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Math Help - The complex exponential function

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

    Hi,

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

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

    Now we can write this as e^{a}*(cos(b) + i sin(b))
    Also i know that the modulus is \sqrt{2}
    But that's about it really..
    Last edited by Jones; August 22nd 2009 at 12:18 PM.
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi,

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

    So let  z =a+bi then 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 \sqrt{2}
    But that's about it really..
    Well, that's all you need! If z= a+bi, then e^z= e^{a+ bi}= e^acos(b)+ i e^asin(b))= 1+ i so you must have e^a cos(b)= 1 and e^a sin(b)= 1. Solve for a and b.
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    Quote Originally Posted by Jones View Post
    Hi,

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

    So let  z =a+bi then 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 \sqrt{2}
    But that's about it really..
    z = 1 + i<br />

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

    equate the real and imaginary parts ...

    e^a \cos{b} = 1

    e^a \sin{b} = 1

    \tan(b) = 1

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

    e^a = \sqrt{2}

    a = \frac{\ln{2}}{2}
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  4. #4
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    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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