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GRE Math Subject Practice Test #43

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  1. Sep 14, 2008 #1
    First of all, I hope I'm posting this in the appropriate forum and let me know if I'm not.

    I did pretty okay on this practice test, but I never took complex analysis, though I've tried to teach myself the rudiments. Maybe that's not even the problem, but this seems kind of basic, something I shouldn't be missing. So I'm looking for help with how to solve this problem, and also suggestions for preparing for the exam so that I don't miss this category of problem come test day. Here it is:

    If z = e^(2*pi *i / 5), then 1 + z + z^2 + z^3 +5*z^4 + 4*z^6 + 4*z^7 +4*z^8 +5*z^9 =

    a)0

    b) 4*e^(3*pi*i / 5)

    c) 5*e^(4*pi*i / 5)

    d) -4*e^(2*pi*i / 5)

    e) -5*e^(3*pi*i / 5)

    Thank you everyone!
     
  2. jcsd
  3. Sep 15, 2008 #2
    [tex] e^{xi}=cosx+isinx[/tex] NOw u have:

    [tex] z= e^{{2\pi}\frac{i}{5}}=cos\frac{2\pi}{5} +isin\frac{2\pi}{5}[/tex]

    [tex] z^2=e^{2*{2\pi}\frac{i}{5}}=cos\frac{4\pi}{5}+isin\frac{4\pi}{5}[/tex]

    Now u can play a lill bit with trig functions of a double angle, so some things will cancel out. Also

    [tex] 1=e^{{2\pi}i}=cos2\pi+isin{2\pi}[/tex]

    i think that doing this for the whole powers of z and looking for a pattern of how things will cancle out, you should be able to get to the result.
     
  4. Sep 15, 2008 #3
    There might be shortcuts though, but none of which i can think at the moment!
     
  5. Sep 15, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I think that's the hard way to do it. It is much simpler to do the multiplications in exponential form than "cos + i sin".

    If [tex]z= e^{2\pi i/5}[/itex] then
    [tex]z^2= e^{4\pi i/5}[/tex]
    [tex]z^3= e^{6 pi i/5}= e^{pi i}e^{\pi i/5}= -e^{\pi i/5}[/tex]
    [tex]z^4= e^{8\pi i/5}= -e^{3\pi/5}[/tex]
    [tex]z^5= e^{10\pi i/5}= e^{2\pi i}= 1[/tex]
    [tex]z^6= e^{12\pi i/5}= e^{2\pi i}e^{2\pi i/5}= e^{2\pi i/5}[/tex]
    [tex]z^7= e^{14\pi i/5}= e^{2\pi i}e^{4\pi i/5}[/tex]
    [tex]z^8= e^{16\pi i/5}= e^{3\pi i}e^{\pi i/5}= -e^{\pi i/5}[/tex]
    and
    [tex]z^9= e^{18\pi i/5}= e^{3\pi i}e^{3\pi i/5}= -e^{3\pi i/5}[/tex]

    You should be able to put those into your formula and come up with an answer.
     
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