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Homework Help: Complex locus

  1. Dec 9, 2008 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    Given [tex]Z=\frac{z-2}{z}[/tex], if [tex]|z|=1[/tex] prove that the locus of Z is another circle whose centre and radius must be determined. Also describe the direction of Z as z describes the unit circle in an anticlockwise direction.


    2. Relevant equations
    [tex]z=x+iy[/tex]


    3. The attempt at a solution
    [tex]Z=\frac{x+iy-2}{x+iy}(\frac{x-iy}{x-iy})[/tex]

    expanded and simplified: [tex]\frac{x^2-2x+y^2+i2y}{x^2+y^2}[/tex]

    I think since [tex]|z|=1[/tex] then [tex]x^2+y^2=1[/tex]?

    This leaves [tex]-2x+i2y[/tex] and I am completely lost at this point...
     
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  3. Dec 9, 2008 #2

    gabbagabbahey

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    Doesn't it leave you with [tex]Z=(1-2x)+(2y)i[/tex]?....What does that say about [tex]|Z|[/tex]?
     
  4. Dec 10, 2008 #3

    Mentallic

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    Ahh yes that was a typo.

    [tex]\frac{x^2-2x+y^2+i2y}{x^2+y^2}[/tex]

    [tex]|z|=1, x^2+y^2=1[/tex]

    therefore simplified to: [tex]Z=1-2x+i2y[/tex]

    [tex]|Z|=x^2+y^2=(1-2x)^2+(2y)^2[/tex]

    Rearranged to: [tex]|Z|=4(x-\frac{1}{2})^2+4y^2[/tex]

    This is in the form of a circle (in cartesian form) but I don't know what it has to do with complex numbers, since on the argand diagram [tex]|z|=r[/tex] is a circle radius r.
     
  5. Dec 10, 2008 #4

    gabbagabbahey

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    For a circle centered at 0+0i, yes. Since |Z| is not of this form, you know it's center must be at a different point in the complex plane. If its center is at the point [itex]\alpha=a+bi[/itex], then you know [itex]|Z-\alpha|=r[/itex]....correct?

    So...what is [itex]|Z-\alpha|[/itex]? In particular, what is [itex]|Z-1|[/itex]? :wink:
     
  6. Dec 10, 2008 #5

    Mentallic

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    Ahh [tex]|Z-a|=r[/tex] would be a circle radius r centred at (a,bi)

    So [tex]|Z-1|=r[/tex] is a circle centre (1,0i) radius r.

    But how can [tex]|Z|=4(x-\frac{1}{2})^2+4y^2[/tex] be transformed into this form? r is real number that is more than 0 is it not?
     
  7. Dec 10, 2008 #6

    gabbagabbahey

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    Well, what is Z-1? So what is |Z-1|?
     
  8. Dec 10, 2008 #7

    Mentallic

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    I'm starting to lose the plot here :cry:
    Are we talking about z or Z now? Remember from the question that [tex]Z=\frac{z-1}{z}[/tex] where [tex]z=x+iy[/tex]
     
  9. Dec 10, 2008 #8

    gabbagabbahey

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    We're talking about capital Z now. If its Locus is a circle with radius r, centered at the point [itex]\alpha=a+bi[/itex] in the complex plane, then [itex]|Z-\alpha|=r[/itex]

    Do you understand that part?


    ....Now, you've already calculated that [itex]Z=(1-2x)+(2y)i[/itex], so what is [itex]Z-1[/itex]?
     
  10. Dec 10, 2008 #9

    Mentallic

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    Yes, but I just hope I can apply it to the question.


    Well, if [tex]Z=1-2x+i2y[/tex] then [tex]Z-1=-2x+i2y[/tex] (1)
    but I also calculated it from scratch ~

    [tex]Z-1=\frac{z-1}{z}-1[/tex]

    [tex]=\frac{z-1-z}{z}[/tex]

    [tex]=\frac{-1}{z}[/tex]

    [tex]=\frac{-1}{x+iy}(\frac{x-iy}{x-iy})[/tex]

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

    therefore, [tex]Z-1=-x+iy[/tex] (2)

    which is not the same to (1)?
     
