Fool Proof Mathematics/CP1/Complex numbers

Consider a mathematics which only allows for positive numbers. This drastically restricts the solutions obtainable such as being unable to answer "What is $$5 + x = -17$$?". A similar problem arises with polynomials, as without expanding our system of numbers some polynomials have no solution. A complex number encapsulates all real numbers as well as introducing the imaginary unit, $$i = \sqrt{-1}$$, such that a complex number has 2 components:

$$z = a + bi$$where $$a,b \in \R$$. $$bi$$is an imaginary number. Two functions are introduced to distinguish between the 2 componentsː $$Re(z) = a, Im(z) = b$$. Addition and subtraction of complex numbers works the same way as algebraic and root manipulation, the only difference to bear in mind is thatː $$i^2 = -1$$which is a real number, allowing further simplification.

As previously alluded to, quadratics now have solutions for when the discriminant is less than 0ː $$b^2 - 4ac < 0$$means the quadratic has 2 distinct complex solutions.

Worked Examples
$ ː$$\begin{align} z^2 &= -9 \\ z &= \pm \sqrt{-9} = \pm \sqrt{-1}\sqrt{9} = \pm 3i \end{align} $$
 * 1) $$i^3 = i^2 i = -i$$
 * 2) Solve the equation $z^2 + 9 = 0

The complex conjugate of a complex number has the same real part but the inverse of its' imaginary part. This allows the us to utilize the difference of two squares identity, making the product of 2 complex numbers realː$$\begin{align} z = a + bi \\ z^* = a - bi \\ z z^* = (a + bi)(a - bi) = a^2 - b^2 i^2 = a^2 + b^2 \end{align}$$We can use an already learnt trick of rationalizing the denominator to divide 2 complex numbers by multiplying the fraction by the denominators' complex conjugate, i.e: $$\dfrac{z_1}{z_2} \times \dfrac{z_2^*}{z_2^*}$$.