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

In trigonometric form $Z = \sqrt{2} \left[\cos 188.13 + i \sin 188.13\right]$
Let$Z = \frac{2 i - 6}{- 2 i + 4} = \frac{\left(2 i - 6\right) \left(4 + 2 i\right)}{\left(4 + 2 i\right) \left(4 - 2 i\right)} = \frac{4 {i}^{2} - 4 i - 24}{16 - 4 {i}^{2}} = \frac{- 4 i - 28}{20} = - \frac{7}{5} - \frac{1}{5} i$ [since ${i}^{2} = - 1$] Modulus $Z = \sqrt{{\left(- \frac{7}{5}\right)}^{2} + {\left(- \frac{1}{5}\right)}^{2}} = \sqrt{2}$ Argument Z:theta=tan^-1(1/5/7/5)=tan^-1(1/7)=8.13^0+180^0=188.13^0[180^0 is added as it is on 3rd quadrant] Hence in trigonometric form $Z = \sqrt{2} \left[\cos 188.13 + i \sin 188.13\right]$[Ans]