  11. Dec 10, 2008 #10

    gabbagabbahey

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    That's because [tex]Z=\frac{z-2}{z}[/tex] not (z-1)/z.
     
  12. Dec 10, 2008 #11

    Mentallic

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    oh lol how embarrassing :blushing:

    ok [tex]Z-1=-2x+i2y[/tex]

    hence, [tex]|Z-1|=(-2x)^2+(2y)^2=4x^2+4y^2[/tex]

    um.. I'm taking a stab at this but does this mean [tex]|Z-1|=4(x^2+y^2)=4[/tex]?

    In which case we have a circle centre (1,0i) radius 4?
     
  13. Dec 10, 2008 #12

    gabbagabbahey

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    Looks good to me :smile:

    That's why I chose alpha=1, to get Z-alpha into the above form.

    Edit- shouldn't it be [itex]|Z-1|=\sqrt{4(x^2+y^2)}=2[/itex] ?:wink:
     
  14. Dec 10, 2008 #13

    Mentallic

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    Awesome! But what method or how did your intuition tell you to use alpha=1?

    For this part I'm unsure what it is asking and how to find it mathematically. i.e. I am completely stumped.
     
  15. Dec 10, 2008 #14

    Mentallic

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    Ahh yes. For all the questions I've been provided to date, it has been with |z|=1. Totally forgot about the pythagorean method lol.
     
  16. Dec 10, 2008 #15

    gabbagabbahey

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    Well, I can't explain why my intuition was what it was, but I can give you a logical step by step reasoning for choosing alpha=1:

    (1)Start by assuming a general alpha [itex]\alpha=a+bi[/itex]
    (2)Calculate [tex]|Z-\alpha|=\sqrt{[(1-2x-a)+(2y-b)i][(1-2x-a)-(2y-b)i]}=\sqrt{(1-2x-a)^2+(2y-b)^2}[/tex]
    (3) You're told to show that the Locus of Z is a circle, so you know you want [itex]|Z-\alpha|[/itex] to be some constant radius 'r'; and the only restriction you have regarding x and y is that x^2+y^2=1, so you need to choose a and b in such a way that [itex]|Z-\alpha|=f(x^2+y^2)[/itex]. Choosing a=1 and b=0 clearly accomplishes that.

    I think you need to express both z and Z in polar form. If you travel around the circle |z|=1 counter clockwise, is the phase of z increasing or decreasing? What happens to the phase of Z as the phase of z increases?
     
  17. Dec 10, 2008 #16

    Mentallic

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    That works by inspection, but I'll have to see how 'inspecting' works for the more difficult ones.


    Sorry I've never learnt or even heard of the term 'phase'. It could just be that you are using different terminology to what our class is accustomed to.

    gabbagabbahey I really appreciate your help. If you are capable of continuing with it for this next part, then please, I would love to have a taste of your wisdom :smile: if however, you aren't sure about what is required, then all you need to do is say so.

    I graphed these circles in the cartesian plane. Maybe it would make more sense by visualising it? http://img380.imageshack.us/img380/2742/complexcirclelocusxc0.png [Broken]
    http://g.imageshack.us/img380/complexcirclelocusxc0.png/1/ [Broken]
     
    Last edited by a moderator: May 3, 2017
  18. Dec 10, 2008 #17

    gabbagabbahey

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    Well, I assume you have at least been taught that any complex number z can be represented not only in the form x+yi, but also in polar form: [itex]z=|z|e^{i\phi}[/itex]?

    [itex]\phi[/itex] is called the phase angle of z. What happens to [itex]\phi[/itex] as you go counterclockwise around the unit circle?
     
  19. Dec 10, 2008 #18

    Mentallic

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    No sorry, we've only been taught [tex]z=x+iy=rcis\theta[/tex].

    I might have to wait till we have class again then. Thank you so much for your help gabbagabbahey!
     
