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Complex locus

  1. Dec 12, 2008 #1


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    1. The problem statement, all variables and given/known data
    Given [tex]Z^2=\frac{1+z}{1-z}[/tex] if [tex]|z|=1[/tex], find the locus of Z

    2. Relevant equations
    let [tex]Z=X+iY[/tex]

    3. The attempt at a solution





    and since [tex]x^2+y^2=1[/tex]


    But [tex]Z^2=(X+iY)^2=X^2-Y^2+i2XY[/tex]

    Therefore, [tex]X^2-Y^2=0[/tex]

    and [tex]2XY=\frac{y}{1-x}[/tex]

    The problem is that I can't rearrange these two equations to express the locus in terms of x and y. Any help is greatly appreciated as always.
  2. jcsd
  3. Dec 12, 2008 #2
    This problem requires a geometrical/graphical understanding before you can use any algebraic methods of solving it. I'm going to give you this insight, and the rest you can solve yourself. Your original try at a solution is not wrong, it just lacks an additional constraint.

    The complex number z can also be interpreted as a 2 dimensional vector (x,y) where x is the real part and y the imaginary part. 1 + z and 1 - z are complex numbers (or vectors) where there is an offset of ±1 in the x-direction. So you have for example, for z = 1 + i, the numbers: 2 + i and - i and their vectors (2,1) and (0,-1). Because |z| = 1, both 1 + z and 1 - z describe a circle with radius 1 and with respectively an offset of -1 and +1.

    Now capital z or Z is a complex or real number. It describes a ratio between 1 + z and 1 - z, where these numbers or vectors must lie on their respective circle. You should be able to figure out on your own that [tex]|Z^2|[/tex] must equal 1 (or the ratio of the lengths of the vectors is 1). You can use this to solve the rest of the problem. In fact, you already have the necessary parts to find the loci.
  4. Dec 12, 2008 #3


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    Sorry, but I couldn't follow this part, and onwards.
    For [tex]z+1[/tex] it will be a circle with radius 1 (since [tex]|z|=1[/tex]) and centre (-1,0)?
    And for [tex]1-z[/tex] it will be a circle centre (1,0)?
  5. Dec 13, 2008 #4
    I must apologize, I was completely wrong, offset values, etc. and there is one thing that must be wrong with this assignment or else it wouldn't make much sense.

    To correct:
    With 1 + z and 1 - z, the offsets are the same, +1, but they still describe a circle because |z| = 1. So basically, the vectors I described my previous reply, lie on the same circle with an offset of +1 and they 'rotate' in the same direction, however, they have a phase difference.

    Algebraic proof of the phase difference:
    [tex]1 + z = 1 + re^{i\phi} = 1 + e^{i\phi} [/tex]

    [tex]- e^{i\phi} = e^{i\pi}e^{i\phi} = e^{i(\phi + \pi)}[/tex]

    [tex]1 - z = 1 - re^{i\phi} = 1 - e^{i\phi} = 1 + e^{i(\phi + \pi)}[/tex]

    The phase difference is [tex]\pi[/tex].

    I was wrong to say that [tex]|Z^2|[/tex] must equal 1, because I claimed you could derive it from a geometrical interpretation. But [tex]Z^2[/tex] is definitely a ratio and [tex]|Z^2|[/tex] is definitely the ratio between the lengths of two complex numbers. The main problem is that Z is a free variable. However, [tex]|Z^2|[/tex] is a real number and must be equal or greater than zero, because it's a ratio of two semi-positive numbers.

    You can find the locus by solving:
    [tex]|Z^2|^2 = k = |\frac{1+z}{1-z}|^2[/tex]

    with k a semi-positive number. The only freedom in the solution is k. It will show that for varying k from 0 to infinity that 1 + z and 1 - z both describe separate half-circles.
  6. Dec 13, 2008 #5


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    What exactly is wrong with the assignment?

    What are you implying that he solve this for? It's clearly true for all k, all z and all Z. :confused:

    Remember that |z|=1, so you can write z in the polar form as [itex]z=cis(\theta)[/itex]. That allows you to express x and y as [itex]\cos(\theta)[/itex] and [itex]\sin(\theta)[/itex] respectively.

    Now, let's look at your first equation: [tex]X^2-Y^2=0[/tex]....what do the solutions to this equation look like?

    Given that; if X and Y satisfy both equations, what are the possible values of theta for which any solution to [tex]2XY=\frac{\sin(\theta)}{1-\cos(\theta)}[/tex] will also satisfy [tex]X^2-Y^2=0[/tex]?
  7. Dec 13, 2008 #6
    It's me that's confused by this assignment. But nvm, ignore my replies. They make no sense.
  8. Dec 13, 2008 #7


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    Sorry D.R. but I just can't seem to follow where you are trying to go with it all. Also, there are too many terms being used that I'm not familiar with and it makes it quite unbearable to comprehend.

    ok well [tex]X^2-Y^2=0 \Rightarrow X^2=Y^2[/tex] therefore the lines [tex]Y=\pm X[/tex].

