You have no doubt heard about the complex numbers $\mathbb{C}$. The word “complex” is an intimidating word. Surely it must be a complex task to learn about the complex numbers. That may be true in general, but not if you know about vectors. Complex numbers are similar to two-dimensional vectors $\vec{v} \in \mathbb{R}^2$. We add and subtract complex numbers like vectors. Complex numbers also have components, length and “direction”. If you understand vectors then you will understand complex numbers at almost no additional mental cost.

Suppose you are asked to solve the following quadratic equation: $$x^2 + 1 = 0.$$ You are looking for a number $x$, such that $x^2=-1$ so that adding one to $x^2$ we get zero. If you are only allowed to give real answers (the set of real numbers is denoted $\mathbb{R}$), then there is no answer to this question. In other words, this equation has no solution. This is because the quadratic function $f(x)=x^2 + 1$ does not cross the $x$ axis.

We are not going to take that though. We will imagine a new number called $i$ which fits this requirement. By definition $i^2=-1$. And we call $i$ the unit imaginary number. The solutions to the equation are going to be $x=i$ and $x=-i$. Remember that a quadratic equation has two solutions. We can check $i^2 + 1 = -1 +1 = 0$ and also $(-i)^2 +1 = (-1)^2i^2 + 1 = i^2 +1 = 0$.

• $i$: The unit imaginary number $i \equiv \sqrt{-1}$ or $i^2 = -1$.
• $bi$: An imaginary number that is equal to $b$ times $i$.
• $\mathbb{R}$: The set of real numbers.
• $\mathbb{C}$: The set of complex numbers $\mathbb{C} = \{ a + bi \ | \ a,b \in \mathbb{R} \}$.
• $z=a+bi$: A complex number (has a real part and an imaginary part).
• $\textrm{Re}\{ z \}=a$: The real part of $z$.
• $\textrm{Im}\{ z \}=b$: The imaginary part of $z$.
• $\bar{z}$: The complex conjugate of $z$. If $z=a+bi$, then $\bar{z}=a-bi$.

NOINDENT When using the polar representation of complex numbers we have:

• $z= |z|\angle \phi_z= |z|\cos\phi_z + i|z|\sin\phi_z$.
• $|z|=\sqrt{ \bar{z}z }=\sqrt{a^2+b^2}$: The magnitude of $z=a+bi$.
• $\phi_z=\tan^{-1}(b/a)$: The phase of $z=a+bi$.
• $\textrm{Re}\{ z \} = |z|\cos(\phi_z)$.
• $\textrm{Im}\{ z \} = |z|\sin(\phi_z)$.

Just like the addition of vectors is done component wise, the addition of complex numbers is done reals-part-with-real-part and imaginary-part-with-imaginary-part: $$(a+bi) + (c+di) = (a+c) + (b+d)i.$$

A geometrical interpretation of the complex numbers is to extend the real number line that stretches from $-\infty$ to $\infty$ into a two-dimensional plane. The horizontal axis (where the $x$-axis is usually) will measure the real part of the number. The vertical axis will measure the imaginary component. Complex numbers are vectors in the complex plane.

It is possible to represent any complex number $z=a+bi$ in terms of length and direction notation: $$z= |z|\angle \phi_z = (|z|\cos\phi_z) + (|z|\sin\phi_z)i.$$

The magnitude of a complex number $z=a+bi$ is $$|z|=\sqrt{a^2+b^2}.$$ This corresponds to the distance of from the origin. The formula is obtained by using Pythagoras theorem.

The phase of the complex number is $$\phi_z=\tan^{-1}(b/a).$$ This corresponds to the angle that $z$ makes with the real axis.

The product of two complex numbers is computed using the usual rules of algebra: $$(a+bi)(c+di) = (ac - bd) + (ad + bc)i.$$ In the polar representation, the product is $$(p\angle \phi)(q\angle \psi) = pq \angle (\phi + \psi).$$

One of the first examples of reasoning using complex numbers was given by Jerome Cardan in his Ars Magna. “If someone says to you, divide 10 into two parts, one of which multiplied into the other shall produce 40, it is evident that this case or question is impossible.” What is required is to find wo numbers $x_1$ and $x_2$ such that $x_1+x_2=10$ and $x_1x_2=40$. This sounds kind of impossible. Or is it?

