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Thread: The complex exponential function

  1. #1
    Member Jones's Avatar
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    The complex exponential function

    Hi,

    I need to find all $\displaystyle z~~\text{such that}~~ e^{z} = 1+i$

    So let $\displaystyle z =a+bi$ then$\displaystyle e^{a}*e^{bi}$

    Now we can write this as $\displaystyle e^{a}*(cos(b) + i sin(b))$
    Also i know that the modulus is $\displaystyle \sqrt{2}$
    But that's about it really..
    Last edited by Jones; Aug 22nd 2009 at 11:18 AM.
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi,

    I need to find all $\displaystyle z~~\text{such that}~~ e^{z} = 1+i$

    So let $\displaystyle z =a+bi$ then$\displaystyle e^{a}*e^{bi}$

    Now we can write this as [tex]e^{a}*(cos(b) + i sin(b))
    Also i know that the modulus is $\displaystyle \sqrt{2}$
    But that's about it really..
    Well, that's all you need! If z= a+bi, then $\displaystyle e^z= e^{a+ bi}= e^acos(b)+ i e^asin(b))= 1+ i$ so you must have $\displaystyle e^a cos(b)= 1$ and $\displaystyle e^a sin(b)= 1$. Solve for a and b.
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  3. #3
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    Quote Originally Posted by Jones View Post
    Hi,

    I need to find all $\displaystyle z~~\text{such that}~~ e^{z} = 1+i$

    So let $\displaystyle z =a+bi$ then$\displaystyle e^{a}*e^{bi}$

    Now we can write this as [tex]e^{a}*(cos(b) + i sin(b))
    Also i know that the modulus is $\displaystyle \sqrt{2}$
    But that's about it really..
    $\displaystyle z = 1 + i
    $

    $\displaystyle z = e^a(\cos{b} + i\sin{b})$

    equate the real and imaginary parts ...

    $\displaystyle e^a \cos{b} = 1$

    $\displaystyle e^a \sin{b} = 1$

    $\displaystyle \tan(b) = 1$

    $\displaystyle b = \frac{\pi}{4} + 2k\pi$

    $\displaystyle e^a = \sqrt{2}$

    $\displaystyle a = \frac{\ln{2}}{2}$
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    $\displaystyle z=\log(1+i)=\frac{\ln(2)}{2}+i\left(\frac{\pi}{4} +2n\pi\right).$
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  5. #5
    Member Jones's Avatar
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    Thank you, i honestly thought it was much harder than this
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