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

1. $i^3=i^{2}i=-i$
2. Solve the equation $z^2+9=0$ː$$\begin{array}{cc}{z}^2& =-9\\ z& =\pm \sqrt{-9}=\pm \sqrt{-1}\sqrt{9}=\pm 3i\end{array}$$

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$$\begin{array}{c}z=a+bi\\ z^{*}=a-bi\\ z{z}^{*}=(a+bi)(a-bi)=a^2-b^2{i}^2=a^2+b^2\end{array}$$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: $\frac{z_1}{z_2}\times \frac{z_2^{*}}{z_2^{*}}$.