# How do you divide ( 3i-7) / ( 2 i -1 ) in trigonometric form?

Sep 18, 2017

Divide the moduli (distances to the origin) and subtract the arguments (polar angles) to get $\frac{13}{5} + \frac{11}{5} i$.

#### Explanation:

The moduli of the numerator and denominator are $| 3 i - 7 | = \sqrt{{3}^{2} + {\left(- 7\right)}^{2}} = \sqrt{9 + 49} = \sqrt{58}$ and
$| 2 i - 1 | = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$. Therefore, the modulus of $\frac{3 i - 7}{2 i - 1}$ is $\frac{\sqrt{58}}{\sqrt{5}} = \frac{\sqrt{58} \sqrt{5}}{5} = \frac{\sqrt{290}}{5}$.

Both the numerator and the denominator sit in the 2nd quadrant of the complex plane. We can use the inverse (arc) cosine function to find their arguments (polar angles) most directly (we could also use the inverse (arc) tangent function but we'd have to add $\pi$ to its output).

The argument of $3 i - 7 = - 7 + 3 i$ is ${\cos}^{- 1} \left(- \frac{7}{\sqrt{58}}\right)$ and the argument of $2 i - 1 = - 1 + 2 i$ is ${\cos}^{- 1} \left(- \frac{1}{\sqrt{5}}\right)$. Therefore, the argument of $\frac{3 i - 7}{2 i - 1}$ is ${\cos}^{- 1} \left(- \frac{7}{\sqrt{58}}\right) - {\cos}^{- 1} \left(- \frac{1}{\sqrt{5}}\right)$.

Now $\cos \left({\cos}^{- 1} \left(- \frac{7}{\sqrt{58}}\right)\right) = - \frac{7}{\sqrt{58}}$, $\sin \left({\cos}^{- 1} \left(- \frac{7}{\sqrt{58}}\right)\right) = \sqrt{1 - {\left(- \frac{7}{\sqrt{58}}\right)}^{2}} = \sqrt{\frac{9}{58}} = \frac{3}{\sqrt{58}}$,
$\cos \left({\cos}^{- 1} \left(- \frac{1}{\sqrt{5}}\right)\right) = - \frac{1}{\sqrt{5}}$, and
$\sin \left({\cos}^{- 1} \left(- \frac{1}{\sqrt{5}}\right)\right) = \sqrt{1 - {\left(- \frac{1}{\sqrt{5}}\right)}^{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.

Furthermore, $\cos \left(\alpha - \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) + \sin \left(\alpha\right) \sin \left(\beta\right)$, which implies that
cos(cos^{-1}(-7/sqrt(58))-cos^{-1}(-1/sqrt(5)))= cos(cos^{-1}(-7/sqrt(58)))cos(cos^{-1}(-1/sqrt(5)))+sin(cos^{-1}(-7/sqrt(58)))sin(cos^{-1}(-1/sqrt(5)))

$= - \frac{7}{\sqrt{58}} \cdot \left(- \frac{1}{\sqrt{5}}\right) + \frac{3}{\sqrt{58}} \cdot \frac{2}{\sqrt{5}} = \frac{13}{\sqrt{290}}$.

In a similar way, $\sin \left(\alpha - \beta\right) = \sin \left(\alpha\right) \cos \left(\beta\right) - \cos \left(\alpha\right) \sin \left(\beta\right)$, which implies that
sin(cos^{-1}(-7/sqrt(58))-cos^{-1}(-1/sqrt(5)))= sin(cos^{-1}(-7/sqrt(58)))cos(cos^{-1}(-1/sqrt(5)))-cos(cos^{-1}(-7/sqrt(58)))sin(cos^{-1}(-1/sqrt(5)))

$= \frac{3}{\sqrt{58}} \cdot \left(- \frac{1}{\sqrt{5}}\right) - \frac{- 7}{\sqrt{58}} \cdot \frac{2}{\sqrt{5}} = \frac{11}{\sqrt{290}}$.

Finally, Euler's formula allows us to put all this together to say that

$\frac{3 i - 7}{2 i - 1} = | z | \cos \left(\theta\right) + i | z | \sin \left(\theta\right)$

$= \frac{\sqrt{290}}{5} \cdot \frac{13}{\sqrt{290}} + i \cdot \frac{\sqrt{290}}{5} \cdot \frac{11}{\sqrt{290}} = \frac{13}{5} + \frac{11}{5} i$.