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Complex numbers - need confirmation

  1. Mar 16, 2010 #1

    Mentallic

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    Homework Helper

    1. The problem statement, all variables and given/known data
    Given [tex]P(z)=4z^3-2z+1[/tex] where [tex]z=cost+isint[/tex], find the maximum and minimum modulus on the argand diagram for the graph as t moves from 0 to [itex]2\pi[/itex].
    I want to check if my solution is valid, and if there is an easier approach to it because I do somewhat answer the question, but I skip all the nitty-gritty arithmetic at the end since it's too overwhelming for me to solve by hand.

    3. The attempt at a solution

    Substituting z(t) into P(z) gives:

    [tex]P(z)=4cos(3t)-2cost+1+i(4sin(3t)-1)[/tex]

    [tex]P(z)=16cos^3t-14cost+1+isint.(10-16sin^2t)[/tex]

    [tex]|z|=\left(\left(4cos(3t)-2cost+1\right)^2+\left(4sin(3t)-2sint\right)^2\right)^{\frac{1}{2}}[/tex] --- (1)

    [tex]\frac{d|z|}{dt}=\frac{1}{2}\left(\left(4cos(3t)-2cost+1\right)^2+\left(4sin(3t)-2sint\right)^2\right)^{\frac{-1}{2}}[/tex]
    [tex]\left(2\left(4cos(3t)-2cost+1\right)\left(-12sin^3t+2sint\right)+2\left(4sin(3t)-2sint\right)\left(12cos(3t)-2cost\right)=0[/tex]


    We're only interested in the numerator to equal zero, therefore,

    [tex]\left(4cos(3t)-2cost+1\right)\left(-12sin^3t+2sint\right)+\left(4sin(3t)-2sint\right)\left(12cos(3t)-2cost\right)=0[/tex]

    Expanding and simplifying, which has nice cancelling and so makes this seemingly difficult task into something much easier.

    Expanded: [tex]-48cos(3t)sin(3t)+8sintcos(3t)+24costsin(3t)-4costsint-12sin(3t)+2sint=0[/tex]

    Simplified: [tex]=-12sin(3t)+16sin(2t)+2sint[/tex] --- (2)

    By De'Moivres Theorem: [tex]sin(3t)=-4sin^3t+3sint; sin(2t)=2sintcost[/tex]

    From (2), converting into trig powers: [tex]=48sin^3t-36sint+32costsint+2sint[/tex]

    [tex]2sint\left(24cos^2t-16cost-7\right)=0[/tex]

    Hence, [tex]sint=0, t=2k\pi[/tex], [tex]k \in Z[/tex]

    and, [tex]24cos^2t-16cost-7=0[/tex]

    [tex]cost=\frac{4\pm \sqrt{58}}{12}[/tex]

    Finding out the answer graphically (because it's easier on me),

    [tex]t=cos^{-1}\left(\frac{4-\sqrt{58}}{12}\right)[/tex]

    Hence, by graphically means again,

    min |z| occurs when [tex]t=\pi(2k+1)[/tex]

    max |z| occurs when [tex]t=(-1)^k\left(cos^{-1}\left(\frac{4-\sqrt{58}}{12}\right)-\frac{\pi}{2}\right)+\frac{\pi}{2}(2k+1)[/tex]

    Now, substituting t into (1) to find min |z|, we get |z|=1

    However, for max |z| the arithmetic is too tricky.

    [tex]cost=\frac{4-\sqrt{58}}{12}[/tex]; [tex]sint=\frac{\sqrt{70+8\sqrt{58}}}{12}[/tex]

    Therefore, [tex]|z|^2=\left(16\left(\frac{4-\sqrt{58}}{12}\right)^3-14\left(\frac{4-\sqrt{58}}{12}\right)+1\right)^2+\left(\frac{70+8\sqrt{58}}{144}\right)\left(10-\frac{70+8\sqrt{58}}{9}\right)^2[/tex]

    I don't want to expand these trinomials, make an arithmetic error in the process early on, then finally after all the
    sweat and tears find the answer is incorrect.
    There must be an easier way!
     
