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Complex number problem

  1. Apr 24, 2004 #1
    Hi All,

    I've been asked to determine the values of z that obey the equation

    e^z = 1 + sqrt(3)i

    I'm still not sure the concept of this question. Could someone point me in the right direction?

  2. jcsd
  3. Apr 24, 2004 #2

    matt grime

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    z=x+iy, put it in, e^x is then the modulus, and is uniquely determined, y isn't, but you can work out all the possibilities.
  4. Apr 24, 2004 #3
    well you can use the folowing direction:

    e^(a+bi)=(e^a)*( cos(b)+sin(b)i)

    Now you need to find a and b according to the result you want.
    please notice that there are infinity number of solution to that !

  5. Apr 24, 2004 #4
    Thanks for the info...

    So based on the first 2 replies:

    e^(x+iy) = 1 + sqrt(3)i
    e^x.e^iy = e^x*(cos(y)+sin(y)i) = 1 + sqrt(3)i

    therefore e^x = sqrt(1*1 + sqrt(3)*sqrt(3))

    so e^x = 2. which is unique.

    Now I just need to find y right?

    so am I on the right track?

  6. Apr 25, 2004 #5

    matt grime

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    Yes, you're on the right track, you need to find the arguments.
  7. Apr 25, 2004 #6
    ok so next ...

    tan(y) = b/a
    tan(y) = sqrt(3)/1
    y= PI/3

    you talked about infinite solutions, so then

    y = PI/3 + (PI*n) where n is an integer.

    so now is have

    x = ln 2
    y= PI/3 + (PI*n)

    so now do I just subsitute x and y into this eqn?
    e^x*(cos(y)+sin(y)i) = 1 + sqrt(3)i
  8. Apr 25, 2004 #7


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    And after all that work, you arrive at: The solution to e^z = 1 + sqrt(3)i

    is z= log(1+ sqrt(3)) since log z= log|z|+ arg(z)+ 2πn

    That is, 1+ sqrt(3)i, in "polar form" is 2 e^(2πi/3) so
    z= log(2)+ 2π/3+ 2πn where n is any integer.
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