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 $a,b\in \mathbb {R}$. $bi$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 ${\textstyle z^{2}+9=0}$ː$${\begin{aligned}z^{2}&=-9\\z&=\pm {\sqrt {-9}}=\pm {\sqrt {-1}}{\sqrt {9}}=\pm 3i\end{aligned}}$$

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$${\begin{aligned}z=a+bi\\z^{*}=a-bi\\zz^{*}=(a+bi)(a-bi)=a^{2}-b^{2}i^{2}=a^{2}+b^{2}\end{aligned}}$$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: ${\dfrac {z_{1}}{z_{2}}}\times {\dfrac {z_{2}^{*}}{z_{2}^{*}}}$.