# How do you divide (9i-5) / (-2i+6) in trigonometric form?

Apr 23, 2018

$\frac{- 5 + 9 i}{6 - 2 i} = \frac{- 12 + 11 i}{10}$ but I couldn't finish in trigonometric form.

#### Explanation:

These are nice complex numbers in rectangular form. It's a big waste of time to convert them to polar coordinates to divide them. Let's try it both ways:

 frac{-5 +9i}{6-2i} cdot {6+2i}/{6+2i} = {-48 +44i }/{40} = {-12+11i}/10

That was easy. Let's contrast.

In polar coordinates we have

-5 + 9i = \sqrt{5^2+9^2} \ e^{i text{ atan2}(9,-5) }

I write $\textrm{a \tan 2} \left(y , x\right)$ as the correct two parameter, four quadrant inverse tangent.

 6-2i = \sqrt{6^2+2^2} \ e^{i text{ atan2}(-2, 6) }

$\frac{- 5 + 9 i}{6 - 2 i} = \frac{\setminus \sqrt{106} {e}^{i \textrm{a \tan 2} \left(9 , - 5\right)}}{\setminus \sqrt{40} \setminus {e}^{i \textrm{a \tan 2} \left(- 2 , 6\right)}}$

$\frac{- 5 + 9 i}{6 - 2 i} = \setminus \sqrt{\frac{106}{40}} {e}^{i \left(\textrm{a \tan 2} \left(9 , - 5\right) - \textrm{a \tan 2} \left(- 2 , 6\right)\right)}$

We can actually make progress with the tangent difference angle formula, but I'm not up for that. I suppose we could get the calculator out, but why turn a nice exact problem into an approximation?

Uncle.