    I honestly don't know what the possible solutions are to the last part.
  9. Dec 13, 2008 #8
    Hey, no problem. It's my mistake screwing up the hints for not understanding the concept of this question.
  10. Dec 13, 2008 #9


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    Hmmm...I think my question may have been worded poorly...let's try it again... You have a system of two equations and two variables (X,Y) which you wish to find solutions for (your solutions will depend on theta--the phase angle of little z--)...You know that the only possible solutions to the 1st equation are the lines [itex]Y=\pm X[/itex], so the only possible solution to the system of equations will come from substituting [itex]Y=\pm X[/itex] into your second equation ans solving for X and then plugging your X solutions back into [itex]Y=\pm X[/itex] to find your possible (X,Y) pairs.

    What do you get for possible solutions (you should get 4 of them) when you do that?
  11. Dec 13, 2008 #10


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    Well then that makes two of us :biggrin:

    ok I will try that:

    [tex]Y=\pm X[/tex] (1)

    [tex]2XY=\frac{sin\theta}{1-cos\theta}[/tex] (2)

    Substituting (1) into (2) [tex]\pm 2X^2=\frac{sin\theta}{1-cos\theta}[/tex]

    Therefore, [tex]X=\pm \sqrt{\pm \frac{sin\theta}{2-2cos\theta}}[/tex] (3)

    Substituting (3) back into (1) [tex]Y=\pm \sqrt{\pm \frac{sin\theta}{2-2cos\theta}}[/tex] (4)

    I'm unsure if these equations are completely correct so please correct me if any of my steps seem to be wrong. But now what do I do? This doesn't seem to be the locus yet :yuck:
  12. Dec 13, 2008 #11


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    Okay good, so your possible solution pairs are


    Now, remember your definition [tex]Z=X+iY[/tex] where X and Y are real....That tells you that for a given value of theta, only solutions that have real values for X and Y are valid.

    So, let's see what happens when you slowly go counterclockwise around the circle |z|=1; by starting with [itex]\theta=0[/itex] and slowly increasing it to [tex]\theta=2\pi[/tex]....What are your real solutions for [itex]\theta=0[/itex]? How about [itex]\theta=\frac{\pi}{4}[/itex]? How about [itex]\theta=\frac{\pi}{2}[/itex], [itex]\theta=\frac{3\pi}{4}[/itex], [itex]\theta=\pi[/itex], [itex]\theta=\frac{5\pi}{4}[/itex], [itex]\theta=\frac{3\pi}{2}[/itex] and [itex]\theta=\frac{7\pi}{4}[/itex]?...what do these solutions look like?

    Apparently your locus is the two lines [itex]Y=\pm X[/itex], and as you go counterclockwise around the circle |z|=1, You start at the points [itex](X,Y)=(\infty,\infty)[/itex] and [itex](X,Y)=(-\infty,-\infty)[/itex] and move inwards along the line [itex]Y=X[/itex] from both directions until you reach the origin at [itex]\theta=\pi[/itex]. Then as you continue along the unit circle, you find that you move from the origin outwards in both directions along the line [itex]Y=-X[/itex].

    Make sense?
  13. Dec 13, 2008 #12


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    I can see how as we travel anticlockwise around [tex]|z|=1[/tex], the locus of Z travels along the lines [tex]Y=\pm X[/tex], but these multiple solutions are getting to be too much. I understand that X and Y need to be real values, so any negatives within the surd are discarded. So for e.g.

    For [tex] \theta=0, (X,Y)=(0,0)[/tex]

    For [tex]\theta=\frac{\pi}{4}, (X,Y)=(\pm \sqrt{\frac{1}{2\sqrt{2}-2}},\pm \sqrt{\frac{1}{2\sqrt{2}-2}})[/tex]

    For [tex]\theta=\frac{\pi}{2}, (X,Y)=(\pm \sqrt{\frac{1}{2\sqrt{2}}},\pm \sqrt{\frac{1}{2\sqrt{2}}})[/tex]

    For [tex]\theta=\frac{3\pi}{4}, (X,Y)=(\pm \sqrt{\frac{1}{2\sqrt{2}+2}},\pm \sqrt{\frac{1}{2\sqrt{2}+2}})[/tex]

    For [tex]\theta=\pi, (X,Y)=(0,0)[/tex]


    Is this the correct locus of Z? It just 'moves' up and down along the lines [tex]Y=\pm X[/tex] while the locus of z moves anticlockwise around [tex]|z|=1[/tex]?
  14. Dec 13, 2008 #13


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    Are you sure about this one....what is 0/0?

