Question

Express the complex number in trigonometric form.

-3 + 3 √3i

Answer 1

you need two numbers, \(r\) and \(\theta\)

Answer 2

\(r=\sqrt{a^2+b^2}\) which in your case is \[r=\sqrt{3^2+(3\sqrt3)^2}\]
what do you get?

Answer 3

huh

Answer 4

how do i square 3 square root of 3

Answer 5

HI!!

Answer 6

square each one

Answer 7

what is \(3^2\)?

Answer 8

9

Answer 9

and what is \(\sqrt3^2\)?

Answer 10

3?

Answer 11

yes so \[(3\sqrt3)^2=9\times 3=27\]

Answer 12

now you have \[r=\sqrt{9+27}\] compute that (don't use decimals)

Answer 13

square root of36, 6

Answer 14

ok good
so you have one of the two numbers you needed \(r=6\)

Answer 15

now you need \(\theta\)

Answer 16

you know how to find it?

Answer 17

how do i get it

Answer 18

nope

Answer 19

hmm must be a trig class right?

Answer 20

pre calc :/

Answer 21

same thing

Answer 22

use \[\cos(\theta)=\frac{a}{r}\\
\sin(\theta)=\frac{b}{r}\]

Answer 23

in your case \[\cos(\theta)=\frac{-3}{6}=-\frac{1}{2}\\
\sin(\theta)=\frac{3\sqrt3}{6}=\frac{\sqrt3}{2}\]

Answer 24

in other words, find the point on the unit circle \[(-\frac{1}{2},\frac{\sqrt3}{2})\] and that will give you \(\theta\) and you are done

Answer 25

pi/3 ?

Answer 26

no

Answer 27

oh

Answer 28

okay one second

Answer 29

it would be \(\frac{\pi}{3}\) except the first coordinate is \(-\frac{1}{2}\) not \(\frac{1}{2}\)

Answer 30

so the cosine has to be negative but the sin is positive, second quadrant?

Answer 31

yes

Answer 32

2pi/3

Answer 33

that is the one yes

Answer 34

what do i do with r and theta

Answer 35

put them in to \[r\left(\cos(\theta)+i\sin(\theta)\right)\]

Answer 36

thank you!!!!!!!!!!!!!!

Answer 37

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