?? where does this come from

lorr3

hi hi in complex number doin de moivres when you are changing lets say

1 + √3i into r(cosӨ +isin Ө) where do you get the angles from??? i know how to get r but this is SUPPOSED to be 2 π
3


i dont get why????
please help.
thats 2 pie over 3 wont write right!!

Amirani

Tan Teta = y/x = √3/1

Therefore teta is equal to inverse tan of √3. Then move around the unit circle.

1 is the real part of the complex number, and sqrt(3) is the imaginary part. In an Argand diagram the i part is your "y" part, and the real part is your "x" part.

Imagine a triangle placed onto this Argand diagram, it's right-angled and has an angle theta. You know that the opposite side is sqrt(3) (as that's the "y" or i height), and the adjacent side is 1 units long (as that's the real part of your complex number).

Therefore, you get the angle (theta) by using the tan function. tan(theta) = opp/adj = sqrt(3)/1 = sqrt(3). So, the angle is the tan^-1 of sqrt(3) = 60 degrees. So the angle you want is 60.

PurpleFistMixer

A complex number is in the form a + bi
The angle theta is equal to arctan (b/a) (that's inverse tan).
If you think about it like on an x y plane, what you have is basically a point (a,b), and the angle is the angle between the line joining that point with (0,0) and the x axis.

EDIT: Jaysus, this thread is hopping today. 2 responses before I even finished typing...

lorr3

ok i get gettin th 60 degrees bit thats complex over real...grand. but thats only pie over 3. why do you move roound argand??? and when should you not?? im sorry if i sound silly.:(

likely_lass

draw a rough diagram for yourself - much easier to understand

measure a related angle if the angle is not to the positive sign of the x axis

your angle should be between 0-180 for the positive and 0 to -180 if its below the x axis

lorr3

lorr3 wrote: »

ok i get gettin th 60 degrees bit thats complex over real...grand. but thats only pie over 3. why do you move roound argand??? and when should you not?? im sorry if i sound silly.:(

It's because tan(60) is positive in other quadrants too; so the angle could be different and that would result in the shape of your "triangle" being different. Tan is positive in the third quadrant as well as the first. Third quadrant is (theta + 180). So, for you it's (60 + 180) = 240. That, in radian measure, is 4pi/3.

Zonda999

After getting the inverse tan of y/x, if the angle is;

In the first quadrant, just leave it as it is

In the second quadrant, subtract the angle from 180 degrees

In the third quadrant, add the angle to 180 degrees

In the fourth quadrant, subtract the angle from 360 degrees

O ya then convert it into radians!

lorr3

-JammyDodger- wrote: »

It's because tan(60) is positive in other quadrants too; so the angle could be different and that would result in the shape of your "triangle" being different. Tan is positive in the third quadrant as well as the first. Third quadrant is (theta + 180). So, for you it's (60 + 180) = 240. That, in radian measure, is 2pi/3.

ok i still follow...except to convert it to radians...calculator???..divine inspiration???:D i know this is crazy with the exam 2 mo.just cant get my head around it.

lorr3

lorr3 wrote: »

ok i still follow...except to convert it to radians...calculator???..divine inspiration???:D i know this is crazy with the exam 2 mo.just cant get my head around it.

No bother.

180 degrees is pi radians. So, 360 = 2pi; 90 = pi/2; 60 = pi/3 etc. etc.

They're easy to work out when they're simple fractions.

For your question, we had to add 180 degrees to 60 degrees, that's adding pi to pi/3, which gives 4pi/3. Think I made a mistake in my last one by saying it was 2pi/3.

lorr3

I GET IT!!!!! you were right the first time its 2pi/3...3rd quadrant!!! THANK YOU SO MUCH!!

lorr3

I GET IT!!! you were right its 2pi/3 in second quadrant! THANK YOU SO SO MUCH!!! forgive the caps but i have a big face right now!

lorr3

lorr3 wrote: »

I GET IT!!! you were right its 2pi/3 in second quadrant! THANK YOU SO SO MUCH!!! forgive the caps but i have a big face right now!

But, ya see tan isn't positive in the second quadrant. Only in the first and the third. To get the third you add one pi, which gives 4pi/3.

Either that, or I'm starting to confuse myself!

lorr3

your makin sese but its the 2002 exam and it says 2pi/3??? i did get it....for a minute...till you said that bit??? if you looked at the solution would you know....
http://www.examinations.ie/archive/markingschemes/2002/LC003ALP1EV.pdf

lorr3

lorr3 wrote: »

your makin sese but its the 2002 exam and it says 2pi/3??? i did get it....for a minute...till you said that bit??? if you looked at the solution would you know....
http://www.examinations.ie/archive/markingschemes/2002/LC003ALP1EV.pdf

Ah yah, I get it now. Basically ignore all of that stuff about quadrants.

Alpha is the smallest angle: the angle that makes up the triangle. (You said +1 instead of -1 in your original post).

Theta is the angle between the hypotenuse of the triangle and the positive x axis, the axis that points to the three o'clock position.

You always use that angle in de Moivre's formula.

It's gotten by taking pi/3 radians (60 degrees) from pi radians (180 degrees), which gives you 2pi/3 radians.

It's a pity we didn't just look at the question before all of this confusion.:pac:

lorr3

sorry my fault!!!! thanks so much.. so do i have it right
1.find the original angle in the 1st quad.
2. decide what quad it should be in
3.add or take away radian angle
4.sub in
????:D

Geog

A few years since I did my Leaving. It's all coming back!!!!