How do you divide $\frac{- 1 + 5 i}{3 - i}$ $(-1+5i)/(3-i)$ in trigonometric form?

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Answer 1 · 2016-12-04T07:07:41

In Trigonometric form
$\frac{\sqrt{13}}{\sqrt{5}} [cos (- 60.26) + i sin (97.12)]$ $sqrt13/sqrt5 [cos (-60.26) +i sin (97.12)]$

In trigonometric form consider Z1 = (1+5i) and Z2=(3-i) and A1 = inv tan (- 5/1) = -78.69 and A2 = inv tan (- 1/3) = -18.43

$mod z1 = sqrt26 and mod z2 = sqrt10$

the number would be
z1/z2 [cos (A1 -A2) + i sin (A1+A2)]

$(sqrt13sqrt2)/(sqrt5sqrt2) [cos (-60.26) + i sin (9.12)]$
$(sqrt13)/(sqrt5) [cos (-60.26) + i sin (9.12)]$

How do you divide (-1+5i)/(3-i) −1+5i3−i (-1+5i)/(3-i) in trigonometric form?