Question

I need help with this practice problem solving The subject is trigonometry Make sure to read the instructions

Answer 1

Solution:

A complex number of the form:

$z=x+iy\text{ ---- equation 1}$

is expressed in polar form as

$\begin{gathered} z=r(cos\theta+isin\theta)\text{ ---- equation 2} \\ where \\ r=\sqrt{x^2+y^2}\text{ ---- equation 3} \\ \theta=\tan^{-1}(\frac{y}{x})\text{ ----- equation 4} \end{gathered}$

Given the complex number:

$z=-3+i3\sqrt{3}$

This implies that

$\begin{gathered} x=-3 \\ y=3\sqrt{3} \end{gathered}$

To express in polar form as shown in equation 2,

step 1: Evaluate the value of r.

From equation 3,

$\begin{gathered} r=\sqrt{x^2+y^2} \\ x=-3,\text{ y=3}\sqrt{3} \\ thus, \\ r=\sqrt{(-3)^2+(3\sqrt{3})^2} \\ =\sqrt{9+27\text{ }} \\ =\sqrt{36} \\ \Rightarrow r=6 \end{gathered}$

step 2: Evaluate the positive value of θ.

From equation 4,

$\begin{gathered} \theta=\tan^{-1}(\frac{y}{x}) \\ =\tan^{-1}(\frac{3\sqrt{3}}{-3}) \\ \theta=-60 \\ From\text{ the second quadrant, the value of }\theta\text{ is also negative.} \\ Thus,\text{ the smallest angle will be} \\ \pi-\frac{\pi}{3}=\frac{2}{3}\pi \\ \end{gathered}$

step 3: Substitute the values of r and θ into equation 2.

Thus,

$z=6(cos\text{ }\frac{2\pi}{3}\text{+isin }\frac{2\pi}{3}\text{\rparen}$

Thus, the polar form of the complex number is expressed as

$z=6\text{ cis\lparen}\frac{2\pi}{3})$