Fool Proof Mathematics/CP1/Argand diagrams

Considering that a complex number is made of 2 parts, we can represent each complex number as coordinates, or vectors, on 2 planes: a real number, x-axis, and imaginary, y-axis, planes:

Using vector notation the above complex number can be represented as the vector $a \choose b$. Therefore we can add and subtract complex numbers as vectors. Alternatively, we can represent each complex number by the angle it makes with the positive real axis and its' magnitude. Firstly, the angle is known as the argument: Argument Complex Analysis, the principal argument being in the region: $-\pi <\arg z\leq \pi$. The magnitude can be calculated as:$|z|={\sqrt {a^{2}+b^{2}}}$. We can derive the modulus-argument form of a complex number given these facts and using basic trigonometric functions:

$${\begin{aligned}|z|&=r,\arg z=\theta \\z&=a+bi\\\cos \theta &={\dfrac {a}{r}},\sin \theta ={\dfrac {b}{r}}\\z&=r(\cos \theta +i\sin \theta )\end{aligned}}$$

Several identities for complex numbers can be derived by using this form, which are displayed below. Due to the triviality of the derivations they have been omittedt:

$${\begin{aligned}|z_{1}z_{2}|&=|z_{1}||z_{2}|\\\arg(z_{1}z_{2})&=\arg z_{1}+\arg z_{2}\\|{\dfrac {z_{1}}{z_{2}}}|&={\dfrac {|z_{1}|}{|z_{2}|}}\\\arg({\dfrac {z_{1}}{z_{2}}})&=\arg z_{1}-\arg z_{2}\end{aligned}}$$

Worked Examples:

${\begin{aligned}z_{1}=3(\cos {\dfrac {5\pi }{12}}+i\sin {\dfrac {5\pi }{12}})\\z_{2}=4(\cos {\dfrac {\pi }{12}}+i\sin {\dfrac {\pi }{12}})\\\end{aligned}}$

Find $|z_{1}z_{2}|$, $\arg(z_{1}z_{2})$

${\begin{aligned}|z_{1}z_{2}|&=|z_{1}||z_{2}|\\|z_{1}|&=3,|z_{2}|=4\therefore |z_{1}z_{2}|=12\\\arg(z_{1}z_{2})&=\arg z_{1}+\arg z_{2}\\\arg z_{1}&={\dfrac {5\pi }{12}},\arg z_{2}={\dfrac {\pi }{12}}\therefore \arg(z_{1}z_{2})={\dfrac {\pi }{2}}\end{aligned}}$

Hence write $z_{1}z_{2}$in modulus-argument form then Cartesian formː

${\begin{aligned}z_{1}z_{2}&=12(\cos {\dfrac {\pi }{2}}+i\sin {\dfrac {\pi }{2}})\\z_{1}z_{2}&=12(0+i)=12i\end{aligned}}$

We can construct a triangle on the complex plane using 2 complex numbers where ${\textstyle z_{2}-z_{1}}$makes the 3rd length.