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Homework Help: Complex number generalizations

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Use de Movire to find solutions for the following:

    z^5 = i
    z^4 = i
    z^3 = i

    Find generalization for z^n = x+iy,
    where modulus of x+iy is 1

    Explore when |x+iy| is not equal to 1

    2. Relevant equations

    [rcis(theta)] = (r^n)cis(theta*n)

    r = /sqrt(y^2 + x^2)

    theta = arctan(y/x)

    Finding the roots of complex numbers: (r^1/n)cis(2kpi/n), where k = 0,1,...n-1

    3. The attempt at a solution

    I tried using moivre's theorem to find the roots of z, but I ended up with theta=arctan(1/0), so it is undefined. Yes, and now I'm stuck. Could the answer possibly be undefined/no solution?
  2. jcsd
  3. Nov 18, 2011 #2

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    Welcome to PF, 314Jason! :smile:

    arctan(1/0) is indeed undefined, but that only means that theta is either pi/2 or -pi/2.
    In your case, for the complex number "i", the corresponding theta is pi/2.

    Your problem is arctan. It is not quite the right function to find theta.
    What you actually need is atan2(x, y) or atan2(y, x) to find theta.
    See http://en.wikipedia.org/wiki/Atan2 for details.
    Or otherwise, you could write:
    [tex]\theta = \textrm{Arg}(x+iy)[/tex]
    See http://en.wikipedia.org/wiki/Argument_(complex_analysis)" [Broken] for details.
    Last edited by a moderator: May 5, 2017
  4. Nov 19, 2011 #3
    Oh okay, so atan2(y, x) is the angle between the point (x,y) and the real axis. And when y>0 and x=0, theta = pi/2. YAY!

    I've done the all z^3, z^4 and z^5 = i. Turns out, the conjecture for the distance between two adjacent roots (when z^n = i) is the same as when z^n = 1. Which is 2|sin(pi/n)|

    Now I have to explore when |x+iy| ≠ 1. On the graph, this the modulus of z only adds to the radius (when mod(z) > 1) or subtracts to the radius (when mod(z) < 1).

    In de moivre's theorem, z = r^(1/n) cis(ø + 2kpi)
    r = mod(z)
    ø = arg(z)
    n = number of roots / exponent of z
    k = 0, 1, ..., n-1
    r^(1/n) is equal to the radius of the circle: x^2 + y^2 = r^(2/n)

    THEN I wanted to find the conjecture interms of r, ø, n and k

    z = r^(1/n)cis(ø+2kπ) <------------ de moirve
    where r = mod(z) = √(x^2 + y^2)
    where ø = arg(z) = atan2(x,y) = arctan (y/x)
    where n = number of roots / exponents of z
    where k = 0,1,...,n-1
    distance formula = √(x2+y2)
    Point 1(r^(1/n)cos(ø), r^(1/n)sin(ø))
    Point 2(r^(1/n)cos(ø+2pi), r^(1/n)sin(ø+2pi))

    I plugged the two points into the distance formula and got an answer of 0. I even put it through Wolframalpha (without the r^(1/n)) and got 0.


    ^ this is after I applied trig's sum-to-product formulas and extracted r^(1/n) since they're common.

    THE DISTANCE CANNOT BE ZERO! What's wrong!? :(
  5. Nov 19, 2011 #4

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    I think you've got the wrong distance between 2 adjacent roots.
    It's not 2|sin(pi/n)|.
    It should be |cis(2pi/n)-1|.

    As for the distance when the mod is not 1.
    Note that the distances between successive adjacent roots will not be the same.

    What you have is the first 2 roots, but you've got their argument (angle) wrong.
    Apparently you forgot to divide the angle by n.
    It should be:
    Point 1(r^(1/n)cos(ø/n), r^(1/n)sin(ø/n))
    Point 2(r^(1/n)cos((ø+2pi)/n), r^(1/n)sin((ø+2pi)/n))
  6. Nov 20, 2011 #5
    I don't understand how :cry:

    Here's how I got the generalization:

    When z^n = 1, z = cis(2kπ/n), where k = 0,...,n-1

    distance formula = √(x^2+y^2)
    P1(1, 0)
    P2(cos(2pi/n), sin(2pi/n)

    d = √[ (cos(2pi/n)-1)^2 + (sin(2pi/n)-0)^2 ]
    = √[ cos^2(2pi/n) - 2cos(2pi/n) + 1 + sin^2(2pi/n) ]
    = √[-2cos(2pi/n)+2]
    = √[-2 ( 1-2sin^2(pi/n) ]+2]v
    = √[-2+4sin^2(pi/n)]+2]
    = √[4sin^2(pi/n)]
    = ± 2[sin(pi/n)] <---- CONJECTURE
    Since distance cannot be negative,d = 2|sin(pi/n)|

    When z^n = i, z =cis[(π + 4kπ)/2n], where k = 0,...,n-1

    distance formula = √(x2+y2)
    P1(cos(pi/2n), sin(pi/2n))
    P2(cos(5pi/2n), sin(5pi/2n))

    d = √[ (sin(5pi/2n)-sin(pi/2n))^2 + (cos(5pi/2n)-cos(pi/2n))^2 ]

    Using trig's sum-to-product formula...