  20. Dec 10, 2008 #19

    gabbagabbahey

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    cis(theta) is another way of writing [itex]e^{i\theta}[/tex]....Using you definition of polar form, theta is what I am calling the phase angle.

    What happens to theta as you go counterclockwise around the unit circle? Does it increase or decrease?

    If [tex]z=cis(\theta)[/itex], what is [tex]Z[/tex]?
     
  21. Dec 10, 2008 #20

    Mentallic

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    Thats a funny way of expressing the same thing :smile:

    Well theta obviously increases as you go counter-clockwise around the unit circle. For Z, I'm not sure if I calculate theta from the centre of the circle it defines (1,0i) or from the origin, but nonetheless theta still increases as we travel counterclockwise.
     
  22. Dec 10, 2008 #21

    gabbagabbahey

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    I meant what is Z in terms of the phase angle of z?

    As you said, as you go counterclockwise around the unit circle, the phase angle of z will increase; so the idea is to see what happens to Z as that phase angle increases.
     
  23. Dec 10, 2008 #22

    Mentallic

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    Are we trying to express this relationship in terms of the phase angle theta and Z=x+iy?

    This will be another guess ~ let [tex]z=cis\theta[/tex]

    therefore, [tex]Z=\frac{cis\theta -2}{cis\theta}[/tex]

    [tex]=\frac{cos\theta -2+isin\theta}{cos\theta +isin\theta}(\frac{cos\theta -isin\theta}{cos\theta -isin\theta})[/tex]

    [tex]=\frac{cos^2\theta -2cos\theta +i2sin\theta -isin\theta cos\theta +isin\theta cos\theta +sin^2\theta}{cos^2\theta +sin^2\theta}[/tex]

    Simplified: [tex]Z=1-2cos\theta +i2sin\theta[/tex]

    Here I am stuck again. Am I at least on the right track?
     
  24. Dec 10, 2008 #23

    gabbagabbahey

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    Yup, that looks real good :smile:

    Now notice that [itex]cis(-\theta)=\cos\theta-i\sin\theta[/itex] so [itex]Z=1-2cis(-\theta)[/itex]

    ...follow?

    Now, since we know that Z travels in a circle with radius 2 around the point (1,0i), let's look at (Z-1)/2: [tex]\frac{Z-1}{2}=-cis(-\theta)=cis(-\theta+\pi)[/tex]

    ....follow?

    What does that tell you about the direction of travel around the circle |Z-1|=2 as you go counterclockwise around |z|=1?

    What does it tell you about the relative phase of the two motions?
     
  25. Dec 10, 2008 #24

    Mentallic

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    With a bit of my own slower step-by-step process, I was able to follow where you went with this.

    From looking at [tex]\frac{Z-1}{2}=cis(-\theta +\pi}[/tex] I am guessing that [tex]|Z-1|=2[/tex] is moving clockwise (from [tex]-\theta[/tex]) beginning at -1 (from [tex]+\pi[/tex]). This again is another guess :cry:

    I know how frustrating this must be for you, as I've been in your shoes quite a few times before trying to tutor some in my class and when they don't get the point even after I take steps very slowly, it can get really annoying.
     
  26. Dec 10, 2008 #25

    gabbagabbahey

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    Yes, as you move counterclockwise around |z|=1, You move Clockwise around |Z-1|=2.

    Another way to see this is to call the phase of Z-1 [itex]\phi[/itex]. Since |Z-1|=2, expressing Z-1 in polar form gives: [itex]Z-1=2cis(\phi)[/itex].When you compare that to our previous result, you see clearly that [itex]\phi=-\theta+\pi[/itex].
    That means that as the phase of z (theta) increases, the phase of Z-1 (phi) decreases. So as you go counterclockwise around |z|=1, you go clockwise around |Z-1|=2.

    It also means that the two motions are 180 degrees out of phase, since there is a phase difference of pi radians (i.e. [itex]\phi-(-\theta)=\pi[/itex])

    ....Does that make it a little clearer?
     
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