How do you write 1-i in trigonometric form?

Oct 23, 2017

$\sqrt{2} \cdot c i s \left(\frac{7 \pi}{4}\right)$

Explanation:

First, recall the basic conversion formulas to convert from a complex number in $a + b i$ format into trigonometric $r \cdot c i s \theta$ format. These come mainly from the rectangular to polar conversion methodology:

$r = \sqrt{{a}^{2} + {b}^{2}}$
 tan theta_{ref} = |b/a| color(white)("aaaaaa")"Reference Angle"

In this problem, $a = 1$ and $b = - 1$. We can find $r$ readily:

$r = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{1}^{2} + {\left(- 1\right)}^{2}} = \sqrt{1 + 1} = \sqrt{2}$

To find the proper $\theta$ angle, we begin with finding the reference angle ${\theta}_{r e f}$ first:

$\tan {\theta}_{r e f} = | \frac{- 1}{1} |$
$\tan {\theta}_{r e f} = 1$
${\theta}_{r e f} = \frac{\pi}{4}$

(This could have been readily apparent by noticing that both $a$ and $b$ are the same absolute value (1).)

If we consider that the complex number $1 - i$ is graphed in Quadrant IV of the complex plane, and we recall how to use reference angles, we can see that the proper $\theta$ in this case is:

$\theta = 2 \pi - \frac{\pi}{4} = \frac{7 \pi}{4}$

Thus, the complex number $1 - i$ in trigonometric form is $\sqrt{2} \cdot c i s \frac{7 \pi}{4}$, also written as $\sqrt{2} \left(\cos \left(\frac{7 \pi}{4}\right) + i \cdot \sin \left(\frac{7 \pi}{4}\right)\right)$