How do you write the trigonometric form of -3-i?

Aug 1, 2017

The trigonometric form is $= \sqrt{10} \left(\cos {198.4}^{\circ} + i \sin {198.4}^{\circ}\right)$

Explanation:

Let $z = - 3 - i$

The trigonometris form is

$z = r \left(\cos \theta + i \sin \theta\right)$

If $z = a + i b$

$z = | z | \left(\frac{a}{|} z | + \frac{b}{|} z | i\right)$

The modulus is

$| z | = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{10}$

Therefore,

$z = \sqrt{10} \left(- \frac{3}{\sqrt{10}} - \frac{1}{\sqrt{10}} i\right)$

$\cos \theta = - \frac{3}{\sqrt{10}}$

and

$\sin \theta = - \frac{1}{\sqrt{10}}$

We are in the quadrant $I I I$

$\theta = {198.4}^{\circ}$

Therefore,

$z = \sqrt{10} \left(\cos {198.4}^{\circ} + i \sin {198.4}^{\circ}\right) = {e}^{i {198.4}^{\circ}}$