    Approximate numerical values would probably be more enlightening.

    You seem to be confusing the 'locus of Z' with the particular value of Z at a given theta. The locus of Z is all possible values of Z; which in this case is every point on the lines [itex]Y=\pm X[/itex]. Therefor the locus of Z are the entire [itex]Y=\pm X[/itex] lines.

    It's the value of Z that changes as you go around the circle |z|=1. It's the value of Z that depends on the phase of z. The locus of Z is the set of all of those possible values.
  15. Dec 13, 2008 #14


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    Oh yes I completely skipped the denominator and assumed 0. I hope I express this correctly: So Z is undefined when the phase of z is 0.

    True, but to get a clear picture of how Z moves as the phase of z rotates anticlockwise, I would need a few more angles, mainly those closer to [tex]0^+[/tex] and [tex]0^-[/tex].

    Thankyou, this is the sort of clarification I'm looking for, as since I've yet to grip my mind around this topic, using the correct terminology and understanding what I'm saying are still very fuzzy.

    Ok so the locus of Z is [tex]Y=\pm X[/tex].
    since Z is satisfied by [tex]X^2-Y^2=0[/tex] and [tex]2XY=\frac{y}{1-x}[/tex] where [tex]Z=X+iY[/tex] and [tex]z=x+iy[/tex], then as the phase of z rotates anticlockwise around the unit circle (since [tex]|z|=1[/tex]), as the phase of [tex]z=0^+[/tex] the movement(?) of Z is from [tex](\pm \infty, \pm \infty)[/tex] towards the origin, as [tex]z\rightarrow \pi[/tex]. As z rotates from [tex]\pi \rightarrow 2\pi[/tex], the locus of Z moves towards infinite but backwards? (this part I cannot seem to visualize very clearly).

    I hope I'm on the right track. Thanks again gabbagabbahey.
  16. Dec 13, 2008 #15


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    If that's what you feel you need to get a clear picture, then by all means plug some values closer to 0 into the equation.

    You should find that for every value of theta, there are two possible values of Z. This should not come as a big surprise since Z was defined by a condition on Z^2, not a condition on Z.

    Yes. :smile:

    Not really. For each value of little z, there are two values of big Z. If you start with the phase of little z equal to zero, and slowly increase its phase you will see the following:

    --Z is undefined at [itex]\theta=0[/itex], but at [itex]\theta=0^{+}[/itex], it takes on two values [itex]Z=\infty+\infty i[/itex] and [itex]Z=-\infty-\infty i[/itex]. As you increase theta, the two values of Z move closer and closer to the origin along the line [itex]Y=X[/itex] until finally at [itex]\theta=\pi[/itex] the two points coincide at Z=0. Then as you continue to increase theta, you see the two values of Z move away from each other along the line [itex]Y=-X[/itex], outward to infinity.

    The locus of Z doesn't move. The locus of Z is the collection of all possible values of Z, and all possible values of Z in this case, are all possible values along the lines [itex]Y=\pm X[/itex]. Therefor, the locus of Z is the two lines [itex]Y=\pm X[/itex].

    If you are still having trouble visualizing it, I suggest you plot a whole bunch of points in the complex plane.
  17. Dec 13, 2008 #16


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    Oh no I just meant that it should probably be shown that as [tex]\theta[/tex] is just slightly positive, Z is close to [tex]\pm \infty[/tex].

    It was quite confusing when the 2nd [tex]\pm[/tex] came into play. I wasn't sure if the x-coordinate of Z was going to be positive/negative, or the y-coordinate, or both :confused:

    So thats how the [tex]\pm[/tex]'s worked for the values of Z... I had no idea what to do with it all!

    I'm trying to solve complex locus problems before I even know the meaning of the term 'locus' :biggrin:

    Your explanation was very clear and concise, I was able to follow it easily and its great that I can skip this tedious task! Thanks!

    ok well I think this problem was put to rest. Gabbagabbahey, thank you once again! Your help is very much appreciated!
  18. Dec 13, 2008 #17


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    The easiest way to keep the +/- 's straight is to treat two different cases: (1) solve the second equation using Y=+X and (2)Solve using Y=-X.

    You should find that case (1) gives real solutions for X and Y only for [itex]0<\theta\leq\pi[/itex] and that case (2) only give real solutions for [itex]\pi\leq\theta <2\pi[/itex].
  19. Dec 14, 2008 #18


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    lol oh yeah. Its been 2 years since I began using [tex]\pm[/tex]'s and I quickly 'grew out' of considering them as 2 cases and working with them in that way, since I didn't need to do so from experience. Its weird that when there are multiple [tex]\pm[/tex]'s in an equation, that they aren't distinguished between each other with a number or something else.
    Thanks gabbagabbahey, your help was awesome.
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