    = √[ (2cos(3pi/n)sin(2pi/n))^2 + (-2sin(3pi/n)sin(2pi/n))^2 ]
    = √[ 4cos^2(3pi/n)sin^2(2pi/n) + 4sin^2(3pi/n)sin^2(2pi/n) ]
    = √[ 4(1-sin^2(3pi/n))(sin^2(2pi/n))+ 4sin^2(3pi/n)sin^2(2pi/n) ]
    = 2 √ [ (sin^2(2pi/n)-sin^2(2pi/n)sin^2(3pi/n) + sin^2(2pi/n)sin^2(3pi/n) ]
    = 2 √ [sin^2(2pi/n)]
    = ± 2sin(2pi/n)
    Since distance cannot be negative,d = 2|sin(2pi/n)|

    Oh you're right! As n increases, the circle formed from the polygon becomes smaller as well. So the distance is decreasing faster than when |x+iy| = 1. How do I use this information?

    Okay, here's the fix:

    d^2 = (r^(1/n)cos((ø+2pi)/n)-r^(1/n)cos(ø/n)^2 + (r^(1/n)sin((ø+2pi)/n-r^(1/n)sin(ø/n))^2

    u + v / 2
    (ø+2π + ø)/2n = 2ø+2π)/2n = (ø+π)/n

    u - v / 2
    (ø+2π - ø)/2n = π/n

    d^2 = (r^(1/n) * 2cos((ø+π)/n)sin(π/n))^2 + (-r^(1/n) * 2sin(ø+π)sin(π))^2
    = 4r^(2/n) [ cos^2((ø+π)/n)sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
    = 4r^(2/n) [ (1-sin^2((ø+π)/n))sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
    = 4r^(2/n) [ sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) - sin^2((ø+π)/n)sin^2(π/n) ]
    = 4r^(2/n) [ sin^2(π/n) - 2sin^2((ø+π)/n)sin^2(π/n) ]
    = 4r^(2/n) [ sin^2(π/n) (1 - 2sin^2((ø+π)/n) ]
    = 4r^(2/n) [ sin^2(π/n) (cos^2((2ø+2π)/n) ]
    = 4r^(2/n) [ sin^2(π/n) (cos^2((2ø+2π)/n) ]
    d = 2r(1/n)|sin(π/n)cos((2ø+2π)/n)|

    Is this right? And how do I prove if its true?? :confused:

    Btw, THANK SO MUCH for all this, I would have gotten this one wrong if you hadn't pointed it out.
  7. Nov 20, 2011 #6

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    My bad.
    Your distance of 2|sin(pi/n)| is right.

    There it is: d = |cis(2pi/n)-1|
    But nevermind. Your distance is right.

    And nice work!

    Btw, there's also a geometrical interpretation and proof.
    If you draw the roots in the complex plane, it should become obvious that the distance between the roots is the same for z^n=1 and for z^n=i.

    Never mind.
    The distance between the roots is equal after all.

    FYI, the distance between root k and (k+1) is:
    |r^(1/n)\cis((ø+2(k+1) pi)/n) - r^(1/n)cis((ø+2k pi)/n)|
    But if you work it out, you'll see it's the same distance.

    I'm afraid you have the wrong trig identities here.
    You should use:

    cos u - cos v = -2sin[(u+v)/2] sin[(u-v)/2]
    sin u - sin v = 2cos[(u+v)/2] sin[(u-v)/2]

    What you have is a proof.
    What you need is a way to verify independently there are no calculation mistakes.

    Geometrically (that is, if you draw it), you can tell that the distance must be:

    No problem! :smile:
  8. Nov 20, 2011 #7
    I noticed a mistake. For z^n = i, I got 2|sin(2pi/n)| but for z^n = 1, I got 2|sin(pi/n)|. I'm sure the later is correct. I've tried and tried again, but I just can't find the mistake. And I KNOW it must be the same length in either casses.

    Also, for z^n = x+iy, when |x+iy| ≠ 1, I got d = 2r^(1/n)|sin(pi/n)|. But, I don't know how to show my generalization on a graph. I don't know what to do with the "r" value :( ...Is there any other way I can prove this is true? I need to explore a variety of different methods to make sure.
  9. Nov 20, 2011 #8

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    The mistake is right here.
    You forgot to divide (u+v) and (u-v) by 2.

    If you draw a graph for z^n=1, you find a regular polygon.
    For z^n=i, you find the same polygon, only rotated by 90 degrees.

    For z^n=x+iy, again you find the same polygon, but rotated over Arg(x+iy) and scaled by r^(1/n).

    And if you're interested, here's another way to show your result.
    It depends on Euler's formula, which is the generalization of De Moivre's theorem.

