Finding the loci represented by arg(z1/z2) = pi/2

  • Thread starter thomas49th
  • Start date
In summary: ThanksThomasThat's a good start! Now you just need to figure out the values for x and y that will make \tan(\theta_+ - \theta_-) = 0. This will give you the equation for the loci in terms of x and y.
  • #1
thomas49th
655
0
Find the loci represented by

[tex] \arg(\frac{z+1}{z-1}) = \frac{\pi}{2}[/tex]

Working from inside the arg operator (it is an operator right?):

let z = x + iy
multiply num and denom by z+1

seperate into real and imag bits and you should get to

[tex] \frac{(x^{2}-1 + y^{2})-2iy}{(x-1)^{2}+y^{2}}[/tex]

Call this w

as arg(w) = arctan(Im(w)/Re(w))

[tex]\frac{2y}{x^{2}+y^{2}-1} = \tan(\frac{\pi}{2})[/tex]

but now I reach a point where my equation equals an undefined number

What should I do (or should of done)?

Thanks
Thomas
 
Last edited:
Physics news on Phys.org
  • #2
Think more geometrically. For some complex numbers [tex]u, v[/tex], what does it mean for [tex]\arg(u/v) = \pi/2[/tex]? Express [tex]u/v[/tex] in polar form and see what you get.
 
  • #3
u = a + ib
v = c + id
[tex]\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{c^{2}+d^{2}}} (cos(\theta_{1} + \theta_{2}) + sin(\theta_{1} + \theta_{2})[/tex]

which means

[tex]\frac{\sqrt{c^{2}+d^{2}}}{\sqrt{a^{2}+b^{2}}} = \tan{\frac{\pi}{2}[/tex]

Is that what you mean. I cannot see anything new from this :( Any more pointers :)

Thanks
Thomas
 
  • #4
thomas49th said:
seperate into real and imag bits and you should get to

[tex] \frac{(x^{2}-1 + y)-2iy}{(x-1)^{2}+y}[/tex]

There's some algebra mistakes here, it should be

[tex] \frac{(x^{2}-1 + y^2)-2ixy}{(x-1)^{2}+y^2}[/tex]

However, it might be easier to use

[tex] \arg(z/w) = \arg(z) - \arg(w) (\mod 2\pi). [/tex]
 
  • #5
fzero said:
There's some algebra mistakes here, it should be

[tex] \frac{(x^{2}-1 + y^2)-2ixy}{(x-1)^{2}+y^2}[/tex]

However, it might be easier to use

[tex] \arg(z/w) = \arg(z) - \arg(w) (\mod 2\pi). [/tex]

Hi, sorry about the y², but I cannot see where the 2xy is coming from?

Okay using the argument rules:

arg(2xy) - arg(x² + y² - 1) = (pi/2)

I still don't feel any closer to the solution

If I try using the arg rules from the start of the question

arg(z+1) - arg(z-1) = pi /2

(x+1) +iy - [(x-1) + iy] = tan(pi/2)

2 = tan(pi/2)

How can this be?
Thanks
Thomas
 
Last edited:
  • #6
I suppose also

from arg rules:

arctan(y/(x+1)) arctan(y/(x-1)) = pi/2

but you've still got tan(pi/2) when you take the tangent! arghh
 
  • #7
thomas49th said:
Hi, sorry about the y², but I cannot see where the 2xy is coming from?

Oh, sorry that's my mistake, it is -2 i y.

Okay using the argument rules:

arg(2xy) - arg(x² + y² - 1) = tan(pi/2)

I still don't feel any closer to the solution

If I try using the arg rules from the start of the question

arg(z+1) - arg(z-1) = pi /2

(x+1) +iy - [(x-1) + iy] = tan(pi/2)

2 = tan(pi/2)

How can this be?
Thanks
Thomas

You are misusing the arg(w) = arctan(Im(w)/Re(w)) definition here in both places.

I found it useful to take a limit to make sense of the tan sum identity for

[tex] \tan ( \theta_+ - \theta_- - \pi/2 ) =0,[/tex]

where

[tex]\theta_\pm = \arg(z\pm 1).[/tex]

There's probably a cleaner way though.
 
  • #8
fzero said:
Oh, sorry that's my mistake, it is -2 i y.



You are misusing the arg(w) = arctan(Im(w)/Re(w)) definition here in both places.

