Finding in a+ib form

[Wynters_X]

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

 

[Deleted member 4993]

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?

 

[Steven G]

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!

 

[Dr.Peterson]

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].

 

[Wynters_X]

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.

 

[Deleted member 4993]

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.

 

[pka]

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 \(\displaystyle a+bi\) it means that \(\displaystyle \bf a ~\&~b\) are real numbers.
So \(\displaystyle \frac{2-i}{3+4i}\) is not in \(\displaystyle \bf a+bi\).
But \(\displaystyle \frac{2-i}{3+4i}=\frac{2}{25}-\frac{11}{25}\bf i\) is in \(\displaystyle \bf a+bi\) form SEE HERE
Where did that come from? Well, if each of \(\displaystyle w~\&~z\) is a complex number then \(\displaystyle \frac{w}{z}=\frac{w}{z}\frac{\overline{z}}{\overline{z}}=\frac{w\overline{z}}{|z|^2}\)
Here is a very useful fact: \(\displaystyle \frac{1}{z}=\frac{\overline{z}}{|z|^2}\).

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