Need a review on Complicated explonential math

1. Sep 23, 2007

Oblio

I completely forget how to do this... can someone give me a review on how to do this?

For example:
z=8e^(i(pi)/3)
w=4e^(i(pi)/6)

z+w= ?

This is an example in my book but no steps are given for this, the answer they give is
(4 + 2sqrt[3]) + (4sqrt[3] + 2) i

I have virtually no memory of this...
I'd greatly appreciate any guidance I can get with this.
Thanks!

2. Sep 23, 2007

ice109

this isn't log rules, this is euler's identity : $e^{i\theta}=cos{\theta}+isin{\theta}$

i don't think you can generally add thing with different exponents, i mean you can't simplify $x^2 + x^3$ right

3. Sep 23, 2007

mathman

Use Euler's identity to get z=8cos(pi/3)+8isin(pi/3) and w=4cos(pi/6)+4isin(pi/6).

cos(pi/3)=sin(pi/6)=1/2, cos(pi/6)=sin(pi/3)=[sqrt(3)]/2. You can finish up to get the book answer.

4. Sep 24, 2007

Oblio

Can you assist me in how to do this step?
I can't seem to 'get', for example, z=8cos(pi/3)+8isin(pi/3) using that identity.

5. Sep 24, 2007

Oblio

How did you manage to get rid of the pesky e using that identity?

6. Sep 24, 2007

learningphysics

z=8e^(i(pi)/3)

From ice109's post: Euler's identity is

$e^{i\theta}=cos{\theta}+isin{\theta}$

directly applying this integral to z=8e^(i(pi)/3)... here theta = pi/3

so z = 8(cos(pi/3) + isin(pi/3))

7. Sep 24, 2007

Oblio

I'm blind.

Ok, that's not hard. What about the next step of getting the pair of three equalities?

cos(pi/3)=sin(pi/6)=1/2, cos(pi/6)=sin(pi/3)=[sqrt(3)]/2

8. Sep 24, 2007

learningphysics

pi/3 = 60. pi/6=30

cos(60) = sin(30) (for any angle x<90, cos(x) = sin(90-x), sin(x) = cos(90-x)

so, cos(60) = sin(30) = 1/2, cos(30) = sin(60) = sqrt(3)/2

Last edited: Sep 24, 2007
9. Sep 24, 2007

Oblio

Ok I can see they are equal.

So, is the logic that you simplify each to 1/2 and sqrt[3]/2 and then add those together?

10. Sep 24, 2007

learningphysics

gather the complex terms (the terms that are multiplied by i) and the real terms together.

11. Sep 24, 2007

Oblio

I have z+w

z+w = 8cos(pi/3) + 8isin(pi/3) + 4cos(pi/6) + isin(pi/6)

Gather them.....

12. Sep 24, 2007

learningphysics

yup... gather the "i" terms together... I mean like this:

z+w = 8cos(pi/3) + 4cos(pi/6) + i[8sin(pi/3) + sin(pi/6)]

now substitute in the values for cos(pi/3) etc...

13. Sep 24, 2007

Oblio

is...

4 + 4sqrt[3]/2 + i[8sqrt[3]/2 + 1/2]

14. Sep 24, 2007

learningphysics

That should be:

z+w = 8cos(pi/3) + 8isin(pi/3) + 4cos(pi/6) + 4isin(pi/6)

15. Sep 24, 2007

Oblio

That gives me the right answer now :P

(4 + 2sqrt[3]) + i(4sqrt[3] + 2)

16. Sep 24, 2007

learningphysics

yeah, except that should have been 4isinpi/6... so the answer is actually:

4 + 4sqrt[3]/2 + i[8sqrt[3]/2 + 4(1/2)]

so when you simplify... you get

4 + 2sqrt[3] + i(4sqrt[3] + 2)

17. Sep 24, 2007

Oblio

yeah I saw your correction. I was right wasnt i?

18. Sep 24, 2007