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Finding real part of complex eqn

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data

    for z= 3e[itex]^{(-5\pi i/6)}[/itex] and w=1+i

    find Re(iw+[itex]\overline{z}[/itex][itex]^{2}[/itex])



    2. Relevant equations



    3. The attempt at a solution

    [itex]\overline{z}[/itex]= 3e[itex]^{5\pi i/6}[/itex]

    so |z|=3 and Arg(z)= 5[itex]\pi[/itex][itex]/6[/itex] which lies in the second quad. so we have z= -1+ i[itex]\sqrt{3}[/itex]

    so squaring z I have 1-2i[itex]\sqrt{3}[/itex] -3
    so Re(z) = -2.

    and for the w once I times through by i we have w=i-1

    so I thought it should be something like Re(iw+[itex]\overline{z}[/itex][itex]^{2}[/itex])=
    -1+-2=-3

    But the answer says 7/2

    Im sure I have gone wrong somewhere please help!
     
    Last edited: Apr 10, 2012
  2. jcsd
  3. Apr 10, 2012 #2

    HallsofIvy

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    .
    That's impossible. That has absolute value [itex]\sqrt{1+ 3}= 2, not 3.. [itex]cos(5\pi/6)= -\sqrt{3}/2[/itex] so [itex]3cos(5/pi/6)= -3\sqrt{3}/2[/itex], not -1. [itex]sin(5/p/6)= 1/2[/itex] so [itex]3sin(5/pi/6)= 3/2[/itex], not [/itex]\sqrt{3}[/itex].

    [itex]\overline{z}[/itex] (not z) is equal to [itex]-3\sqrt{3}/2+ (3/2)i[/itex]

     
  4. Apr 10, 2012 #3

    Ray Vickson

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    So you want the real part of [itex] i + i^2 + (3 e^{\,i 5 \pi/ 6})^2.[/itex]

    RGV
     
  5. Apr 10, 2012 #4
    Yes which the first part is equal to i-1 its just the second part I am confused with.
     
  6. Apr 10, 2012 #5

    Ray Vickson

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    So, are you saying that you don't know how to calculate [itex] (r e^{i \theta})^2 [/itex] for real [itex]r \text{ and } \theta?[/itex] Have you tried looking in your textbook, or looking on-line?

    RGV
     
  7. Apr 10, 2012 #6
    Yes but I cant find any similar examples on line and I only have a calculus textbook.
    But I think I am confused about what happens when I square the term.
    Should I convert it to a+ib form and then square it or just square it in exponent form because If I do that I get e^10Pi*i/6 once I have squared and changed it into its conjugate form.
    But that means I have an angle of 300 deg. Which I got stuck on because I did not know what quad it should be in on the argand diagram since it only uses the interval [-pi,pi], so where does 300 deg fit in there? do you got to 180 then go back in the opp direction or do you keep going around to the 4th quad but then that should give a negative 'a' and negative 'bi' ?? Im really confused on this.
     
  8. Apr 10, 2012 #7

    Ray Vickson

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    Google is your friend. See, eg., http://en.wikipedia.org/wiki/Complex_number and look about 2/5 of the way down. Frankly, I am very surprised that you have not seen this material before, since it is so very fundamental.

    RGV
     
  9. Apr 10, 2012 #8
    OK so basically if its out of the range of [-pi,pi] just add 2kpi to make it in the right interval?
     
  10. Apr 11, 2012 #9
    [itex]\overline{z}[/itex][itex]^{2}[/itex] = 9([itex]{{\rm e}^{5/6\,i\pi }}[/itex])^2

    So you essentially get 9(cos([itex]\frac{5\pi}{6}[/itex])+isin([itex]\frac{5\pi}{6}[/itex]))^2

    Once you simplify that trigonometric expression, you should be able to calculate the real part of that fairly easily, once you have that, you simply add on the real part of (i*w)

    Remember Euler's formula:

    http://en.wikipedia.org/wiki/Euler's_formula
     
    Last edited: Apr 11, 2012
  11. Apr 11, 2012 #10
    Yes, as you got (10*Pi)/6 - simplify that, and get it into range by adding or subtracting 2kPi

    Since you did the squaring directly in the exponential, this will save you some work when you convert it into trig form, then it's pretty much solved.
     
  12. Apr 11, 2012 #11
    Ok Thanks!!
     
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