Question

Express the complex number in trigonometric form.

-2 + 2(radical3)i

Answer 1

@zepdrix do you mind helping me with this one too?

Answer 2

#1 - You should be very well-acquainted with this circle.

http://www.snow.edu/jonathanb/Courses/Math1060/unit_circ_trig.html

Scroll down about a one screen.

Answer 3

yes but i dont understand how to relate that to the given question

Answer 4

Ok so we're somewhere around here.
Do you understand how to find this radial length?

Answer 5

no could you explain how you got there?

Answer 6

We have a complex number in the form:\[\Large\rm z=a+b\mathcal i\]

Answer 7

\[\Large\rm z=-2+2\sqrt3 \mathcal i\]

Answer 8

So we'll move -2 along the real axis (horizontal axis),
and 2sqrt(3) along the imaginary axis (vertically)

Answer 9

oh okay i see that

Answer 10

If you think of it as a triangle,
then that radial length is simply the hypotenuse, yes?

Answer 11

yes

Answer 12

but is that all the question is asking to find?

Answer 13

No, this is like... almost half of the question, not quite though.

Answer 14

We're trying to rewrite our point:\[\Large\rm z=a+b\mathcal i\]In the form:\[\Large\rm z=r(\cos \theta+\mathcal i \sin \theta)\]

Answer 15

To do so, we need two pieces of information,
the correct angle,
and the correct length r.

Answer 16

oh ok

Answer 17

after we find r how do we find the angle?

Answer 18

Recall:\[\Large\rm \tan \theta=\frac{y}{x}\]\[\Large\rm \theta= \arctan\left(\frac{y}{x}\right)\]For our problem, the b is the vertical component while the a is the horizontal,\[\Large\rm \theta^*= \arctan\left(\frac{b}{a}\right)\]

Answer 19

I put a little star on theta because uhhh.....
The inverse tangent will only give us angles in the 1st and 4th quadrants.
Ahh ignore all that.. we're going to get a special angle anyway....

Answer 20

\[\Large\rm \tan \theta=\frac{b}{a}\]\[\Large\rm \tan \theta=\frac{2\sqrt3}{-2}\]What special angle does this correspond to?\[\Large\rm \tan \theta=-\sqrt3\]

Answer 21

ok so im confused with the side and the angle how do we get the numbers for the equation

Answer 22

Remember that our complex number has this form:\[\Large\rm a+b\mathcal i\]A real part and an imaginary part.
And we have:\[\Large\rm a^2+b^2=r^2\]\[\Large\rm \tan \theta=\frac{b}{a}\]The b is always coming from the one that has the \(\Large\rm \mathcal i\) attached to it.

Answer 23

\[\Large\rm \color{orangered}{-2}+\color{royalblue}{2\sqrt3}\mathcal i\]\[\Large\rm \color{orangered}{(-2)}^2+\color{royalblue}{\left(2\sqrt3\right)}^2=r^2\]\[\Large\rm \tan \theta=\frac{\color{royalblue}{2\sqrt3}}{\color{orangered}{-2}}\]

Answer 24

ok i understand that so is that it?

Answer 25

Yes, but there is something important going on with the angle,
so i'll ask again:

Do you know what special angle(s) correspond to this?\[\Large\rm \tan \theta=-\sqrt3\]

Answer 26

no special angles?