# How do you divide ( 2i+9) / ( 5 i +12 ) in trigonometric form?

Jun 5, 2018

$\frac{\sqrt{85}}{13} \cdot {e}^{- {\tan}^{- 1} \left(\frac{21}{118}\right) i}$

with $r = \frac{\sqrt{85}}{13} , \theta = - {\tan}^{-} 1 \left(\frac{21}{118}\right)$

#### Explanation:

First, express $\left(2 i + 9\right) , \left(5 i + 12\right)$ in polar form.

### Converting to Polar Form

For a complex number $a + b i$, the polar form is $r \left(\cos \theta + i \sin \theta\right) = r {e}^{i \theta}$ (Euler's Formula), hence

$r \cos \theta = a$ (Equation 1)

$r \sin \theta = b$ (Equation 2)

Squaring both equations and adding them together, we have

${r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = {a}^{2} + {b}^{2}$

Since ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

${r}^{2} = {a}^{2} + {b}^{2}$

$\textcolor{red}{r = \sqrt{{a}^{2} + {b}^{2}}}$

To solve for $\theta$, we can divide Equation 2 by Equation 1, yielding

$\frac{r \sin \theta}{r \cos \theta} = \tan \theta = \frac{b}{a}$

Hence $\textcolor{red}{\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)}$

Substituting $\left(2 i + 9\right) , \left(5 i + 12\right)$ for $a + b i$, we have their polar forms

$2 i + 9 = \sqrt{{9}^{2} + {2}^{2}} {e}^{{\tan}^{-} 1 \left(\frac{2}{9}\right) i} = \sqrt{85} {e}^{{\tan}^{-} 1 \left(\frac{2}{9}\right) i}$

$5 i + 12 = \sqrt{{12}^{2} + {5}^{2}} {e}^{{\tan}^{-} 1 \left(\frac{5}{12}\right) i} = 13 {e}^{{\tan}^{-} 1 \left(\frac{5}{12}\right) i}$

### Division in Polar Form

Now $\frac{2 i + 9}{5 i + 12}$ becomes

$\frac{\sqrt{85} {e}^{{\tan}^{-} 1 \left(\frac{2}{9}\right) i}}{13 {e}^{{\tan}^{-} 1 \left(\frac{5}{12}\right) i}}$

$= \frac{\sqrt{85}}{13} \cdot {e}^{{\tan}^{-} 1 \left(\frac{2}{9}\right) i} / {e}^{{\tan}^{-} 1 \left(\frac{5}{12}\right) i}$

$= \frac{\sqrt{85}}{13} \cdot {e}^{{\tan}^{-} 1 \left(\frac{2}{9}\right) i - {\tan}^{-} 1 \left(\frac{5}{12}\right) i}$ (1)

$= \frac{\sqrt{85}}{13} \cdot {e}^{- {\tan}^{- 1} \left(\frac{21}{118}\right) i}$ (2)

(1) Exponent Properties ${e}^{a} / {e}^{b} = {e}^{a - b}$

(2) Tangent Properties ${\tan}^{-} 1 \left(a\right) - {\tan}^{-} 1 \left(b\right) = {\tan}^{-} 1 \left(\frac{a - b}{1 + a b}\right)$ (derive by substituting a=tan(x), b=tan(y) into tangent addition formula)