“Nevertheless,” Cardan says, “we shall solve it in this fashion: $$x_1 = 5 + \sqrt{15}i, \ \ \text{and} \ \ x_2 = 5 - \sqrt{15}i.$$ When you add $x_1 + x_2$ you get 10, and when you multiply them you get $$x_1x_2=\left(5 + \sqrt{15}i\right)\left(5 - \sqrt{15}i\right) = 25 - \sqrt{15}^2i^2 = 25 + 15 = 40.$$ Hence this product is 40.”[1]

Both $i$ and $-1$ have a magnitude of $1$, but $i$ has an argument of $\pi/2$ ($90^\circ$) while $-1$ has an argument of $\pi$ ($180^\circ$). $$(i)(-1) = (1\angle \frac{\pi}{2})(1 \angle \pi) = 1 \angle \frac{3\pi}{2} = -i.$$ Effectively, multiplication by $i$ is like rotation by $90$ degrees leftward.

Let me now show you the procedure for dividing complex numbers: $$\frac{(a+bi)}{(c+di)} = \frac{(a+bi)}{(c+di)}\frac{(c-di)}{(c-di)} = (a+bi)\frac{(c-di)}{(c^2+d^2)} = (a+bi)\frac{\overline{c+di}}{|c+di|^2}.$$ In other words, if you want to divide the number $z$ by the complex number $s$, you should compute $\bar{s}$ and $|s|^2=s\bar{s}$ and then use: $$z/s = z\frac{ \bar{s} }{ |s|^2 }.$$ You can think of $\frac{ \bar{s} }{ |s|^2 }$ as being equivalent to $s^{-1}$.

The solutions to any polynomial equation $a_0 + a_1x + a_2x^2 + \cdots a_nx^n=0$ are of the form: $$z=a+bi.$$ In other words, any polynomial $P(x)$ of $n^\textrm{th}$ degree can be written as $$P(x) = (x - z_1)(x - z_2) \cdots (x-z_n),$$ where $z_i \in \mathbb{C}$ are its complex roots.

Before today, you would say that the equation $x^2 + 1 =0$ had no solutions. Now you know that actually it has solutions, but the solutions are complex numbers: $x_1=i$ and $x_2=-i$.

We know that $\sin(\theta)$ is just a shifted version of $\cos(\theta)$, so clearly these two functions are related. The exponential function, however, seems kind of unrelated to $\sin$ and $\cos$. Lo and behold Euler's formula: $$\exp(i\theta) \equiv e^{i\theta} = \cos(\theta) + i\sin(\theta),$$ where $i = \sqrt{-1}$ is the unit imaginary number. An imaginary input number to the exponential function produces a complex number as output which contains both $\cos$ and $\sin$. Euler's formula gives us a more natural notation for the polar representation of complex numbers: $z=|z|\angle\phi_z = |z|e^{i\phi_z}$

If you want to impress your friends with you math knowledge, you can plug $\theta=\pi$ into the above equation to get $$\exp(i\pi) = \cos(\pi) + i\sin(\pi)= -1,$$ which can be rearranged into this form: $e^{\pi i} + 1 = 0$. This equation shows a relationship between the five most important numbers in all of mathematics: Euler's number $e=2.71828...$, $\pi=3.14…$, the imaginary number $i$, one and zero.

We can replace $\theta$ in Euler's formula with $n\theta$, we obtain de Moivre's theorem: $$\left( \cos \theta + i \sin \theta \right)^n = \cos n\theta + i \sin n\theta.$$

De Moivre's Theorem seems obvious if you think of the multiplication law in the polar representation of the complex number as $z=|z|e^{i\theta}$ which is simply raised to the $n$th power $(\cos \theta + i \sin \theta)^n=z^n = (|z|e^{i\theta})^n = |z|^ne^{in\theta}=\cos n\theta + i \sin n\theta$.

Using $n=2$ in de Moivre's formula, we can derive the double angle formulas as the real an imaginary part of the following equation: $$(\cos^2 \theta - \sin^2 \theta) + (2\sin \theta \cos \theta )i = \cos(2\theta) + \sin(2\theta)i.$$

[ Mini tutorial ]
http://paste.lisp.org/display/133628

[ Pretty pictures of the Mandelbrot set ]
https://christopherolah.wordpress.com/2011/08/08/the-real-3d-mandelbrot-set/

NOINDENT [1] Girolamo Cardan, The Great Art or The Rules of Algebra, trans. and ed. by T. Richard Widmer (Cambridge: Massachusetts Institute of Technology Press, 1968), pp. 219–20.