How do you write the complex number in trigonometric form $- 8 - 5 \sqrt{3} i$ $-8-5sqrt3i$ ?

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Answer 1 · 2017-12-14T09:41:53

$z = \sqrt{139} (cos 47.3 + i sin 47.3)$ $z=sqrt(139)(cos47.3+isin47.3)$

If $z = a + b i$ $z=a+bi$ then $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$

$r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$

$θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^(-1)(b/a)$

$a = - 8$ $a=-8$
$b = - 5 \sqrt{3}$ $b=-5sqrt(3)$

$r = \sqrt{{(- 8)}^{2} + {(- 5 \sqrt{3})}^{2}} = \sqrt{139}$ $r=sqrt((-8)^2+(-5sqrt(3))^2)=sqrt(139)$

$θ = {tan}^{- 1} (\frac{- 5 \sqrt{3}}{- 8}) \approx = {47.3}^{\circ}$ $theta=tan^(-1)((-5sqrt(3))/(-8))~~=47.3^circ$

$z = \sqrt{139} (cos 47.3 + i sin 47.3)$ $z=sqrt(139)(cos47.3+isin47.3)$

How do you write the complex number in trigonometric form −8−5√3i-8-5sqrt3i?