Question

Express the complex number in trigonometric form.

-6 + 6\sqrt(3) i

Answer 1

Answer:

12(cos120°+isin120°)

Step-by-step 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 $\theta = tan^{-1}\frac{y}{x}$

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, $z = -6+6\sqrt{3} i$. To express in trigonometric form, we need to get the modulus and argument of the complex number.

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

For the argument;

$\theta = tan^{-1} \frac{6\sqrt{3} }{-6} \\\theta = tan^{-1}-\sqrt{3} \\\theta = -60^{0}$

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.