I found it useful to take a limit to make sense of the tan sum identity for

[tex] \tan ( \theta_+ - \theta_- - \pi/2 ) =0,[/tex]

where

[tex]\theta_\pm = \arg(z\pm 1).[/tex]

There's probably a cleaner way though.

arctan(y/(x+1)) - arctan(y/(x-1)) = pi/2

y/(x+1) - y/(x-1) = tan(pi/2)

I'm afraid I don't understand the use of a limit as you've presented it :( This can't be that hard :S
 
  • #9
Try to decide if

[tex]\lim_{a\rightarrow \pi/2} \tan (x-a)[/tex]

is defined. If it is, it can be written in terms of [tex]\tan x[/tex].

Edit: or just use trig identities, I made this harder than it was.
 
Last edited:
  • #10
hmmmm. Right I'll use [tex]tan\theta = \frac{2tan\frac{\theta}{2}}{1-2tan^{2}\frac{\theta}{2}}[/tex]

so tan(pi/2) = using magic identity = -2. I don't know how you can do that

y/(x+1) - y/(x-1) = -2

y disappears, and x is a quadratic giving me x = 1, -1.

What on Earth have I done?

Thanks
Thomas
 
  • #11
thomas49th said:
hmmmm. Right I'll use [tex]tan\theta = \frac{2tan\frac{\theta}{2}}{1-2tan^{2}\frac{\theta}{2}}[/tex]

so tan(pi/2) = using magic identity = -2. I don't know how you can do that

y/(x+1) - y/(x-1) = -2

y disappears, and x is a quadratic giving me x = 1, -1.

What on Earth have I done?

Thanks
Thomas

Your double-angle formula is incorrect, it should be

[tex]tan\theta = \frac{2tan\frac{\theta}{2}}{1-tan^{2}\frac{\theta}{2}}.[/tex]

Just use (double-check the sign)

[tex] \tan (x - \pi/2) = -\cot x[/tex]
 
Last edited:
  • #12
fzero said:
Your double-angle formula is incorrect, it should be

[tex]tan\theta = \frac{2tan\frac{\theta}{2}}{12tan^{2}\frac{\theta}{2}}.[/tex]

Just use (double-check the sign)

[tex] \tan (x - \pi/2) = -\cot x[/tex]

huh? Where did 12 come from??

[tex] \tan (x - \pi/2) = -\cot x[/tex] How do I use this? I'm completely lost :(

Thanks
Thomas
 
  • #13
thomas49th said:
huh? Where did 12 come from??

Sorry, I wanted to get rid of the 2 in the denominator, not the minus sign. I went back and fixed it. The point is that both sides diverge as [tex]\theta\rightarrow \pi/2[/tex].

[tex] \tan (x - \pi/2) = -\cot x[/tex] How do I use this? I'm completely lost :(

Thanks
Thomas

You show that the loci are described by

[tex]\cot (\theta_+ - \theta_-) =0[/tex]

After another trig identity, we find that this reduces to a very simple condition on x and y.
 
  • #14
sorry still not seeing.

cos(large theta - baby theta) = 0. Well there could be loads of stuff there. cos(180-90) for starters? Where does this lead to

Thanks
Thomas
 
  • #15
thomas49th said:
sorry still not seeing.

cos(large theta - baby theta) = 0. Well there could be loads of stuff there. cos(180-90) for starters? Where does this lead to

Thanks
Thomas

It's probably easier to use the other formula and write everything in terms of tangents, since the formulas for those in terms of x and y will be a bit simpler. But just use the sum rules for trig functions and find the expression in terms of x and y.
 
  • #16
errr still not liking this.

Can I explain it like this:

if [tex]\tan \frac{\pi}{2} = \frac{1}{0}[/tex]

[tex]\frac{2y}{x^{2}+y^{2}-1} = \frac{1}{0}[/tex]

And here comes the dogdy bit

[tex]\frac{2y}{\frac{1}{0}} = x^{2}+y^{2}-1 [/tex]

[tex]\frac{0 \cdot 2y}{1} = x^{2}+y^{2}-1 [/tex]

[tex]0 = x^{2}+y^{2}-1 [/tex]

so the loci is a circle with radius 1, centre 0,0

What do we think?

