argz

[kingjewel1]

Hi guys!

I'm told arg( z/z-2)=pi/4

How do i find the locus of this?

so far i did. but i don't know if this is correct.

 

from argz - argz-2=pi/4

[math]\frac {\frac{y}{x}-\frac {y}{(x-2)}}{1+\frac{y}{x}\frac{y}{x-2}}[/math]=[math]1[/math]

 

then

[math]=> (xy-2y-yx)(x-2)=(2x-2)x(x-2)[/math]

therefore [math]-2y=2x-2[/math]

so y=1-x

 

Thanks in advance

[kingjewel1]

fixed it! mistake in the algebra

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

centre (1,-1) radius 4

correct?

[matt grime]

are you claiming that the locus is the circle of center (1,-1) and radius 4? because there is certainly a real number w on that locus and thus w/(w-2) is real and its argument therefore isn't pi/4

[kingjewel1]

no. you're right, my mistake radius should be 2, should it not?

because sqrt(1^2+1^2-(-2)=2

[kingjewel1]

What would this locus look like?

l

[DQW]

no. you're right' date=' my mistake radius should be 2, should it not?

because sqrt(1^2+1^2-(-2)=2[/quote']This still doesn't fix the problem matt pointed out. There must be a mistake in the algebra...will look at it sometime later.

[matt grime]

just have a google for locus points complex plane, i'm sure that you'll find a page explaining how to do these locus questions (it isn#t a circle)

[DQW]

(it isn#t a circle)

It isn't ???

 

I get the same locus that the OP has in the second post. Only it's a circle of radius [imath]\sqrt{2} [/imath]. The two real points on the circle are at (0,0) (for which, z/(z-2) is the origin and hence has any argument you choose to give it) and at (2,0) (which must be excluded from the domain for z/(z-2) to belong in C).

 

Surely I'm making some stupid mistake ...

[matt grime]

no, perhaps i am misremembering a course a took several years ago, though as you yourself state it isn't a circle so i don't see the issue.

[DQW]

...as you yourself state it isn't a circle ...

I don't recall doing this. In any case, I think it very much is a circle - and exactly the one whose equation is written in post #2. It is a circle centered on (1,-1) with radius \sqrt{2}.

 

To the OP : An alternate approach (there's nothing wrong with what you've done) is to write z = x+iy and hence express z/(z-2) as some a+ib and find b/a.

[matt grime]

You state that the some ponts of said circle are not on the locus. thus you have stated locus is a circle with some points removed. a circle with 2 points ommited is not a circle.

[matt grime]

consider the transform M(z)=z/(z-2) this is a Mobius map: it takes circles and straight lines to circles and straight lines. The locus L you want satisfies z in L iff arg(M(z))=pi/4, thus M transrforms L into the half line starting at the origin and going at an angle of pi/4 to the real axis. L is then the inverse image of this half line under M. L is therefore either another half line or *some* segment of a circle. clealry it is only some arc of the circle you have given, not all of it. it will be one of the two arcs from the real points on the circle.

[DQW]

You state that the some ponts of said circle are not on the locus.

Yes, I don't listen to myself, do I ?

 

consider the transform M(z)=z/(z-2) this is a Mobius map: it takes circles and straight lines to circles and straight lines. The locus L you want satisfies z in L iff arg(M(z))=pi/4, thus M transrforms L into the half line starting at the origin and going at an angle of pi/4 to the real axis. L is then the inverse image of this half line under M. L is therefore either another half line or *some* segment of a circle. clealry it is only some arc of the circle you have given, not all of it. it will be one of the two arcs from the real points on the circle.

Okay, clearly the lower arc works (test with z=-2i). Naturally, the upper arc maps onto arg(M(z))=5pi/4

[kingjewel1]

Cheers guys!

I have to admit, I don't follow all your workings, as I'm not familiar with transformations.... (looking them up in a min).

 

But as you both say: the locus is two segments of a circle, which seems right to me.

The first line from (0,0) with arg pi/4, okay, but its inverse would not meet the original line apart from at (0,0) So how do i know where these two converge on the edge of the circle?

[matt grime]

it is one of two segments of a circle.

[DQW]

The first line from (0,0) with arg pi/4, okay, but its inverse would not meet the original line apart from at (0,0) So how do i know where these two converge on the edge of the circle?

I can't follow any part of this. What is it that you want to have "converge" on the circle ? And why ?

[kingjewel1]

Sorry.

I've got to admit I don't know what it's supposed to look like exactly, but as far as i can see, we're creating a cirleshape with a segment missing, ie between the two z points.

But i cant find an example method to use(not in my book or net) to find the proper segment and so i can draw it. Any ideas of where i could look? Cheers that's great!

[DQW]

Yes, essentially your circle is divided into 2 segments by the two real points (0,0) and (2,0). One of these two segments is the required locus (the other segment contains points which map onto a slope of 5pi/4; note that tan(5pi/4) = 1 as well). You can determine which one of the two segments it is by simply choosing a convenient point from each segment and substituting into the given function to see if the equation is satisfied.

 

I've actually done this for you in post #13.