Question

Express the number in trigonometric form. -2 + 2 √3 i

Answer 1

Given:

z = -2 + 2 √3 i​

The objective is to write the number in trigonometric form.

The trigonometric form can be represented as,

`z=r(\cos \theta+i\sin \theta)`

The general form of complex number is, z = x+iy.

By comparing with the given equation.

x = -2

y = 2 √3

The value of r can be calculated as,

`\begin{gathered} r=\sqrt[]{x^2+y^2} \\ r=\sqrt[]{(-2)^2+(2\sqrt[]{3})^2} \\ r=\sqrt[]{4+4\cdot3} \\ r=\sqrt[]{4+12} \\ r=\sqrt[]{16} \\ r=4 \end{gathered}`

The value of theta can be calculated as,

`\begin{gathered} \theta=\tan ^(-1)((y)/(x)) \\ \theta=\tan ^(-1)(\frac{2\sqrt[]{3}}{-2}) \\ \theta=\tan ^(-1)(-\sqrt[]{3}) \\ \theta=-^{}60^0 \\ \theta=-(\pi)/(3) \end{gathered}`

Now, substitute the obtained values of r and theta in the general equation of trigonometric form.

`z=4\lbrack\cos (-(\pi)/(3))+i\sin (-(\pi)/(3))\rbrack`

Since,

`\begin{gathered} \cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta \end{gathered}`

Then,

`z=4\lbrack\cos (\pi)/(3)-i\sin (\pi)/(3)\rbrack`

Hence, the required trigonometric form is obtained.