# π Complex Numbers

The emoji for this section is a helicopter, because helicopters rotate, which we will see is the way to understand multiplication by complex numbers.

Complex Number Intuitive Introduction: Take the equation y=aΓx^n we want to be able to solve ALL variants of this equation, no exceptions. To βSolveβ an equation means undoing mathematical steps (βperform the right sequence of INVERSE OPERATIONSβ), to arrive at a number (might need to AUGMENT OUR NUMBER SET to have a place to land).

$\begin{array} {lrclccccll}\mathbb{N} & y & = & 2\times(8)^3 & & & y & = & \phantom{-}1024 & \\ \mathbb{Z} & y & = & 2\times(-8)^3 & & & y & = & -1024 & \text{Inverse of }+\text{ requires Negative Numbers} \\ \mathbb{Q} & y & = & 2\times(-8)^{-3} & & & y & = & -\frac{1}{512} & \text{Inverse of }Γ\text{ requires Rational Numbers} \\ \mathbb{R} & y & = & 2\times(8)^{\frac{1}{6}} & & & y & = & 2\sqrt{2} & \text{Inverse of }\hat{} \: \text{ requires Real Numbers} \\ \mathbb{C} & -2 & = & 2\times(x)^2 & & & y & = & \sqrt{-1} & \sqrt{\text{Negatives}} \text{ requires Complex Numbers} \end{array}$

We can take radicals of whole numbers, fractions, and other radicals, so why keep negatives as an exception? If we give ourselves permission to use a mathematical object with the new property iΒ²=-1, then whenever we happen to get radicals of negatives, we re-write them using βiβ for example β5+β-18 = β5+3β2Γi (from now on itβll just be written 3β2i , but be clear that βiβ isnβt inside the square root, 3β2 is a coefficient multiplying the front of βiβ).

Complex Number Arithmetic radicals of positives: We are used to keeping radical parts separate from non-radical parts e.g. (2+β3)(3-β3)+β3 = 6-2β3+3β3-3+β3 = 3+2β3 Notice that if it was (a+b)(a-b)=aΒ²+bΒ² then the βββ would eliminate entirely

Complex Number Arithmetic radicals of negatives: Keep the βiβ parts separate e.g. (2+β[-3])(3-β[-3])+β3 = (2+β3i)(3-β3i)+β3 = 6-2β3i+3β3i+3+β3 = 9+β3+β3i Notice that if it was (a+b)(a-b)=aΒ²+bΒ² then the βββ would eliminate entirely

Complex Solutions to $$f(z) = az^2+bz+c$$

Inputs and outputs for complex functions: $$\; e^z, z^2, \sqrt{z}, z^4$$