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

[cooljosh2k2]

Homework Statement

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

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? I am stuck here. Please help

 

[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

 

[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?

 

[cooljosh2k2]

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?

Is this what I am 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 I am wrong, what am i doing wrong?

 

[micromass]

Is this what I am 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 I am wrong, what am i doing wrong?

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??

 

[cooljosh2k2]

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??

Ok, i get: (a²+b²+2bi-1)/(a²+b²+2a+1)

Is that right?

 

[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?

 

[cooljosh2k2]

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?

Wow, am i rusty. I am so embarrassed to say that i don't know. I may be wrong, but i think i faintly remember $a^2+b^2$= 1. But i don't think its right. As for your other questions, i don't know. I am sorry, I am an idiot.

 

[micromass]

Wow, am i rusty. I am so embarrassed to say that i don't know. I may be wrong, but i think i faintly remember $a^2+b^2$= 1. But i don't think its right. As for your other questions, i don't know. I am sorry, I am an idiot.

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

 

[cooljosh2k2]

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

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

Thanks for helping me through it btw.

 

[micromass]

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

Thanks for helping me through it btw.

OK, so it is

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

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

 

[cooljosh2k2]

OK, so it is

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

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

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

 

[micromass]

Well, what is the argument of

$0+ci$

?

 

[cooljosh2k2]

Well, what is the argument of

$0+ci$

?

0? My textbook barely even touches on the argument, that's why I am so lost.

 

[cooljosh2k2]

Are you still there? I am so confused.

 

[micromass]

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

 

[cooljosh2k2]

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

"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 = I am z/|z| ."

 

[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?

 

[cooljosh2k2]

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?

I don't know what "c" is, am i supposed to come up with a number? Does ci = sinθ?

 

[micromass]

c is just a real number.

What angle does i make with the x-axis?

 

[cooljosh2k2]

c is just a real number.

What angle does i make with the x-axis?

I don't know. You make it sound so simple, but I am so confused. I am sorry.

arcsin(Im(z)/|z|)?

 

[micromass]

Pick a piece of Paper. Draw an x-axis and a y-axis on it. Find the point represented by i. Connect i to 0. What is the angle between the line segment that you've drawn and the x-axis?? Just look at it.

 

[cooljosh2k2]

Pick a piece of Paper. Draw an x-axis and a y-axis on it. Find the point represented by i. Connect i to 0. What is the angle between the line segment that you've drawn and the x-axis?? Just look at it.

Isnt i on the y-axis? which would make the angle 0.

 

[micromass]

What is the angle between the x-axis and the y-axis?

 

[cooljosh2k2]

What is the angle between the x-axis and the y-axis?

OMG, i forgot you said in relation to the x axis. 90 degrees. *hits head*

 

[cooljosh2k2]

OMG, i forgot you said in relation to the x axis. 90 degrees. *hits head*

or pi/2

 

[micromass]

Yes!

The argument of i is just the angle between i and the x-axis. So, what is the argument of i??

 

[cooljosh2k2]

Yes!

The argument of i is just the angle between i and the x-axis. So, what is the argument of i??

pi/2!

 

[micromass]

And what is the argument of -i??

 

[cooljosh2k2]

And what is the argument of -i??

-pi/2! but how do i go from (2bi)/(2a+1)? do i just ignore the (2b/2a+1)?

 

[micromass]

OK, now the $\frac{2b}{2a+1}$ is just a real number, so we're dealing with a multiple of i.

So our value could be 3i, 4i, 6i or some other multiple of i. What is the argument of a multiple of i?

 

[cooljosh2k2]

OK, now the $\frac{2b}{2a+1}$ is just a real number, so we're dealing with a multiple of i.

So our value could be 3i, 4i, 6i or some other multiple of i. What is the argument of a multiple of i?

pi/2, right?

 

[micromass]

pi/2, right?

Not really, because we can also have negative multiples. -i and -3i are also multiples.

 

[cooljosh2k2]

Not really, because we can also have negative multiples. -i and -3i are also multiples.

well, pi/2 for positive multiples of i or I am > 0, or -pi/2 for negative multiples of i, or I am < 0

 

[micromass]

Indeed! That's it!