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

[thomas49th]

Find the loci represented by

\arg(\frac{z+1}{z-1}) = \frac{\pi}{2}

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

\frac{(x^{2}-1 + y^{2})-2iy}{(x-1)^{2}+y^{2}}

Call this w

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

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

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

What should I do (or should of done)?

Thanks
Thomas

 

[ystael]

Think more geometrically. For some complex numbers u, v, what does it mean for \arg(u/v) = \pi/2? Express u/v in polar form and see what you get.

 

[thomas49th]

u = a + ib
v = c + id
\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{c^{2}+d^{2}}} (cos(\theta_{1} + \theta_{2}) + sin(\theta_{1} + \theta_{2})

which means

\frac{\sqrt{c^{2}+d^{2}}}{\sqrt{a^{2}+b^{2}}} = \tan{\frac{\pi}{2}

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

Thanks
Thomas

 

[fzero]

seperate into real and imag bits and you should get to

\frac{(x^{2}-1 + y)-2iy}{(x-1)^{2}+y}

There's some algebra mistakes here, it should be

\frac{(x^{2}-1 + y^2)-2ixy}{(x-1)^{2}+y^2}

However, it might be easier to use

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

 

[thomas49th]

There's some algebra mistakes here, it should be

\frac{(x^{2}-1 + y^2)-2ixy}{(x-1)^{2}+y^2}

However, it might be easier to use

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

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

 

[thomas49th]

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

 

[fzero]

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

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

where

\theta_\pm = \arg(z\pm 1).

There's probably a cleaner way though.

 

[thomas49th]

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

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

where

\theta_\pm = \arg(z\pm 1).

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

 

[fzero]

Try to decide if

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

is defined. If it is, it can be written in terms of \tan x.

Edit: or just use trig identities, I made this harder than it was.

 

[thomas49th]

hmmmm. Right I'll use tan\theta = \frac{2tan\frac{\theta}{2}}{1-2tan^{2}\frac{\theta}{2}}

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

 

[fzero]

hmmmm. Right I'll use tan\theta = \frac{2tan\frac{\theta}{2}}{1-2tan^{2}\frac{\theta}{2}}

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

tan\theta = \frac{2tan\frac{\theta}{2}}{1-tan^{2}\frac{\theta}{2}}.

Just use (double-check the sign)

\tan (x - \pi/2) = -\cot x

 

[thomas49th]

Your double-angle formula is incorrect, it should be

tan\theta = \frac{2tan\frac{\theta}{2}}{12tan^{2}\frac{\theta}{2}}.

Just use (double-check the sign)

\tan (x - \pi/2) = -\cot x

huh? Where did 12 come from??

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

Thanks
Thomas

 

[fzero]

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 \theta\rightarrow \pi/2.

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

Thanks
Thomas

You show that the loci are described by

\cot (\theta_+ - \theta_-) =0

After another trig identity, we find that this reduces to a very simple condition on x and y.

 

[thomas49th]

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

 

[fzero]

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.

 

[thomas49th]

errr still not liking this.

Can I explain it like this:

if \tan \frac{\pi}{2} = \frac{1}{0}

\frac{2y}{x^{2}+y^{2}-1} = \frac{1}{0}

And here comes the dogdy bit

\frac{2y}{\frac{1}{0}} = x^{2}+y^{2}-1

\frac{0 \cdot 2y}{1} = x^{2}+y^{2}-1

0 = x^{2}+y^{2}-1

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

What do we think?

Thanks
Thomas

 

[thomas49th]

Your double-angle formula is incorrect, it should be

tan\theta = \frac{2tan\frac{\theta}{2}}{1-tan^{2}\frac{\theta}{2}}.

Just use (double-check the sign)

\tan (x - \pi/2) = -\cot x

ALSO (after my previous post), according to Stroud

Let \theta = \frac{\phi}{2}
2 \sin \frac{\phi}{2} \cos \frac{\phi}{2}
\cos \phi = \cos^{2} \frac{\phi}{2} - \sin^{2}\frac{\phi}{2}
= 1 - 2 \sin^{2}\frac{\phi}{2} = \cos^{2}\frac{\phi}{2}

\tan \phi = \frac{2 \tan \frac{\phi}{2}}{1 - 2 \tan ^{2} \frac{\phi}{2}}

Also check the previous post

 

[fzero]

errr still not liking this.

Can I explain it like this:

if \tan \frac{\pi}{2} = \frac{1}{0}

\frac{2y}{x^{2}+y^{2}-1} = \frac{1}{0}

And here comes the dogdy bit

\frac{2y}{\frac{1}{0}} = x^{2}+y^{2}-1

\frac{0 \cdot 2y}{1} = x^{2}+y^{2}-1

0 = x^{2}+y^{2}-1

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.

ALSO (after my previous post), according to Stroud

Let \theta = \frac{\phi}{2}
2 \sin \frac{\phi}{2} \cos \frac{\phi}{2}
\cos \phi = \cos^{2} \frac{\phi}{2} - \sin^{2}\frac{\phi}{2}
= 1 - 2 \sin^{2}\frac{\phi}{2} = \cos^{2}\frac{\phi}{2}

\tan \phi = \frac{2 \tan \frac{\phi}{2}}{1 - 2 \tan ^{2} \frac{\phi}{2}}

Also check the previous post

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

\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}}