Finding in a+ib form

Wynters_X

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Given that z=cos(theta) + isin(theta), express (1-z^2)/(1+z^2) in the form a+ib giving answer in terms of theta.

I can't find a way to simplify it
 
Given that z=cos(theta) + isin(theta), express (1-z^2)/(1+z^2) in the form a+ib giving answer in terms of theta.

I can't find a way to simplify it
Do you know that:

z = cos(theta) + i * sin(theta) = r * e^(i * theta)

z2 = [r * e^(i * theta)]2 = r2 * e^(i * 2 * theta)

How can this knowledge help you?
 
Do you know that:

z = cos(theta) + i * sin(theta) = r * e^(i * theta)

z2 = [r * e^(i * theta)]2 = r2 * e^(i * 2 * theta)

How can this knowledge help you?
Excuse me, but you introduced a new variable without defining it!
 
Given that z=cos(theta) + isin(theta), express (1-z^2)/(1+z^2) in the form a+ib giving answer in terms of theta.

I can't find a way to simplify it
One direct way is to write [MATH]z = x + iy[/MATH], where [MATH]x^2 + y^2 = 1[/MATH]. Do you see why the latter is true?

Then just simplify [MATH]\frac{1-z^2}{1+z^2}[/MATH] in terms of x and y, then at the end replace x with [MATH]\cos(\theta)[/MATH] and y with [MATH]\sin(\theta)[/MATH].
 
Thx for the replies i tried replacing z with e^(i*theta) and even divided both numerator and denominator to turn the 1s into e^(-i*theta). From that point i don't know how to proceed.
 
Thx for the replies i tried replacing z with e^(i*theta) and even divided both numerator and denominator to turn the 1s into e^(-i*theta). From that point i don't know how to proceed.
Try to follow suggestions at resoponse #3 and show us what do you get.
 
Given that z=cos(theta) + isin(theta), express (1-z^2)/(1+z^2) in the form a+ib giving answer in terms of theta.
I did not try to answer before this for several reasons. After first reading the post, I started a reply which was interrupted by other replies to gave me pause. Then I took note that this was posted in Beginning Algebra. That really confused the issue.
Normally when the instructions say to express a complex number in a+bi\displaystyle a+bi it means that a & b\displaystyle \bf a ~\&~b are real numbers.
So 2i3+4i\displaystyle \frac{2-i}{3+4i} is not in a+bi\displaystyle \bf a+bi.
But 2i3+4i=2251125i\displaystyle \frac{2-i}{3+4i}=\frac{2}{25}-\frac{11}{25}\bf i is in a+bi\displaystyle \bf a+bi form SEE HERE
Where did that come from? Well, if each of w & z\displaystyle w~\&~z is a complex number then wz=wzzz=wzz2\displaystyle \frac{w}{z}=\frac{w}{z}\frac{\overline{z}}{\overline{z}}=\frac{w\overline{z}}{|z|^2}
Here is a very useful fact: 1z=zz2\displaystyle \frac{1}{z}=\frac{\overline{z}}{|z|^2}.

But as posted the given question is much much more complicated than my example.
 
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