The question is to express (1- i tanx) / (1+ i tanx) in polar form.
First, multiple the whole fraction by cosx. It becomes (cosx - i sin x) / (cosx + i sin x). We can find the modulus and argument easily by using the fact that "if z1=r cis a, z2=r cis b , then z1 / z2 = r1/r2 [cos(a-b) + i sin (a-b)] ". They are 1 and -2x respectively.

However, there exists another way to handle it. By multipling the whole fraction by (cosx - i sin x) / (cosx - i sin x), we can obtain (cos2x-sin2x-2cosxsinx i). The modulus is, of course, found to be 1. However, problem arose when I wanted to find the argument. Letting B be the argument, I set up "tanB = (-2cosxsinx) / (cos2x-sin2x). The equation can be written as tanB=-tan2x. Finally, both -2x and 180o-2x are found as the solutions. However, there should be two arguments for the same "polar form", right? I wonder what is wrong with that.

tiny-tim
Homework Helper
hi abcd8989!
(cosx - i sin x) / (cosx + i sin x)

erm

Euler's equation ?

HallsofIvy
Homework Helper
The question is to express (1- i tanx) / (1+ i tanx) in polar form.
First, multiple the whole fraction by cosx. It becomes (cosx - i sin x) / (cosx + i sin x). We can find the modulus and argument easily by using the fact that "if z1=r cis a, z2=r cis b , then z1 / z2 = r1/r2 [cos(a-b) + i sin (a-b)] ". They are 1 and -2x respectively.

However, there exists another way to handle it. By multipling the whole fraction by (cosx - i sin x) / (cosx - i sin x), we can obtain (cos2x-sin2x-2cosxsinx i). The modulus is, of course, found to be 1. However, problem arose when I wanted to find the argument. Letting B be the argument, I set up "tanB = (-2cosxsinx) / (cos2x-sin2x). The equation can be written as tanB=-tan2x. Finally, both -2x and 180o-2x are found as the solutions. However, there should be two arguments for the same "polar form", right? I wonder what is wrong with that.
When you use $tan(\theta)= a/b$ for individual a and b, you lose track of the signs of a and b separately. That is, if a/b is positive, it may be that a and b are both positive (first quadrant) or it may be that a and b are both negative (third quadrant). Similarly, a/b is negative, it may be that a is negative and b positive (second quadrant) or that a is positive and b negative (fourth quadrant).

Simpler, as tiny-tim suggests, is to write
$$\frac{cos(x)- i sin(x)}{cos(x)+ i sin(x)}= \frac{e^{-ix}}{e^{ix}}= e^{-2ix}$$
to see that the modulus is 1, as you say, and the argument is -2x.