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Scaling of a Circle or a Straight Line Using Complex Numbers

  1. May 20, 2013 #1
    I'm working on an assignment that is due in roughly two weeks and I'm stuck on a problem. I have what I believe may be a solution but am unsure whether or not it is 'complete'. Here is the problem:

    "Let C be a circle or a straight line. Show that the same is true of the locus of points [itex]\alpha[/itex]z, z[itex]\in[/itex] C, and [itex]\alpha[/itex] a fixed nonzero complex number." (Complex Variables 2nd Ed. by Stephen D. Fisher; pg. 22, #34)

    By multiplying each element contained in the set C by [itex]\alpha[/itex] (ie. essentially a "constant") we are scaling the set. (is this the correct interpretation of what occurs?)

    Take, for example, C to be a circle. If [itex]\alpha = 1[/itex] we have the circle C of radius R. If [itex]\alpha > 1[/itex] we have a circle with radius R' > R and if [itex]\alpha < 1 [/itex] we have a circle of radius [itex]R' < R[/itex]. That is, by multiplying each each of C by a constant complex number [itex]\alpha[/itex] that 'shape' of the set is unchanged.

    The same is true if C is a straight line: the value [itex]\alpha[/itex] instead results in a change in the slope of the original line C when [itex]\alpha = 1[/itex].


    Everything above comes simply from me considering the effect of [itex]\alpha[/itex] geometrically in my head without any real 'evidence' involved.

    I have tried to justify that the shape of C is unchanged by showing that the form of the set C is entirely equivalent when z is replaced by [itex]\alpha z[/itex]. That is, given a circle of radius R:

    C[itex]_{R}(z_{o}) = \left| z - z_{o}\right| = R[/itex]

    When z is replaced by [itex]\alpha z[/itex] we can show that the product of two complex numbers [itex]\alpha z[/itex], is itself a complex number. If we let [itex]\alpha z[/itex] = w we find:

    C[itex]_{R}(z_{o}) = \left| \alpha z - z_{o}\right| = \left| w - z_{o}\right| = R[/itex]

    This has exactly the same form of a circle and so the shape is retained. The same argument can be made for a straight line.

    Is this a valid way to show that the shape of the set C is unchanged?

    If not, I was considering another process. Again, for a circle, if we multiply all z by the constant complex number [itex]\alpha[/itex] we would expect the radius to scale by a factor [itex]\alpha[/itex] (to see this imagine a circle of radius 1 centered at the origin and choose [itex]\alpha = 2[/itex]. The result is a circle of radius 2 centered at the origin.) In this case we could write the following:

    C[itex]_{R}(z_{o}) = \left| \alpha z - z_{o}\right| = \alpha R[/itex]

    And transform the equation into the equation of a circle:

    C[itex]_{R}(z_{o}) = \left| z - z_{o}\right| = R[/itex]

    I thought this would be more 'proper' but keep getting stuck. Is my first approach valid or should I follow the second - or am I missing something entirely?


    Lastly, am I getting the question conceptually? The real part of [itex]\alpha[/itex] is a scaling factor and the imaginary part is a rotational factor - is this correct? I feel a bit more confused for the case of C as a straight line, but I'll work on that once I understand it for the circle.

    Any help is greatly appreciated. :) Let me know if anything I have done is unclear and I'll try to clarify it!
     
  2. jcsd
  3. May 20, 2013 #2

    CompuChip

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    Your approach works, but it is not completely rigorous. What you would like is to get the formula after the substitution ##z \to \alpha z## back to the form ##|z - z_0'| = R'## back, where z0' and R' are (different) complex numbers.

    To do this, you may want to use that ##|\alpha w| = |\alpha| |w|## for complex numbers ##\alpha, w##. Fiddle with the algebra a bit, and you will be able to define z0' in terms of z0 and α; and R' in terms of R and α.
     
  4. May 21, 2013 #3
    Thanks a lot, CompuChip! I managed to manipulate it into the form you described above. Is the proof for a straight line similar? A straight line is of the form:

    [itex]0 = Re(az + b)[/itex], where [itex]a = A + iB[/itex] and b is a complex number is the straight line [itex]0 = Ax - By + Re(b)[/itex]

    Should I be replacing [itex]0 = Re(az + b)[/itex] with [itex]0 = Re(a \alpha z + b)[/itex] and trying to manipulate it into the form [itex]0 = Re(az' + b)[/itex]?
     
  5. May 26, 2013 #4
    I apologize for double-posting, but I returned to edit my post and found that option unavailable.

    I have managed to complete the question for the case of a circle but would like to check my solution for the case of a straight line. As I described above, a straight line can be described by complex numbers by writing:

    [itex]0 = Re(az +b)[/itex], where [itex]a = A + iB[/itex] and b [itex]\in[/itex] ℂ. This describes the line [itex]0 = Ax - By + Re(b)[/itex].

    By replacing all instance of [itex]z[/itex] with [itex]\alpha z[/itex] we must manipulate the equation [itex]0 = Re(a\alpha z + b)[/itex] into the form [itex]0 = Re(a'z + b')[/itex]. Without any mathematical effort I predict [itex]a' = a\alpha, b' = b[/itex].

    [itex]0 = Re(a\alpha z + b)[/itex]
    [itex]0 = Re((A+iB)\alpha z + b)[/itex]
    [itex]0 = Re((A\alpha +iB\alpha)z + b)[/itex]
    [itex]0 = Re((A' +iB')z + b)[/itex]
    [itex]0 = Re(a'z + b')[/itex]

    Here we see that [itex]a' = A' + iB' = A\alpha + iB\alpha = (A+iB)\alpha = a\alpha, b' = b[/itex] as I predicted above. Is this a sufficient proof of the claim that if S is straight line then the locus of points [itex]\alpha z[/itex] is also a straight line for any [itex]z \in S[/itex]?
     
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