Homework Help: Complex Analysis - Value of Arg[(z-1)/(z+1)] between -pi and pi

1. Sep 13, 2011

cooljosh2k2

1. The problem statement, all variables and given/known data
Let Arg(w) denote that value of the argument between -π and π (inclusive). Show
that:

Arg[(z-1)/(z+1)] = { π/2, if Im(z) > 0 or -π/2 ,if Im(z) < 0.

where z is a point on the unit circle ∣z∣= 1

3. The attempt at a solution

First, i know that Arg(w) = arctan(b/a) and that Arg(z/w) = Arg(z) - Arg(w).

So Arg[(z-1)/(z+1)] = Arg(z-1) - Arg(z+1).

2. Sep 13, 2011

ehild

Note that z is a point of the unit circle. What are its real and imaginary parts? You can find the real and imaginary parts of (z-1)/(z+1) if you multiply both the numerator and denominator with the conjugate of the denominator: z*+1.

ehild

3. Sep 13, 2011

micromass

Write $z=a+bi$.

Can you calculate

$$\frac{z-1}{z+1}$$

in terms of a and b?? In particular, can you make the denominator real?

4. Sep 13, 2011

cooljosh2k2

Is this what im supposed to do:

(z-1)/(z+1) = (a + bi -1)/(a+bi +1) = (a2 - b2 +2abi -1)/ (a+bi+1)2

Is that right? If so, what do i do next, or if im wrong, what am i doing wrong?

5. Sep 13, 2011

micromass

No, that's not what I get. Just calculate

$$\frac{a-1+bi}{a+1+bi}=\frac{(a-1+bi)(a+1-bi)}{(a+1+bi)(a+1-bi)}$$

What do you get??

6. Sep 13, 2011

cooljosh2k2

Ok, i get: (a2+b2+2bi-1)/(a2+b2+2a+1)

Is that right?

7. Sep 13, 2011

micromass

Yes, and what is $a^2+b^2$??? (hint: z is on the unit circle).

Can you show that $\frac{z-1}{z+1}$ is purely imaginary?? What are the consequences for the arg?

8. Sep 13, 2011

cooljosh2k2

Wow, am i rusty. I am so embarrassed to say that i dont know. I may be wrong, but i think i faintly remember $a^2+b^2$= 1. But i dont think its right. As for your other questions, i dont know. Im sorry, im an idiot.

9. Sep 13, 2011

micromass

Yes, $a^2+b^2=1$. So what does that mean for your equation?

10. Sep 13, 2011

cooljosh2k2

Wow, im shocked i remembered that lol. It would equal 2bi/(2a +1).

Thanks for helping me through it btw.

11. Sep 13, 2011

micromass

OK, so it is

$$\frac{2b}{2a+1}i$$

Now calculate the argument. With the arctangent if you prefer.

12. Sep 13, 2011

cooljosh2k2

How do i do that with 2b and 2a+1?

13. Sep 13, 2011

micromass

Well, what is the argument of

$0+ci$

???

14. Sep 13, 2011

cooljosh2k2

0? My textbook barely even touches on the argument, thats why im so lost.

15. Sep 13, 2011

cooljosh2k2

Are you still there? Im so confused.

16. Sep 13, 2011

micromass

What does your book say that the definition of the Arg is??

17. Sep 13, 2011

cooljosh2k2

"Arg z, the argument of z, defined for z ≠ 0, is the angle which the vector (originating
from 0) to z makes with the positive x-axis. Thus Arg z is defined (modulo
2π) as that number θ for which
cos θ = Re z/|z|
; sin θ = Im z/|z| ."

18. Sep 13, 2011

micromass

OK, so the arg is the angle that the vector makes with the x-axis. Now, what angle does ci make with the x-axis?

19. Sep 13, 2011

cooljosh2k2

I dont know what "c" is, am i supposed to come up with a number? Does ci = sinθ?

20. Sep 13, 2011

micromass

c is just a real number.

What angle does i make with the x-axis?