# Write the complex number in trigonometric form? using both degrees and radians 1 − i

May 11, 2018

$1 - i$

$= \sqrt{2} \left(\cos \left(- {45}^{\circ}\right) + i \setminus \sin \left(- {45}^{\circ}\right)\right)$

$= \sqrt{2} \setminus \textrm{c i s} \left(- {45}^{\circ}\right)$

$= \setminus \sqrt{2} \setminus \textrm{c i s} \left(- \frac{\pi}{4}\right)$

#### Explanation:

Almost every trig and trig-like problem involves the 30/60/90 or 45/45/90 triangle. This one is the latter.

$1 - i$ corresponds to the point $\left(1 , - 1\right)$, right there in the fourth quadrant, magnitude $\sqrt{2} ,$ angle $- {45}^{\circ}$.

Trigonometric form is essentially polar form written rectangularly, as $r \left(\cos \theta + i \sin \theta\right) ,$ often conveniently abbreviated $r \setminus \textrm{c i s} \setminus \theta$.

$1 = \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = \sqrt{2} \setminus \cos \left(- {45}^{\circ}\right)$

$- 1 = \sqrt{2} \left(- \frac{1}{\sqrt{2}}\right) = \sqrt{2} \setminus \sin \left(- {45}^{\circ}\right)$

$1 - i = \sqrt{2} \left(\cos \left(- {45}^{\circ}\right) + i \setminus \sin \left(- {45}^{\circ}\right)\right) = \sqrt{2} \setminus \textrm{c i s} \left(- {45}^{\circ}\right)$

Using our goofy system where the circle constant $\pi$ is half a circle, in radians that's just

$1 - i = \setminus \sqrt{2} \setminus \textrm{c i s} \left(- \frac{\pi}{4}\right)$