# How do you divide  (-4+2i)/(6-2i)  in trigonometric form?

Sep 7, 2016

$\frac{\sqrt{2}}{2} \left(\cos \left(0.412\right) - i \sin \left(0.412\right)\right)$

#### Explanation:

We have: $\frac{- 4 + 2 i}{6 - 2 i}$

First, let's multiply both the numerator and the denominator by the complex conjugate of the denominator:

$= \frac{- 4 + 2 i}{6 - 2 i} \cdot \frac{6 + 2 i}{6 + 2 i}$

$= \frac{\left(- 4\right) \left(6\right) + \left(- 4\right) \left(2 i\right) + \left(2 i\right) \left(6\right) + \left(2 i\right) \left(2 i\right)}{{\left(6\right)}^{2} - {\left(2 i\right)}^{2}}$

$= \frac{- 24 - 8 i + 12 i + 4 {i}^{2}}{36 - 4 {i}^{2}}$

Let's apply the fact that ${i}^{2} = - 1$:

$= \frac{- 24 + \left(4 \cdot \left(- 1\right)\right) + 4 i}{36 - \left(4 \cdot \left(- 1\right)\right)}$

$= \frac{- 24 - 4 + 4 i}{36 + 4}$

$= \frac{- 28 + 4 i}{40}$

$= - \frac{7}{10} + \frac{1}{10} i$

Then, we need to determine the modulus and the argument of this complex number.

Let $z = - \frac{7}{10} + \frac{1}{10} i$:

$\implies | z | = \sqrt{{\left(- \frac{7}{10}\right)}^{2} + {\left(\frac{1}{10}\right)}^{2}}$

$\implies | z | = \sqrt{\frac{49}{100} + \frac{1}{100}}$

$\implies | z | = \sqrt{\frac{50}{100}}$

$\implies | z | = \sqrt{\frac{1}{2}}$

$\implies | z | = \frac{\sqrt{2}}{2}$

and

$\implies a r g \left(z\right) = \arctan \left(\frac{\frac{1}{10}}{- \frac{7}{10}}\right)$

$\implies a r g \left(z\right) = \arctan \left(- \frac{1}{7}\right)$

$\implies a r g \left(z\right) \approx - 0.142$

Now, let's express the complex number in polar form:

$\implies z = \frac{\sqrt{2}}{2} \left(\cos \left(- 0.412\right) + i \sin \left(- 0.412\right)\right)$

$\implies z = \frac{\sqrt{2}}{2} \left(\cos \left(0.412\right) - i \sin \left(0.412\right)\right)$