    Euler's formula is:
    [tex]e^{i \phi} = \cos \phi + i \sin \phi[/tex]
    Which can be rewritten as:
    [tex]\cos \phi = {1 \over 2}(e^{i \phi} + e^{i \phi})[/tex][tex]
    \sin \phi = {1 \over 2i}(e^{i \phi} - e^{i \phi})[/tex]

    Starting with your problem, you have:
    z^n = i
    The solution as you basically already know is:
    [tex]z = (e^{i ({\pi \over 2} + 2k\pi)})^{1 \over n} = e^{i ({\pi \over 2} + 2k\pi) {1 \over n}}[/tex]

    The distance between the first and second root is:

    [tex]d = |e^{i ({\pi \over 2} + 2\pi) {1 \over n}} - e^{i ({\pi \over 2}) {1 \over n}}|[/tex]
    [tex]d = |e^{i ({\pi \over 2} + \pi) {1 \over n}}(e^{i {\pi \over n}} - e^{-i {\pi \over n}})|[/tex]
    [tex]d = 1 \cdot |(e^{i {\pi \over n}} - e^{-i {\pi \over n}})|[/tex]
    [tex]d = |2i \sin{\pi \over n}|[/tex]
    [tex]d = 2 |\sin{\pi \over n}|[/tex]
  10. Nov 20, 2011 #9
    Oooo. Is there a way to prove this? I drew the graphs and they same to be rotating all over the place.

    If I were to make this the "original polygon" (z^n = 1 with ø = 0)
    Then this (z^n = i with ø = π/2) would rotate it 90/n degrees anti-clockwise.

    E.g. When z^3 = x+iy, the "original polygon" would rotate 90/3 = 30 degrees anti-clockwise.

    I feel like I need to adress this pattern ("rotated over Arg(x+iy) and scaled by r^(1/n)") but I don't know how I can show it or prove it mathematically! :frown:

    Wow this is great! Thanks!

    But I don't quite understand what you did here:

    The distance between the first and second root is:
    [tex]d = |e^{i ({\pi \over 2} + 2\pi) {1 \over n}} - e^{i ({\pi \over 2}) {1 \over n}}|[/tex]

    Is there a rule to find the distance between the two points with Euler? I haven't learned Euler.

    Also, I happen to study Mathematical Induction in class, and that's pretty much the only proving I know how to do. But it doesn't seem to work here! Since you must have a lot more experience with this, from the methods listed in the link below, what do you think I can use?

  11. Nov 20, 2011 #10

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    The "original" polygon has its points on the unit circle.
    The angle at the origin between successive points is 2pi/n.
    Therefore the polygon is regular, meaning all its edges have the same length.

    When x+iy has an angle ø, the first root is at ø/n, and on the circle with radius r^(1/n).
    Each successive root is on the same circle and has an angle 2pi/n further along.
    Therefore the polygon is regular, meaning all its edges have the same length.

    I believe this would suffice as a proof.

    A complex number represents a point in the complex plane, meaning it has an x-coordinate (the real part) and an y-coordinate (the imaginary part).
    The distance between 2 of these points is simply the absolute value of the difference of the 2 complex numbers.

    The exponential expression is simply another way to write a complex number, using its distance to the origin and the angle with the real axis.
    The exponential expression is also another way to express De Moivre's theorem.

    In practice I basically use only 3 proofs: forward proof, proof by contradiction, and full induction.
    Actually I rarely see full induction except in a math context that is dedicated to it.
    I guess I do use the others too, but I'm not really aware that I'm doing that.
    I haven't studied how you would call the specific ways of proving things.
    Last edited: Nov 20, 2011
  12. Nov 27, 2011 #11
    Hi! Sorry for the late reply, I've been busy completing a ton of assignments.

    I talked to my teacher about proving the conjecture and he said he expects me to use mathematical induction, and that its the "best, valid proof".

    I just have no idea how I could apply that here. According to this: http://people.richland.edu/james/lecture/m116/sequences/induction.html

    I need to make the left hand side equal the right hand side. But my conjecture is (the distance between 2 adjacent roots of z^n=1) = 2|sin(pi/2)|

    From the examples I found online, its generally listing out the values (in this case the distances) and seeing the pattern, then forming a conjecture from the values observed. And proving it by making the expressions equal.

    For example, I have to find the distances of n=3, n=4, n=5, etc.

    n distance
    3 √3
    4 √2
    5 1.755
    6 1

    And then see from the values above that its 2|sin(pi/2)|. THEN I'll prove it by simplifying the distance formula. But the thing is, I can't form a conjecture just by looking at those values!
  13. Nov 27, 2011 #12

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    Uum... didn't you already proof this for the general case with Direct proof?

    I'm not sure what else there is to do.
    I do not see how to proof this by full induction.

    Listing specific cases for a sequence of n-values makes the conjecture likely.
    I guess this is also a form of induction, but it is no proof.
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