How do you find the trigonometric form of the complex number $3$ $3$ $\sqrt{3}$ $sqrt3$ $+$ $+$ $3 i$ $3i$ ?

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Answer 1 · 2015-10-09T02:23:50

I found: $z = 6 [cos (\frac{π}{6}) + i sin (\frac{π}{6})]$ $z=6[cos(pi/6)+isin(pi/6)]$

To write your number in trigonometric form you need to find the Modulus $ρ$ $rho$ and the Argument $θ$ $theta$ of it:
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How do you find the trigonometric form of the complex number 33√3sqrt3++3i3i?