# 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).

But then what do i do? Im stuck here. Please help

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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
c is just a real number.

What angle does i make with the x-axis?