Question

Express the complex number in trigonometric form.

-3 + 3 square root of 3i.

Answer 1

\(-3+3\sqrt{3}i\) right?

Answer 2

Yes.

Answer 3

Let wait for Zepdrix, he is better than me hihihi

Answer 4

lol XD

Answer 5

@zepdrix ,please

Answer 6

Hehe, its okay. I'm the one coming to you guys afterall.

Answer 7

So you can write it in the form:\[\Large\rm r(\cos \theta+\mathcal i \sin \theta)\]^That is our trig form.
We need to find \(\Large\rm r\) and \(\Large\rm \theta\)

Answer 8

When we have:\[\Large\rm a+b\mathcal i\]Our radius is given by:\[\Large\rm r=\sqrt{a^2+b^2}\]

Answer 9

So what do we get for a radius in this case? :D
Crunch some numbers!!

Answer 10

\[-3^{2} + 3\sqrt3^{2} = 0\]

Answer 11

So, the radius would be 0.

Answer 12

You got zero for a radius? XD

Answer 13

\[\Large\rm -3+3\sqrt{3}\mathcal i\]

\[\Large\rm r=\sqrt{(-3)^2+(3\sqrt3)^2}\]Don't forget to square the negative sign!!

Answer 14

Ah right, I forgot my calculator always gives me negative times negative = negative.

Answer 15

So, it would be 3 times sqrt 2?

Answer 16

Gotta use brackets in calculator:
\(\Large\rm -3^2\) is not the same as \(\Large\rm (-3)^2\) silly gal! :d

Answer 17

You're right. Hehe, I should probably check my other problems too. I keep forgetting, silly me.

Answer 18

3sqrt2? Hmm I came up with something different.
Lemme see if I did that correctly.

Answer 19

\[\Large\rm r=\sqrt{(-3)^2+(3\sqrt3)^2}\]\[\Large\rm r=\sqrt{9+9(3)}\]\[\Large\rm r=\sqrt{35}\]

Answer 20

Did you remember to square both the 3 and the sqrt3 in the second term? :o

Answer 21

Oh! Geeze, this is embarrassing. I didn't.

Answer 22

Err I guess it's sqrt(36) right?
My addition was a bit off.

Answer 23

Yeah, because sqrt 35 gives a totally different answer than the choices given to me. So that eliminates two of the answer choices.

Answer 24

The next thing we're looking for is theta, correct?

Answer 25

\[\Large\rm r=6\]Yes, correct.

Answer 26

So for theta, we have to use.. \[\tan^{-1} \frac{ b }{ a }\]

Answer 27

Yes, good.

I would probably just leave it like this though, and relate it to your unit circle.\[\Large\rm \tan \theta=\frac{b}{a}\]
But I mean, yes that's how you explicitly solve for theta.

Answer 28

so it would look like: \[\tan \theta = 3\sqrt{3}/3 \]

Answer 29

With a negative in there somewhere, yes.\[\Large\rm \tan \theta=-\sqrt3\]Simplify!! :O

Answer 30

Okay, so going off of common sense. The only time that tan is ever negative is in the second and fourth quadrant.

Answer 31

Which angle will we want, the one in the `second quadrant` or the `fourth quadrant` angle?

`Hint`: Look back your at expression to see which quadrant we're in! \(\Large\rm -3+3\sqrt3\mathcal i\)

Answer 32

I'll assume second quadrant.

Answer 33

Real component is negative,
imaginary is positive,

So yah that puts us up in the second quadrant.

Answer 34

Do you remember your tangent special angles? :O
Those are a little harder to remember than sine and cosine I think.

Answer 35

I had to look it up, but 60 degrees which is pi/3 has a tangent of sqrt 3.

Answer 36

So, it would be 2pi/3 for this problem.

Answer 37

Mmm ok good, that sounds right!

Answer 38

Great! Thank you so much!

Answer 39

\[\Large\rm r=6,\qquad \theta=\frac{2\pi}{3}\]Yay team \c:/