# How do you divide ( 12i -3) / ( 7 i -2 ) in trigonometric form?

Mar 11, 2018

color(red)(90/53-(3i)/53 = (3/53)sqrt(901){cos(-a tan(1/30)+2pi}+i sin{-a tan(1/30)+2pi}

#### Explanation:

Given:

Divide the complex numbers in trigonometric form:

color(blue)((12i-3)/(7i-2)

Rewrite in color(green)(z=(a+bi) form:

color(blue)((-3+12i)/(-2+7i)

Multiply and divide by the complex conjugate of the denominator

$\Rightarrow \frac{- 3 + 12 i}{- 2 + 7 i} \cdot \frac{- 2 - 7 i}{- 2 - 7 i}$

rArr {(-3+12i)(-2-7i)}/{(-2+7i)(-2-7i)

$\Rightarrow \frac{\left(- 3 + 12 i\right) \left(- 2 - 7 i\right)}{{\left(- 2\right)}^{2} - {\left(7 i\right)}^{2}}$

Note that $i = \sqrt{- 1} \mathmr{and} {i}^{2} = \left(- 1\right)$ in complex number arithmetic.

rArr (6+21i-24i-84i^2)/(4-(49i^2)

$\Rightarrow \frac{90 - 3 i}{4 - 49 \cdot \left(- 1\right)}$

$\Rightarrow \frac{90 - 3 i}{4 + 49}$

$\Rightarrow \frac{90 - 3 i}{53}$

$\Rightarrow \frac{90}{53} - \frac{3}{53} i$

Next, we will find the Polar form

Important:

For any complex number color(blue)(a+bi,

Polar form is given by

color(blue)(r[cos(theta)+i sin(theta)], where

color(brown)(r = sqrt(a^2+b^2), and

color(brown)(theta = tan^(-1)(b/a)

We have, for $a + b i$

$a = \left(\frac{90}{53}\right) \mathmr{and} b = \left(- \frac{3}{53}\right)$

$r = \sqrt{{\left(\frac{90}{53}\right)}^{2} + {\left(- \frac{3}{53}\right)}^{2}}$

$r = \sqrt{\frac{8100}{2809} + \frac{9}{2809}}$

$r = \sqrt{\frac{8109}{2809}}$

$r = \sqrt{\frac{8109}{53 \cdot 53}}$

$r = \sqrt{\frac{901 \cdot 9}{53 \cdot 53}}$

$r = \sqrt{\frac{901}{53} \cdot \frac{9}{53}}$

color(brown)(r=(3sqrt(901))/53

Also,

$\theta = {\tan}^{-} 1 \left[\frac{- \frac{3}{53}}{\frac{90}{53}}\right]$

$\Rightarrow \theta = - {\tan}^{-} 1 \left(\frac{1}{30}\right)$

Add $2 \pi$, since $\theta$ is negative.

color(brown)[:. theta = -tan^-1(1/30)+2pi

Hence, the final answer is given by

color(blue)(90/53-(3i)/53 = (3/53)sqrt(901){cos(-a tan(1/30)+2pi}+i sin{-a tan(1/30)+2pi}

Hope you find this solution useful.