Need a review on Complicated explonential math

[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!

 

[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

 

[mathman]

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!

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.

 

[Oblio]

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

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.

 

[Oblio]

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

 

[learningphysics]

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

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))

 

[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

 

[learningphysics]

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

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

 

[Oblio]

pi/3 = 60.

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

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?

 

[learningphysics]

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?

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

 

[Oblio]

I have z+w

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

Gather them...

 

[learningphysics]

I have z+w

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

Gather them...

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...

 

[Oblio]

is...

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

 

[learningphysics]

I have z+w

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

Gather them...

That should be:

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

 

[Oblio]

Yep your right.
That gives me the right answer now :P

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

 

[learningphysics]

is...

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

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)

 

[Oblio]

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

 

[learningphysics]

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

yeah, sorry. I replied before seeing your reply. :tongue: