Arithmetic with complex number

The motivating factor for the use of complex numbers is that they form an algebraically closed field. In simpler language any algebraic operation that is proformed on a complex number gives a complex number. This is not true for real numbers, for example, $x^{2}+1=0$ is an entirly real expression but $x$ cannot be a real number. The solution is $x=\pm i$.

For this we will consider two complex numbers, $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$, where $x_{1},x_{2},y_{1},y_{2}\in \mathbb {R}$ and $i^{2}=-1$.

Addition

$z_{1}+z_{2}=(x_{1}+x_{2})+i(y_{1}+y_{2})$

Multiplication

$z_{1}z_{2}=(x_{1}+iy_{1})(x_{2}+iy_{2})=x_{1}x_{2}+ix_{1}y_{2}+iy_{1}x_{2}+i^{2}y_{1}y_{2}=(x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+y_{1}x_{2})$

Division

${\frac {z_{1}}{z_{2}}}={\frac {x_{1}+iy_{1}}{x_{2}+iy_{2}}}={\frac {x_{1}+iy_{1}}{x_{2}+iy_{2}}}\cdot {\frac {x_{2}-iy_{2}}{x_{2}-iy_{2}}}={\frac {(x_{1}x_{2}+y_{1}y_{2})-i(x_{1}y_{2}-y_{1}x_{2})}{(x_{2})^{2}+(y_{2})^{2}}}$

Here we see two important number. The complex conjugate, ${\bar {z}}={\overline {x+iy}}=x-iy$ and the length of a complex number ${||z||}^{2}=z.{\bar {z}}=x^{2}+y^{2}.$