    Last edited: Mar 16, 2010
  2. jcsd
  3. Mar 16, 2010 #2

    Jmf

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    Yikes, that's some scary algebra :)

    I had a go, and the first simplification is to write:

    [tex]z = e^{jt}[/tex]

    then P becomes:

    [tex]P(z) = 4\angle{3t} - 2\angle{t} + 1[/tex]

    (where I'm using the angle notation instead of e to the power of)

    Now this is where I go totally anti-mathematical and lose all pretense of rigor (I'm an engineer, really, not a mathematician):

    If we imagine drawing P(z) as a sum of vectors in the complex plane, we have the sum of a vector magnitude 1 with angle 0 (measured anticlockwise from positive real axis), a vector magnitude 2 with angle [tex]t + \pi[/tex] (in radians, and plus pi thanks to the minus sign in P), and a vector magnitude 4 with angle 3t.

    Intuitively, adding the vectors 'tip-to-tail', we can see that the minimum modulus will be 1, which is when the vectors of magnitude 1 and 2 are pointing the opposite direction to that of magnitude 4. This happens for [tex]t = \pi[/tex].

    Irritatingly, there isn't a value of t for which all three will align in the same direction, which would give the maximum modulus. So perhaps we're stuck with an analytical method to find that. I'll give it some more thought.
     
  4. Mar 16, 2010 #3

    Mentallic

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    Homework Helper

    Anti-mathematical? Blasphemy! :tongue:

    And I'm going to have to give some thought about what you did to find the minimum modulus, seems interesting but hard for me to understand with my choppy vector skills.

    Thanks for the input by the way!
     
  5. Mar 16, 2010 #4

    Jmf

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    You're welcome :)

    Actually, I've an interesting geometrical way of thinking about the maximum modulus problem, though it doesn't really help us get any more than a vague idea of where the solution should lie:

    P(z) is just a polynomial, so we can factorise it into the form:

    [tex]P(z) = (z-\alpha_1)(z-\alpha_2)(z-\alpha_3)[/tex]

    Where [tex]\alpha_n[/tex] are just the roots of P(z) = 0. In this case these are (roughly anyway, the exact solution for them is not very nice, which is why this problem leads to such horrid working out):

    [tex]
    \alpha_1 = 0.442323 + j0.294871[/tex]
    [tex]
    \alpha_2 = 0.442323 - j0.294871[/tex]
    [tex]
    \alpha_3 = - 0.884646 + j0
    [/tex]

    Notice that the first two are complex conjugates, and the third is real. Recall that the modulus of a product is the product of the moduli, and our modulus of P(z) becomes:

    [tex]|P(z)|=|z-\alpha_1||z-\alpha_2||z-\alpha_3|[/tex]

    Each of these product terms is, in the complex plane, the distance between our value of z and the root (a value of alpha). So in maximising the modulus we are looking for the point in the complex plane where the product of these three distances is a maximum. Remember that z is constrained to lie on the unit circle by [tex]z = cos(t) + jsin(t)[/tex], and draw yourself a picture:

    I've attached a horrible mspaint sketch of what I mean. :) What the problem is, in this context, is looking for the point on the unit circle (which I've done in blue) where the product of the distances a b and c is a maximum (as labelled on my sketch in green. The roots (alphas) are in red). By inspection this should happen somewhere around the point which I have those three lines running to.

    Note also that by symmetry, once we've found this point, it's complex conjugate will also be a point for which the magnitude is a maximum. So we can see that we are looking for two solutions, one (by inspection) somewhere near [tex]t = \frac{2\pi}{3}[/tex] and another near [tex]t = \frac{4\pi}{3}[/tex] (such that the two corresponding values of z are conjugates).

    Don't worry if I make no sense, by the way, that's just my way of thinking of the problem. I don't much like algebra. ;-)

    If nothing else, this should at least serve as a way of telling you if your solution above is correct. If you like, you could draw P(z) as a sum of vectors on an argand diagram for t = 2pi/3 and check that it's modulus is quite big.
     

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    Last edited: Mar 16, 2010
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