7. Complex numbers

Part 1.: Problems, solutions.
Part 2.: Problems, solutions.

Complex numbers

To be able to calculate all algebraic operations, we need to be able to calculate roots of negative numbers. Therefore we introduce a new number i, such that i^2=-1.

Complex numbers \mathbb{C} are numbers of form z=a+bi, where a,b\in\mathbb{R}. Reals a and b are called the real and imaginary part of z and we denote them by \text{Re} z and \text{Im} z.

Therefore any complex number a+bi can be shown on a plane as a point with coordinates (a,b). This plane will be usually called the complex plane.

Adding and multiplying

We can add and multiply complex numbers straightforward, but recall that i^2=-1, so: (a+bi)+(c+di)=(a+c)+i(b+d) and (a+bi)(c+di)=(ac-bd)+i(bc+ad).

E.g.: (1+i)(-2-3i)+(-1+2i)=(1-5i)+(-1+2i)=-3i.

Dividing

It is also easy to divide a complex number by a complex number. One shall proceed similarly as when there is an irrational root in the denominator. So:

    \[\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{(ac+bd)+i(bc-ad)}{c^2+d^2}.\]

Polar form

Every non-zero complex number z=a+ib, can be described in a yet another way — it id defined by its distance from zero (its modulus) |z|=\sqrt{a^2+b^2} and the angle between it and the real axis, called the argument \text{Arg} z (\sin \text{Arg} z=\frac{b}{|z|}, \cos \text{Arg} z=\frac{a}{|z|}). Therefore, z=|z|\left(\cos \text{Arg} z+i\sin \text{Arg} x\right).

E.g.: let z=1-i, then |z|=\sqrt{2} and \text{Arg}{z}=\frac{-\pi}{4}. On the other hand, if \text{Arg} x=\frac{-\pi}{6}, |x|=2, then x=2\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)=\sqrt{3}-i.

Geometrical view on addition and multiplication

You can notice immediately, that adding two complex numbers is the same as adding two vectors on the complex plane.

Multiplication is even more interesting. Namely, multiplication of two complex numbers actually multiplies its modules and adds their arguments. This is especially useful when calculating a power of a complex number, and it implies so called de Moivre’s equality:

    \[z^n=|z|^n\left(\cos n\text{Arg} z+i\sin n\text{Arg} z\right).\]

E.g., Let us calculate (1+i)^6(\sqrt{3}-i). We get:

  • modulus of 1+i is \sqrt{2} and argument is \frac{\pi}{4}, so modulus of (1+i)^6 is 8 and argument is \frac{6\pi}{4}, in other words \frac{-\pi}{2}.
  • modulus of \sqrt{3}-i is 2 and argument is \frac{-\pi}{6}, so modulus (1+i)^6(\sqrt{3}-i) is 16 and argument is -\frac{2\pi}{3}.

Therefore (1+i)^6(\sqrt{3}-i)=16\left(\cos \frac{-2\pi}{3}+i\sin \frac{-2\pi}{3}\right)=-8-8i\sqrt{3}.

By the way, de Moivre’s equality implies many trigonometric equalities. E.g., because

    \[(\cos \alpha+i\sin\alpha)^2= \cos 2\alpha +i\sin 2\alpha,\]

and:

    \[(\cos \alpha+i\sin\alpha)^2= \cos^2\alpha-\sin^2\alpha+2i\sin\alpha\cos\alpha,\]

we get that:

    \[\cos 2\alpha =\cos^2\alpha - \sin^2\alpha\]

and

    \[\sin 2\alpha = 2\sin\alpha\cos\alpha.\]

Calculating roots

Notice, that for every complex number z\neq 0 there are always n numbers x such that x^n=z. Indeed the first one can be calculated by calculating the root of the modulus \sqrt[n]{|z|} and by dividing the argument by n. But if we add to this argument any multiple of \frac{2\pi}{n}, then after taking the number to the n-th power we will get the same number, because this additional angle will sum up to some multiple of 2\pi. All those numbers will be called n-th roots of z.

E.g. \sqrt[4]{-16} are 4 numbers. Since |-16|=16, every root will be of modulus 2. Argument 16i is \pi, so the first root will have \frac{\pi}{4} as its argument and we can add to it any multiple of \frac{2\pi}{4}=\frac{\pi}{2}, so we get the following roots:

  • argument: \frac{\pi}{4}, so \sqrt{2}+i\sqrt{2}
  • argument: \frac{3\pi}{4}, so -\sqrt{2}+i\sqrt{2},
  • argument: \frac{-3\pi}{4}, so -\sqrt{2}-i\sqrt{2},
  • argument: \frac{-\pi}{4}, so \sqrt{2}-i\sqrt{2}.

Those root of z, argument of which is in [0,\frac{2\pi}{n}), is called the main root and denoted by \sqrt[n]{_+ z}. So \sqrt[4]{_+ -16}=\sqrt{2}+i\sqrt{2}.

Therefore we can calculated roots of a quadratic equations regardless of the sign of \Delta, or even when \Delta is not real. E.g. x^2+2x+(1-i), \Delta=4i, \sqrt{\Delta} has modulus 2 and argument \frac{\pi}{4}, so it is \sqrt{2}+i\sqrt{2}. And -\sqrt{2}-i\sqrt{2}. Therefore, x_1=-1+\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} and x_2=-1-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}.

Exponential form and calculating complex powers

We will see that for x\in\mathbb{R}, we have: e^x=\sum_{i=0}^\infty \frac{x^n}{n!}. This gives us an idea how to define complex powers. If we substitute i to x we get e^{i}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}, moreover, \sin x= \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!} and \cos x= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}. Therefore, we can define e to a complex power as follows:

    \[e^{a+bi}=e^a(\cos b+i\sin b),\]

Which implies the famous equation of the 5 constants of mathematics e^{i\pi}+1=0. 🙂

And we can write any complex number in an exponential form, which is similar to the polar form. Namely, z=e^{\ln|z|+i\text{Arg} z}.