(z\) as its absolute value

\[|z|:= r = \sqrt{x^2+y^2}\]

and the argument of \(z\) is defined by its angle when \(z\neq 0\)

\[\arg z := \theta = \arctan(y/x), z\neq 0\]

Note that \(\theta\) is periodic with period \(\pi\), which means it's multivalued.

Geometry of Arithmetic

Using the complex plane, we can better understand the "addition" and "multiplication". Let \(z, w\) be two complex variables. Then, addition is the addition of their vectors in the complex plane.

Multiple \(z\) by \(i\) will rotate \(z\) counterclockwise by 90 degrees.

\[iz = i(a+ib) = -b + ai\]

Multiple \(z\) by \(c\in\mathbb R\) simply scales the vector.

\[cz = ca+icb\]

Multiple \(z=a+bi, w=c+di\in\mathbb C\) gives

\[zw = (a+bi)w = aw+biw\]

Geometrically, it scales \(z\) by \(|w|\) and rotates from \(w\) counterclockwise for \(\theta\).

Polar Exponential

Define the polar exponential by

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

So that \(z\) can be written in the polar exponential form as

\[z = r(\cos\theta+i\sin\theta) = re^{i\theta}\]

Note that this notation is so convenient as we can prove exponential properties from trigonometric identities.

Theorem

\[\begin{align*} e^{i\theta_1} e^{i\theta_2} &= (\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2)\\ &= \cos\theta_1\cos\theta_2+i(\cos\theta_1\sin\theta_2 + \cos\theta_2\sin\theta_1) + i^2\sin\theta_1\sin\theta_2\\ &= (cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) + i\sin(\theta_1+\theta_2) + (i^2+1)\sin\theta_1\sin\theta_2\\ &= \cos(\theta_1 + \theta_2) + i\sin(\theta_1+\theta_2) + 0\\ &= e^{(\theta_1+\theta_2)i}\\ (e^{i\theta})^m &= e^{im\theta}\\ (e^{i\theta})^{1/n} &= e^{\frac{i\theta}n} \end{align*}\]

Example complex number to polar exponential form (since \(\theta\) is periodic, we will only take values in \(0<\theta\leq 2\pi\))

\[\begin{align*} 1 &\Rightarrow r = \sqrt{1^2+0^2} = 1, \theta = \arctan(0) = 2\pi \Rightarrow e^{i2\pi}\\ -i &\Rightarrow r = \sqrt{0^2+(-1)^2} = 1, \theta = \arctan(-\infty) = \frac{3\pi}2 \Rightarrow e^{i\frac32\pi}\\ 1+i &\Rightarrow r = \sqrt{1^2+1^2} = \sqrt2, \theta = \arctan(1) = \frac{\pi}4 \Rightarrow \sqrt2e^{i\frac\pi4}\\ \frac12+\frac{\sqrt3}2i &\Rightarrow r = \sqrt{(1/2)^2+(\sqrt{3}/2)^2} = 1, \theta = \arctan(\sqrt3) = \frac{\pi}3 \Rightarrow e^{i\frac\pi3}\\ \frac12-\frac{\sqrt3}2i &\Rightarrow r = \sqrt{(1/2)^2+(-\sqrt{3}/2)^2} = 1, \theta = \arctan(-\sqrt3) = \frac{\pi}3 \Rightarrow e^{i\frac{5\pi}3}\\ \end{align*}\]

Example polar exponential form to \(x+yi\)

\[\begin{align*} e^{2+i\pi/2}&\Rightarrow x^2 + y^2 = e^4, \frac{y}x=\tan(\frac{\pi}2)=\infty\Rightarrow e^2i\\ &\text{or} \Rightarrow e^2e^{i\frac\pi2} = e^2i\\ (1+i)^{-1}&= (e^{i2\pi} + e^{i\pi/2})^{-1} = e^{-i\frac52\pi}=e^{i\frac32\pi} = -i\\ (1+i)^3 &= e^{i\frac{15}{2}\pi} = e^{i\frac32\pi} = -i\\ |3+4i| &= \sqrt{3^2+4^2} = 5\\ \end{align*}\]

Example Define \(\cos z = \frac{e^{iz} + e^{-iz}}2, e^z = e^{x}e^{iy}\), evaluate \(\cos(c+i\frac{\pi}4)\) where \(c\in\mathbb R\)

\[\begin{align*} \cos(c+i\frac{\pi}4) &= \frac12\big(\exp(i(c+i\frac\pi4)) + \exp(-i(c+i\frac\pi4))\big)\\ &= \frac{1}2\big(e^{ic + e^{i^2}\frac{\pi}4} + e^{-ic-i^2\frac\pi4}\big)\\ &= \frac12\big(e^{ic-\pi/4} + e^{\pi/4 - ic}\big)\\ &= \frac12\big(e^{ic}e^{-\pi/4} + e^{-ic}e^{\pi/4}\big)\\ &= \frac12\big((\cos c+i\sin c)e^{-\pi/4} + (\cos(-c) +i \sin(-c))e^{\pi/4}\big)\\ &= \frac12\big(e^{-\pi/4}\cos c+ e^{\pi/4}\cos(-c)\big)+ i\frac12\big(e^{-\pi/4}\sin c + e^{\pi/4}\sin(-c)\big) \end{align*}\]

Powers of Complex Numbers

Consider an equation of the form for \(z, w\in\mathbb C, n\in\mathbb N^+\)

\[z^n = w = |w|e^{i\theta} = |w|e^{i(\theta + 2\pi m)}, m\in\mathbb N\]

so that

\[z = |w|^{1/n} e^{i(\theta + 2\pi m)/n}, m\in\mathbb N\]

Since the period is \(2\pi\), this will yield \(n\) unique roots.

