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Math Help - Can this be writen any differently?

  1. #1
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    Can this be writen any differently?

    (e^z-1)(e^z-i)=0

    Can e^z be isolated in any way?

    I need to find all the solutions for e^z
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Can this be writen any differently?

    Do you mean you want something of the form e^z=...? Yes, you're dealing with an equation of the form a\cdot b=0 which implies that a=0 \ \mbox{or} \ b=0 are the solutions.
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    Re: Can this be writen any differently?

    I need to find the solutions for e^z.
    I know how to find the solutions when it's written as e^z=... but I have a hard time figuring this one out.
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    MHF Contributor Siron's Avatar
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    Re: Can this be writen any differently?

    I gave you a hint:

    Quote Originally Posted by Siron View Post
    you're dealing with an equation of the form a\cdot b=0 which implies that a=0 \ \mbox{or} \ b=0 are the solutions.
    Can you do something with this?
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    Re: Can this be writen any differently?

    Quote Originally Posted by Siron View Post
    I gave you a hint:



    Can you do something with this?
    Hmmmm...
    Well one of the brackets must be zero...
    So e^z has to be either 1 or i, right? But I'm not sure how to use this information.
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  6. #6
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    Re: Can this be writen any differently?

    If I solve for e^z=1 then I get the solution:
    z=2p \pi i

    If I solve for e^z=i then I get the solution:
    z=i(\pi /2 + 2p \pi)
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    Re: Can this be writen any differently?

    Quote Originally Posted by MathIsOhSoHard View Post
    (e^z-1)(e^z-i)=0
    I need to find all the solutions for e^z
    There are many, many solutions.
    e^z=1\text{ or }e^z=i.

    In the first case z=i(2k\pi),~k\in\mathbb{Z}.

    Then z=\frac{i(4k+1)\pi}{2}.
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    Re: Can this be writen any differently?

    Quote Originally Posted by Plato View Post
    There are many, many solutions.
    e^z=1\text{ or }e^z=i.

    In the first case z=i(2k\pi),~k\in\mathbb{Z}.

    Then z=\frac{i(4k+1)\pi}{2}.
    I understand the z=i(2k\pi),~k\in\mathbb{Z}. that's the conclusion I came to too.
    But I'm not sure I understand your second solution.

    My book says the facit is:
    z=2k\pi i
    or
    z=1+2k\pi i
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    Re: Can this be writen any differently?

    Quote Originally Posted by MathIsOhSoHard View Post
    I understand the z=i(2k\pi),~k\in\mathbb{Z}.
    My book says the facit is:
    z=2k\pi i or z=1+2k\pi i
    Check the text again because:
    e^{1+2ki\pi}=e~.
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  10. #10
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    Re: Can this be writen any differently?

    Quote Originally Posted by Plato View Post
    Check the text again because:
    e^{1+2ki\pi}=e~.
    This is taken straight out of my text book word by word (it is scanned):


    Is it incorrect?
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  11. #11
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    Re: Can this be writen any differently?

    Quote Originally Posted by MathIsOhSoHard View Post
    This is taken straight out of my text book word by word (it is scanned):

    Is it incorrect?
    If you have copied it correctly, e to that power is e\ne 1.
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  12. #12
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    Re: Can this be writen any differently?

    Quote Originally Posted by Plato View Post
    If you have copied it correctly, e to that power is e\ne 1.
    So the correct answers are:

    z=2p \pi i for when e^z=1

    and

    z=i(\pi /2 + 2p \pi) for when e^z=i

    Any of these values will make one of the brackets 0, and thus the entire equation will be zero, right?
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  13. #13
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    Re: Can this be writen any differently?

    That is correct.
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