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4 votes
Express the complex number in trigonometric form.

-6 + 6\sqrt(3) i

User DropDrage
by
5.3k points

1 Answer

5 votes

Answer:

12(cos120°+isin120°)

Explanation:

The rectangular form of a complex number is expressed as z = x+iy

where the modulus |r| =
\sqrt{x^(2)+y^(2) and the argument
\theta = tan^(-1)(y)/(x)

In polar form, x =
rcos\theta \ and\ y = rsin\theta


z = rcos\theta+i(rsin\theta)\\z = r(cos\theta+isin\theta)

Given the complex number,
z = -6+6√(3) i. To express in trigonometric form, we need to get the modulus and argument of the complex number.


r = \sqrt{(-6)^(2)+(6√(3) )^(2)}\\r = √(36+(36*3)) \\r = √(144)\\ r = 12

For the argument;


\theta = tan^(-1) (6√(3) )/(-6) \\\theta = tan^(-1)-√(3) \\\theta = -60^(0)

Since tan is negative in the 2nd and 4th quadrant, in the 2nd quadrant,


\theta =180-60\\\theta = 120^(0)

z = 12(cos120°+isin120°)

This gives the required expression.

User Vittoria
by
5.8k points