Example

\[\begin{align*} z^3 &= 4 \\ z^3 &= 4e^{i2\pi} \\ z &= \{4^{1/3}e^{i\frac{2\pi}3}. 4^{1/3}e^{i\frac{4\pi}3}, 4^{1/3}e^{i2\pi}\} \end{align*}\]

\[\begin{align*} z^4 &= -1 \\ z^4 &= e^{i\pi} \\ z &= \{e^{i\frac{\pi}{4}}, e^{i\frac{3\pi}{4}}, e^{i\frac{5\pi}{4}}, e^{i\frac{7\pi}{4}}\} \end{align*}\]

\[\begin{align*} (az+b)^3 &= c;\:a,b,c \in\mathbb R^+\\ (az+b)^3 &= |c|e^{i2\pi}\\ az+b &= \{c^{1/3}e^{i\frac{2k\pi}{3}}: k = 0, 1, 2\}\\ z &= \{\frac{1}a(c^{1/3}e^{i\frac{2k\pi}{3}} -b): k = 0, 1, 2\} \end{align*}\]

\[\begin{align*} z^4+2z^2+2 =0\\ (z^2+1)^2&=-1\\ z^2+1 & =\pm i\\ z^2_1 &= -1 + i\\ z_1^2 &= 2^{1/2}e^{i\frac{3\pi}4}\\ z_1 &= \{2^{1/4}e^{i\frac{3\pi}{8}}, 2^{1/4}e^{i\frac{11\pi}{8}} \}\\ z_2^2 &= -1-i\\ z_2 &= \{2^{1/4}e^{i\frac{\pi}{8}}, 2^{1/4}e^{i\frac{9\pi}{8}} \}\\ \Rightarrow z &= \{2^{1/4}e^{i\frac{\pi}8}, 2^{1/4}e^{i\frac{3\pi}8}, 2^{1/4}e^{i\frac{9\pi}8}, 2^{1/4}e^{i\frac{11\pi}8}\} \end{align*}\]

Arithmetic Operations

Equivalence \(z_1 = z_2\) IFF their real and imaginary parts are respectively equal, i.e.

\[z_1 x_1+iy_1 = x_2 + iy_2=z_2 \iff x_1=x_2\land y_1=y_2\]

Also note that given the definition as \(\mathbb C = \{a+ib:a,b\in\mathbb R\}\), Addition, subtraction, multiplication, division follows the same rules as real numbers. and the commutative, associative, distributive laws of addition and multiplication hold.

Complex Conjugate

For \(z = x+iy =re^{i\theta}\), define its complex conjugate as \(\bar z = x - iy = re^{-i\theta}\). Consequently, we can have the following results,

\[z\bar z = \bar z z= (x+iy)(x-iy) =|z|^2\]

and the division can be better viewed as

\[\begin{align*} \frac{z_1}{z_2} &= \frac{z_1\bar{z_2}}{z_2\bar{z_2}} = \frac{z_1\bar{z_2}}{|z_2|^2} = \frac{x_1x_2+y_1y_2}{x_2^2+y_2^2} + i\frac{x_2y_1-x_1y_2}{x_2^2+y_2^2} \end{align*}\]

Results of Elementary Functions

Let \(z=a+ib, w =c+id\in\mathbb C\),

Theorem \(\overline{z+w} = \bar z + \bar w\)
proof \(z+w=(a+c)+i(b+d), \overline{z+w}= (a+c) -i(b+d) = (a-ib)+(c-id) = \bar z + \bar w\)

Theorem \(|z-w| \leq |z|+|w|\)
proof \(|z-w|^2 = (a-c)^2 + (b-d)^2 \leq a^2+b^2 + c^2 + d^2 = |z|^2 + |w|^2\),
since the absolute values are all real numbers, it follows triangle inequality that \(|z|^2 + |w|^2 \leq (|z|+|w|)^2\)
Given \(|z-w| \geq 0, |z|+|w|\geq 0, |z-w|\leq |z|+|w|\)

Theorem \(z-\bar z = 2i Im(z)\)
proof \(z-\bar z = a+ib-(a-ib) = 2ib\)

Theorem \(Re(z)\leq |z|\)
proof \(a \leq \sqrt{a^2+b^2}\)

Theorem \(|wz| = |w||z|\)
proof We will equivalently prove \(|wz|^2 = |w|^2|z|^2\)
\(|wz|^2 = (ac-bd)^2 + (ad+bc)^2 = \cdots = (a^2+b^2)(c^2+d^2)\)

Triangular Inequality

\[|z+w|\leq |z|+|w|\]

proof First note that

\[|z+w|^2 = (z+w)(\bar z+ \bar w) = |z|^2+|w|^2 + z\bar w + \bar z w\]

Then,

\[\begin{align*} z\bar w + \bar z w &= (ac+bd) + i(ad-bc) + (ac+bd) + i(bc-ad) \\&= 2(ac+bd)\\ &= 2Re(z\bar w) \end{align*}\]

Using the results above,

\[Re(z\bar w) \leq |z\bar w | = |z||w|\]

So that \(|z+w|^2 \leq |z|^2 + 2|z||w| + |w|^2 = (|z| + |w|)^2\), and take a square root, we have the right inequality.

Note that the reverse triangular inequality is a corollary of triangular inequality, which

\[||z|-|w||\leq |z+w|\leq |z|+|w|\]

Theorem \(|w\bar z + \bar w z| \leq 2|wz|\)
proof First note that \(|z| = |\bar z|\) as \(a^2+b^2 = a^2 + (-b)^2\), then by trig-inequality and the theorem above,
\(|w\bar z + \bar w z| \leq |w\bar z| + |\bar w z| = 2|w||z| = 2|wz|\)