Question

Express the complex number in trigonometric form. -2 + 2square root of three i

Answer 1

4cos(120) + 4isin(120)

Answer 2

Answer:

The required trigonometric form is $4(\cos(60)-i\sin(60))$

Step-by-step explanation:

Given : Complex number $-2+2\sqrt{3}i$

To find : Express the complex number in trigonometric form?

Solution :

The complex number $a+ib$ trigonometric form is $r(\cos\theta+i\sin\theta)$                      

Where, $r=\sqrt{a^2+b^2}$

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

On comparing with given complex number  $-2+2\sqrt{3}i$

a=-2 and $b=2\sqrt{3}$

Substitute the value,

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

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

$r=\sqrt{4+12}$

$r=\sqrt{16}$

$r=4$

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

$\theta=\tan^{-1}(\frac{2\sqrt3}{-2})$

$\theta=\tan^{-1}(-sqrt3)$

$\theta=\tan^{-1}(\tan(-60))$

$\theta=-60$

Substituting all values in the formula,

$r(\cos\theta+i\sin\theta)$

$4(\cos(-60)+i\sin(-60))$

$4(\cos(60)-i\sin(60))$

Therefore, The required trigonometric form is $4(\cos(60)-i\sin(60))$