Thanks
Thomas
 
  • #17
fzero said:
Your double-angle formula is incorrect, it should be

[tex]tan\theta = \frac{2tan\frac{\theta}{2}}{1-tan^{2}\frac{\theta}{2}}.[/tex]

Just use (double-check the sign)

[tex] \tan (x - \pi/2) = -\cot x[/tex]

ALSO (after my previous post), according to Stroud

Let [tex]\theta = \frac{\phi}{2}[/tex]
[tex] 2 \sin \frac{\phi}{2} \cos \frac{\phi}{2}[/tex]
[tex] \cos \phi = \cos^{2} \frac{\phi}{2} - \sin^{2}\frac{\phi}{2}[/tex]
[tex] = 1 - 2 \sin^{2}\frac{\phi}{2} = \cos^{2}\frac{\phi}{2}[/tex]

[tex] \tan \phi = \frac{2 \tan \frac{\phi}{2}}{1 - 2 \tan ^{2} \frac{\phi}{2}}[/tex]

Also check the previous post
 
Last edited:
  • #18
thomas49th said:
errr still not liking this.

Can I explain it like this:

if [tex]\tan \frac{\pi}{2} = \frac{1}{0}[/tex]

[tex]\frac{2y}{x^{2}+y^{2}-1} = \frac{1}{0}[/tex]

And here comes the dogdy bit

[tex]\frac{2y}{\frac{1}{0}} = x^{2}+y^{2}-1 [/tex]

[tex]\frac{0 \cdot 2y}{1} = x^{2}+y^{2}-1 [/tex]

[tex]0 = x^{2}+y^{2}-1 [/tex]

so the loci is a circle with radius 1, centre 0,0

What do we think?

Thanks
Thomas

The loci are a unit circle, but this falls a bit short of a valid proof. It's better to work with well-defined quantities.

thomas49th said:
ALSO (after my previous post), according to Stroud

Let [tex]\theta = \frac{\phi}{2}[/tex]
[tex] 2 \sin \frac{\phi}{2} \cos \frac{\phi}{2}[/tex]
[tex] \cos \phi = \cos^{2} \frac{\phi}{2} - \sin^{2}\frac{\phi}{2}[/tex]
[tex] = 1 - 2 \sin^{2}\frac{\phi}{2} = \cos^{2}\frac{\phi}{2}[/tex]

[tex] \tan \phi = \frac{2 \tan \frac{\phi}{2}}{1 - 2 \tan ^{2} \frac{\phi}{2}}[/tex]

Also check the previous post


You've left some steps out, but you have enough there to show that

[tex]\tan\phi = \frac{\sin\phi}{\cos\phi} = \frac{2 \sin \frac{\phi}{2} \cos \frac{\phi}{2}}{\cos^{2} \frac{\phi}{2} - \sin^{2}\frac{\phi}{2}} = \frac{2 \tan\frac{\phi}{2}}{1-\tan^{2}\frac{\phi}{2}}[/tex]
 

Related to Finding the loci represented by arg(z1/z2) = pi/2

1. What is the meaning of "loci" in this context?

"Loci" refers to the set of points in a coordinate system that satisfy a specific condition or equation. In this case, we are looking for the points that satisfy the equation arg(z1/z2) = pi/2, which represents a specific angle in the complex plane.

2. What does the symbol "z" represent in this equation?

In mathematics, "z" is commonly used to represent a complex number. In this equation, z1 and z2 represent two complex numbers.

3. How is the angle pi/2 represented in the complex plane?

The angle pi/2 (or 90 degrees) in the complex plane is represented by the positive y-axis. This is because in the polar coordinate system, the angle is measured counterclockwise from the positive x-axis.

4. What is the significance of finding the loci represented by arg(z1/z2) = pi/2?

Finding the loci represented by this equation can help us understand the behavior of complex numbers in relation to one another. It can also be useful in solving problems involving complex numbers and their angles.

5. Can this equation be solved algebraically?

Yes, this equation can be solved algebraically by converting the complex numbers z1 and z2 into polar form and using trigonometric identities to simplify the equation. However, it may be easier to visualize and understand the solution by graphing the equation in the complex plane.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
537
  • Calculus and Beyond Homework Help
Replies
4
Views
982
  • Calculus and Beyond Homework Help
Replies
3
Views
919
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
971
  • Calculus and Beyond Homework Help
Replies
1
Views
862
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
810
  • Precalculus Mathematics Homework Help
Replies
10
Views
966
Back
Top