Question

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

Answer 1

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=√(x^2+y^2)\text{ ---- equation 3} \\ \theta=\tan^(-1)((y)/(x))\text{ ----- equation 4} \end{gathered}`

Given the complex number:

`z=-3+i3√(3)`

This implies that

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

step 2: Evaluate the positive value of θ.

From equation 4,

`\begin{gathered} \theta=\tan^(-1)((y)/(x)) \\ =\tan^(-1)((3√(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-(\pi)/(3)=(2)/(3)\pi \\ \end{gathered}`

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

Thus,

`z=6(cos\text{ }(2\pi)/(3)\text{+isin }(2\pi)/(3)\text{\rparen}`

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

`z=6\text{ cis\lparen}(2\pi)/(3))`