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Complex numbers (practice exam questions for exam in 2 days)

  1. Jun 22, 2006 #1
    Hey if anyone could help me to understand wat to do in this question I would be appreciative!

    Find in the complex plane the fourth roots of -64, Use the result to factor
    Z^4+ 64 into
    i) a product of four linear factors
    I kinda thought that you could write something along the lines of this froman example id seen!
    (z+64)(z+b)(z+c)(z+d), but I wouldnt know how to find the rest!
    If I can be helped to work out this answer then I could work out the next most likely!
    ii) a product of two quadratic factors with real coefficients!
     
  2. jcsd
  3. Jun 22, 2006 #2

    nrqed

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    To get the four roots, notice that, written as a complex number, 64 may be written as [itex] 64 e^{i 0} [/itex] but also as [itex] 64 e^{2 \pi i} [/itex] (and of course, there are an infinite number of other possibilities). It is easy to take the fourth root of the second expression, right? (you take the fourth root of 64 and the fourth root of the exponential. To get the fourth root of an exponential is straightforward). Then you may write 64 as [itex] 64 e^{4 \pi i} [/itex] and take the fourth root again, then [itex] 64 e^{6 \pi i} [/itex] and then [itex] 64 e^{8 \pi i} [/itex]. If you would keep going with other expressions of 64 (like with an exponent of [itex] 10 \pi i[/itex] or [itex] -2 \pi i[/itex] ) you would start to obtain results corresponding to the same answers you found with the 4 ones above.
     
  4. Jun 22, 2006 #3
    Judging from your question, I suspect you have not covered complex logarithm/Euler's law yet. If that is the case, then nrged's explanation may not be of much use.

    Simply consider that the equation Z^4 + 64 = 0 can be rewritten as (z^2)^2 - (i8)^2 = 0 and that the 4 solutions to this equation are the 4 fourth roots of -64. In this way you do both parts of the q together.

    Another approach would be to find the square roots of -64 and the square roots of those directly.

    Molu
     
  5. Jun 22, 2006 #4

    HallsofIvy

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    z+ 64 is certainly not a factor of z4+ 64! One way you could do this is write z4+ 64 as (z2)2-(-64) so that z4+ 64= (z2-8i)(z2+8i) and then factor those.

    However, since the problem says to first find the fourth roots of -64, you are obviously intended to use De Moivre's formula:
    [tex](re^{i\theta})^\frac{1}{n}= r^\frac{1}{n}e^{\frac{i\theta}{n}[/tex]
    Of course, [itex]-64= 64e^{i\pi}[/itex].

    (I just noticed that that is essentially what nrqed told you!)
     
  6. Jun 22, 2006 #5
    I aint quite following, its kinda confusing wat I was even gettin at!
    I am really confused, I know De'Moivre's theorem.
    Find the fourth roots! Okay so this is wat I thought or did now!
    1st root is 64e^i*Pi
    2nd root is 8e^i*Pi/2
    3rd root is 4e^i*Pi/3
    4th root is (64^1/4)e^i*Pi/4

    Now then is that right and then I would write something along the lines of
    (z+64e^i*Pi)(z+8e^i*Pi/2)(z+4e^i*Pi/3)(z+(64^1/4)e^i*Pi/4)

    Then thats all right for the first part! The second part would just be
    (z^2+8i)(z^2-8i) as shown in one of the replies or I could just times the first two brackets together and the last 2 to get the quadratic! So if this is right then I should be fine! Thanks for your help! But yeah can you tell me whether I am right!
     
  7. Jun 22, 2006 #6
    wait a second that is wrong anyhow!I would do it like this
    1st root is (64^1/4)e^-i*pi/4= 64^1/4(1/sqrt2+i/sqrt2) and so on!
    then couldnt I say that (z-(64^1/4(1/sqrt2+i/sqrt2))
     
  8. Jun 22, 2006 #7
    this will be the answer to the quadratic part I think
    (z^2-(128^3/4)+4)(z^2+(128^3/4)+4)
     
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