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Thread: Can this be writen any differently?

  1. #1
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    Can this be writen any differently?

    $\displaystyle (e^z-1)(e^z-i)=0$

    Can $\displaystyle e^z$ be isolated in any way?

    I need to find all the solutions for $\displaystyle e^z$
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    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 $\displaystyle e^z=...$? Yes, you're dealing with an equation of the form $\displaystyle a\cdot b=0$ which implies that $\displaystyle a=0 \ \mbox{or} \ b=0$ are the solutions.
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    Re: Can this be writen any differently?

    I need to find the solutions for $\displaystyle e^z$.
    I know how to find the solutions when it's written as $\displaystyle e^z=...$ 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 $\displaystyle a\cdot b=0$ which implies that $\displaystyle a=0 \ \mbox{or} \ b=0$ 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 $\displaystyle e^z$ has to be either 1 or i, right? But I'm not sure how to use this information.
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    Re: Can this be writen any differently?

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

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

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

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

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

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

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

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

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

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

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


    Is it incorrect?
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    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, $\displaystyle e$ to that power is $\displaystyle e\ne 1$.
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    Re: Can this be writen any differently?

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

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

    and

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

    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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