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Math Help - Solve for z in the equation e^z=0

  1. #1
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    Solve for z in the equation e^z=0

    This is a homework problem for my Mathematical Physics course, where we're learning complex numbers and so on right now. I didn't see the subject over in the Physics site, and I wasn't quite sure which category this belonged in over here, so please move it if it if it's more appropriate somewhere else!

    Anyway, on to the problem.

    Either solve for z in the equation e^z = 0, or show that this cannot be done.

    I attempted to show that it could not be done by taking the natural logarithm of both sides:

    ln(e^z) = ln(0)

    z = ln(0)

    I then concluded by saying that, since you cannot take the natural logarithm of 0, the problem is unsolvable, or z is undefined, or you cannot solve for z.

    I wanted to make sure that my reasoning is correct, or if I am actually wrong and it can be done (though I don't want the solution handed to me, if I'm wrong- just point me in the right direction, please!).

    Thank you in advance!
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  2. #2
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    Quote Originally Posted by Physics-is-Phun View Post
    This is a homework problem for my Mathematical Physics course, where we're learning complex numbers and so on right now. I didn't see the subject over in the Physics site, and I wasn't quite sure which category this belonged in over here, so please move it if it if it's more appropriate somewhere else!

    Anyway, on to the problem.

    Either solve for z in the equation e^z = 0, or show that this cannot be done.

    I attempted to show that it could not be done by taking the natural logarithm of both sides:

    ln(e^z) = ln(0)

    z = ln(0)

    I then concluded by saying that, since you cannot take the natural logarithm of 0, the problem is unsolvable, or z is undefined, or you cannot solve for z.

    I wanted to make sure that my reasoning is correct, or if I am actually wrong and it can be done (though I don't want the solution handed to me, if I'm wrong- just point me in the right direction, please!).

    Thank you in advance!
    e^z=e^x(\cos(y)+i\sin(y))

    e^x>0

    When will cos(y) + isin(y) = 0
    Last edited by dwsmith; February 1st 2011 at 07:09 PM. Reason: Misspoke
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  3. #3
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    Not as far as I can tell- so that would indicate that you cannot solve for z and have it be meaningful/possible (although your method probably showed it much more clearly than mine)?
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  4. #4
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    Quote Originally Posted by Physics-is-Phun View Post
    Not as far as I can tell- so that would indicate that you cannot solve for z and have it be meaningful/possible (although your method probably showed it much more clearly than mine)?
    e^x(\cos(y)+i\sin(y))=e^z\neq 0
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  5. #5
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    Thank you very much! That was fast!
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  6. #6
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    Having said that e^x is not 0, you still need to show that cos(y)+ i sin(y) is not 0. That is, there there is no real number, y, such that both cos(y) and sin(y) are equal to 0.
    Last edited by mr fantastic; February 2nd 2011 at 08:13 PM. Reason: Fixed latex tags.
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  7. #7
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    Isn't this already known, though? I thought that:

    sin(y) = cos(y - pi/2)

    If that's true, then if I want to try to get them to be equal:

    y = y - \pi/2

    and then:

    0 = \pi/2

    Since that's not true, they cannot both equal 0... right? Was that what you were thinking of?

    (almost irrelevant- I don't know how to make the "pi" symbol in the post. Does anyone know how, or if it's possible?)

    (edit- thank you, dwsmith, for the tip!)
    Last edited by Physics-is-Phun; February 2nd 2011 at 08:34 PM. Reason: Changing "pi" to "Pi"
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  8. #8
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    Quote Originally Posted by Physics-is-Phun View Post
    Isn't this already known, though? I thought that:

    sin(y) = cos(y - pi/2)

    If that's true, then if I want to try to get them to be equal:

    y = y - pi/2

    and then:

    0 = pi/2

    Since that's not true, they cannot both equal 0... right? Was that what you were thinking of?

    (almost irrelevant- I don't know how to make the "pi" symbol in the post. Does anyone know how, or if it's possible?)
    Pi = \pi

    You could do \displaystyle \cos(y)=0 \ \ \ y=\frac{\pi}{2}+k\pi \ \ k\in\mathbb{Z}

    \displaystyle \sin(y)=0 \ \ \ \ y=k\pi \ \ \  k\in\mathbb{Z}
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  9. #9
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    Quote Originally Posted by Physics-is-Phun View Post
    Isn't this already known, though? I thought that:

    sin(y) = cos(y - pi/2)

    If that's true, then if I want to try to get them to be equal:

    y = y - \pi/2

    and then:

    0 = \pi/2

    Since that's not true, they cannot both equal 0... right? Was that what you were thinking of?

    (almost irrelevant- I don't know how to make the "pi" symbol in the post. Does anyone know how, or if it's possible?)

    (edit- thank you, dwsmith, for the tip!)
    But I had no way of knowing if it was "already known" by you.
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