# How do you divide ( 2i -4) / ( 7 i -2 ) in trigonometric form?

$\frac{2 i - 4}{7 i - 2} = \frac{2 \sqrt{265}}{53} \left[\cos {47.48}^{\circ} + i \cdot \sin {47.48}^{\circ}\right]$

#### Explanation:

The solution:

$2 i - 4 =$
$\sqrt{4 + 16} \left[\cos \left({\tan}^{-} 1 \left(- \frac{1}{2}\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(- \frac{1}{2}\right)\right)\right]$
$\sqrt{20} \left[\cos \left({\tan}^{-} 1 \left(- \frac{1}{2}\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(- \frac{1}{2}\right)\right)\right]$

$7 i - 2 =$
$\sqrt{4 + 49} \left[\cos \left({\tan}^{-} 1 \left(- \frac{7}{2}\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(- \frac{7}{2}\right)\right)\right]$

$\frac{2 i - 4}{7 i - 2} =$
$\frac{\sqrt{20}}{\sqrt{53}} \left[\cos \left({\tan}^{-} 1 \left(- \frac{1}{2}\right) - {\tan}^{-} 1 \left(- \frac{1}{2}\right)\right) + i \cdot \sin \left({\tan}^{-} 1 \left(- \frac{1}{2}\right) - {\tan}^{-} 1 \left(- \frac{1}{2}\right)\right)\right]$

$\frac{2 i - 4}{7 i - 2} = \frac{2 \sqrt{265}}{53} \left[\cos {47.48}^{\circ} + i \cdot \sin {47.48}^{\circ}\right]$

God bless.....I hope the explanation is useful.