Complex numbers simplification

[Mrencko]

Homework Statement

[/B]
Z=((2z1)+(4z2))/(z1)(z2) where Z1=4e^2pi/3
Z2=2/60 degre, z3=1+i
The answer must be in polar form r/theta

Homework Equations

Well in the upper section

The Attempt at a Solution

After do some operations i get to this and unable to convert to polar form... - i((squarert3+1)/2) i need some help, the polar form go to infinite when i apply arctan(y/x)

 

[Dr. Courtney]

I don't see a question.

 

[Mrencko]

The thread implies, the "question" or in this case an issue whit the problem stated at the start, so i need to convert my answer to polar form but i can't because one C/0 so i need to know if i have a good answer and its not possible to convert to polar or if i did some mistake in operating the complex numbers

 

[SammyS]

The thread implies, the "question" or in this case an issue whit the problem stated at the start, so i need to convert my answer to polar form but i can't because one C/0 so i need to know if i have a good answer and its not possible to convert to polar or if i did some mistake in operating the complex numbers

We should not have to guess what your question is. A complete statement of the problem should be given in the body of the post which initiates the thread, no matter what is the thread's title.

I addition to that, It looks as if you have some typographical errors in the statement of your problem.

What do you mean by " Z2=2/60 degre, " ?

What is z3 to be used for?

Homework Statement

Z=((2z1)+(4z2))/(z1)(z2) where Z1=4e^2pi/3
Z2=2/60 degre, z3=1+i
The answer must be in polar form r/theta[/B]

Homework Equations

Well in the upper section

The Attempt at a Solution

After do some operations i get to this and unable to convert to polar form... - i((squarert3+1)/2) i need some help, the polar form go to infinite when i apply arctan(y/x)

 

[Mrencko]

Oh sorry you are right z3 its for (2z1+4z3)/z1z2, and z2 its in the polar form
Where 2=r and 60=theta

 

[SammyS]

z₁ looks like it was written in a form close to being in polar form except that there is an i (imaginary unit) missing from the exponent.

Is it intended that $\displaystyle \ z_1=4e^{2\pi i/3} \$ ?

 

[Mrencko]

yes sorry again i assume the imaginary unit will be implicit in the exponential form of complex numbers, well doing my research, appears to be correct and the polar form will be: in the form Rθ being theta= -90 so tell me if i am correct, or incorrect please, just to finish properly my homework :)

 

[Mark44]

yes sorry again i assume the imaginary unit will be implicit in the exponential form of complex numbers

No, that's not true. $e^{2\pi/3}$ is a real number, and $e^{2i\pi/3}$ is a complex number.

 

[Mrencko]

Sorry for that

 

[SammyS]

The Attempt at a Solution

After do some operations i get to this and unable to convert to polar form... - i((squarert3+1)/2) i need some help, the polar form go to infinite when i apply arctan(y/x)

Is that $\displaystyle \ -i\frac{\sqrt{3}+1}{2}\$, or something else? (You have been a bit careless with parentheses.)

For what values of θ, is tan(θ) undefined (sort of like being infinite)?

 

[Mrencko]

Yes that its my finale answer, and for the indetermination of theta i use the form thetha=arctan(y/x) so if my answer its pure complex number, so y/0 its infinite right? For r the value exist for thetha doesn't exist

 

[Mrencko]

Being x the real part and y the complex in the form x+yi

 

[SammyS]

Being x the real part and y the complex in the form x+yi

How do you convert polar form, let's say 2/60°, to the form, x + yi ?

 

[Mrencko]

Whit the form x=rcosthetha y=rsinthetha
To the form x+yi

 

[SammyS]

Whit the form x=rcosthetha y=rsinthetha
To the form x+yi

Fine.

So what does θ need to be to get x = 0 and get y to be negative (actually -r) ?

 

[Mrencko]

It must be in 90 degres to get rcos90=0, but i don't get the think of - r

 

[SammyS]

It must be in 90 degres to get rcos90=0, but i don't get the think of - r

Well sin(90°) = 1 , so 90° won't do what's needed.

What does θ need to be for sin(θ) = -1 ? What is cosine for this θ ?

 

[Mrencko]

I guess - 90 degrees to get the both answers

 

[SammyS]

I guess - 90 degrees to get the both answers

Correct.

Does that work out for the problem you're working on in this thread?

 

[Mrencko]

Yes then i must say the polar form of my complex number its given by R=√((√3+1)/2)^2 and thetha=-90 degres by definition

 

[SammyS]

Yes then i must say the polar form of my complex number its given by R=√((√3+1)/2)^2 and thetha=-90 degres by definition

Right.

You don't need R to be that complicated.

If $\ a>0\,, \$ then $\ \left(\sqrt a\,\right)^2 =a\ .$

So $\displaystyle \ R=\frac{1+\sqrt 3}{2}\$

 

[Mrencko]

Well i must say thanks, but only one more doubt, its the finale answer right? I mean before the conversion to polar

 

[SammyS]

Well i must say thanks, but only one more doubt, its the finale answer right? I mean before the conversion to polar

Yes. That's what I got.

 

[Mrencko]

Ok thanks for the help, and sorry for the mistakes in the explanation of the problem

 

[SammyS]

Ok thanks for the help, and sorry for the mistakes in the explanation of the problem

You're welcome.