# How do you divide ( -3i+2) / (-i -1 ) in trigonometric form?

Jul 1, 2018

$z = \frac{\sqrt{26}}{2} \left(\cos \left({78.69}^{\circ}\right) + i \sin \left({78.69}^{\circ}\right)\right)$ or
$z = \frac{5}{2} i + \frac{1}{2}$

#### Explanation:

Let's start by solving this in the form given and see what we get:

Multiply the conjugate of the denominator:
$\frac{- 3 i + 2}{- 1 - i} \cdot \frac{- 1 + i}{- 1 + i} = \frac{3 i - 2 - 3 {i}^{2} + 2 i}{1 - {i}^{2}} = \frac{5 i + 1}{2} = \frac{5}{2} i + \frac{1}{2}$

$\frac{1}{2} + \frac{5}{2} i$
We can convert this to trigonometric form by finding $r$:
$r = \sqrt{{a}^{2} + {b}^{2}}$
$\theta = \arctan \left(\frac{b}{a}\right)$

$r = \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{5}{2}\right)}^{2}}$

$r = \frac{\sqrt{26}}{2}$

$\theta = \arctan \left(5\right)$
$\theta \approx {78.69}^{\circ}$

So we can say the solution in trig form is
$z = \frac{\sqrt{26}}{2} \left(\cos \left({78.69}^{\circ}\right) + i \sin \left({78.69}^{\circ}\right)\right)$

Let's double check by solving in trigonometric form:
${z}_{1} = 2 - 3 i$
$r = \sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2}}$
$r = \sqrt{13}$

$\theta = \arctan \left(- \frac{3}{2}\right) \approx - {56.31}^{\circ} \approx {303.69}^{\circ}$

${z}_{2} = - 1 - i$
$r = \sqrt{{\left(- 1\right)}^{2} + {\left(- 1\right)}^{2}}$
$r = \sqrt{2}$
$\theta = \arctan \left(- \frac{1}{-} 1\right) = {45}^{\circ}$, however this in the third quadrant so$+ {180}^{\circ} = {225}^{\circ}$

z_1/z_2= (sqrt13(cos(303.69^@)+isin(303.69^@)))/(sqrt2(cos(225^@)+isin(225^@))

WHEN DIVIDING IN TRIGONOMETRIC FORM:
Divide the r by the other r
And subtract the angles

$\frac{\sqrt{13}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{26}}{2}$

${303.69}^{\circ} - {225}^{\circ} = {78.69}^{\circ}$

So same solution:
$z = \frac{\sqrt{26}}{2} \left(\cos \left({78.69}^{\circ}\right) + i \sin \left({78.69}^{\circ}\right)\right)$