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Complex numbers equation

  1. Sep 8, 2016 #1
    1. The problem statement, all variables and given/known data upload_2016-9-8_16-17-44.png ask to find all the values in z to the equation to be true


    2. Relevant equations
    upload_2016-9-8_16-18-18.png



    3. The attempt at a solution
    upload_2016-9-8_20-32-49.png
    this is my atemp of solution i dont know what else do, because the problem ask for z values
    well i must add that i am thinking about x=0 and y= pi/2 will solve the equation but i dont know if i am right
     

    Attached Files:

  2. jcsd
  3. Sep 8, 2016 #2

    Dr Transport

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    What values of [itex] x [/itex] and [itex]y[/itex] could you come up with? Think about it systematically, [itex] x[/itex] first, do you have an exponential on the right hand side?
     
  4. Sep 8, 2016 #3
    nope i dont have and exp in the right hand side
     
  5. Sep 8, 2016 #4

    fresh_42

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    What can you say about ##x## and ##y## if you know that ##e^{4x} \cdot (\cos (4y)+ i \cdot \sin (4y)) = i##?
     
  6. Sep 8, 2016 #5
    i can say, i need the exp to become 1, and i need the cos4y to become o also i need the sin(4y) become 1 to the equation be true, so i am thinking in x=0 and y=pi/8
     
  7. Sep 8, 2016 #6

    Dr Transport

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    Sounds right......just think methodically about the problem, the solution will drop on out for you.
     
  8. Sep 8, 2016 #7
    tell me please why i am wrong, or exactly what do you mean about methodically
     
  9. Sep 8, 2016 #8

    fresh_42

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    Is this the only solution?
     
  10. Sep 8, 2016 #9
    well that is a good question, i dont know, is there a method to find out if this is the only solution?, i am taking this aproach by the eye method
    z=0+pi/8
     
  11. Sep 8, 2016 #10

    Dr Transport

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    No, [itex]y = \pi / 8 \pm 2\pi[/itex] are all the solutions.... Graph it out and you'll see the other solutions
     
  12. Sep 8, 2016 #11
    i guess the other solution will be -15pi/8?
     
  13. Sep 8, 2016 #12
    well the graph says that e^4z=i and z=(1/8)i(4piN+pi) nEz but i dont know what that means or how to manueally get there
     
  14. Sep 8, 2016 #13

    fresh_42

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    Formally you have ##e^{x} \cdot (\cos (4y)+ i \cdot \sin (4y)) = |i| \cdot (a+ib)##. (The coefficient ##4## at the ##x## isn't really needed.)
    Therefore you have the equations ##1= |i| =e^x ## and ##0 = a = \cos (4y)## and ##1 = b = \sin (4y)## by simply comparing the three terms "length", "real part" and "imaginary part" which you found "by eye". This gave you ##x=0## and ##y=\frac{1}{8} \pi##.
    But ##\cos## and ##\sin## are repeating their values.
     
  15. Sep 8, 2016 #14
    do you mean the natural cicles of sin and cos? i dont get it very clearly
     
  16. Sep 8, 2016 #15

    fresh_42

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    Yes exactly. Every full circle (##2 \pi##) you'll get the same values ##\sin (4y) = 1## and ##\cos (4y) = 0## again. So all ##\frac{1}{8} \pi + 2k\pi \; (k\in \mathbb{Z})## are solutions, which means that every full circle (backwards or forwards) you return to the point ##i = (0,i)## again.

    This is important! Because it shows you, that the complex exponential function (and the logarithms, too) behave differently compared with the real exponential function. They are not unique anymore! Solutions without repetitions are called principal.
     
  17. Sep 8, 2016 #16
    oh thanks i am a beginner in the complex variable and have many doubs about how its behave, many thanks, so this must have infinite ciclical solutions?
     
  18. Sep 8, 2016 #17

    SammyS

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    Actually you need ##\ 4y = \pi / 2 \,,\ ## but then you can add (integer) multiples of 2π to this.

    So you then have: ##\ 4y = \frac \pi 2 +2\pi k \,.\ ## Now divide by 4
     
  19. Sep 8, 2016 #18

    fresh_42

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    Oops.
     
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