# How do you write the complex number in trigonometric form -2?

Dec 28, 2016

$2 \left(\cos \left(\pi\right) + i \sin \left(\pi\right)\right)$

#### Explanation:

$2 = 2 + i \cdot 0 \mathmr{and} \textcolor{red}{2} \left(\textcolor{b l u e}{1} + i \cdot \textcolor{b l u e}{0}\right)$

The general trigonometric for a complex number is
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{r} \left(\cos \left(\textcolor{b l u e}{\pi}\right) + i \cdot \sin \left(\textcolor{b l u e}{\pi}\right)\right)$

So we need a value $\textcolor{b l u e}{\theta}$
such that
$\textcolor{w h i t e}{\text{XXX")cos(color(blue)theta)=1color(white)("XX")andcolor(white)("XX}} \sin \left(\textcolor{b l u e}{\theta}\right) = \textcolor{b l u e}{0}$

The simplest solution that meets these requirements is $\textcolor{b l u e}{\theta} = \textcolor{b l u e}{\pi}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

To be complete, we should possibly note that any
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{\theta} = \textcolor{b l u e}{\pi} + k 2 \pi , k \in \mathbb{Z}$
would also meet the stated requirement
and the "complete" answer would be
color(white)("XXX")2(cos(pi+k2pi)+isin(pi+k2pi))), k in ZZ

While this is correct, I personally find that it obscures the significant components of the answer.