Reduce to x+iy: Solving for z=cos\theta+isin\theta

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Homework Help Overview

The problem involves reducing the expression \(\frac{1+z}{1-z}\) to the form \(x+iy\) where \(z\) is defined as \(z=\cos\theta+i\sin\theta\). The context is complex numbers and their manipulation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the approach of multiplying by the conjugate to simplify the expression. There is a question about whether the correct conjugate has been used, with suggestions to consider the conjugate of \(1-z\) instead of \(1+z\).

Discussion Status

The discussion is ongoing, with participants questioning the initial steps taken in the solution process. Some guidance has been offered regarding the correct conjugate to use, but no consensus has been reached on the next steps.

Contextual Notes

There is a focus on ensuring the correct application of complex conjugates, and participants are clarifying the definitions and forms of the expressions involved.

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Homework Statement



Reduce to x+iy
[tex]\frac{1+z}{1-z}[/tex] where z=cos[tex]\theta[/tex]+isin[tex]\theta[/tex].

Homework Equations





The Attempt at a Solution


[tex]\frac{1+z}{1-z}[/tex]
Multiply by conjugate
[tex]\frac{1+2z+z^{2}}{1-z^{2}}[/tex]
When I plug in the z value nothing seems to cancel out.
 
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You seem to have multiplied both the numerator and denominator by [itex](1+z)[/itex]...but that isn't really the complex conjugate of [itex](1-z)[/itex] is it?:wink:...Don't you actually want to multiply by [itex](1-\bar{z})[/itex] instead?
 
So i should multiply by 1+(costheta+isintheta)?
 
No, [itex]1-z=(1-\cos\theta)-i\sin\theta[/itex], so [itex]\overline{1-z}=[/itex]___?
 

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