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Math Help - Complex Limit (2)

  1. #1
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    Complex Limit (2)

    \lim_{z\to0}\frac{e^z-e^{\bar{z}}}{Im(z)}

    e^z=e^x(cos(y)+isin(y))

    e^{\bar{z}}=e^x(cos(y)-isin(y))

    \lim_{z\to0}e^x\frac{cos(y)+isin(y)-(cos(y)-isin(y))}{y}

    \lim_{z\to0}e^x\frac{2isin(y)}{y}

    Now what?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by dwsmith View Post
    \lim_{z\to0}\frac{e^z-e^{\bar{z}}}{Im(z)}

    e^z=e^x(cos(y)+isin(y))

    e^{\bar{z}}=e^x(cos(y)-isin(y))

    \lim_{z\to 0}e^x\frac{cos(y)+isin(y)-(cos(y)-isin(y))}{y}

    \lim_{z\to0}e^x\frac{2isin(y)}{y}

    Now what?
    \lim_{z\to 0 }e^x\frac{2isin(y)}{y}

    \lim_{(x,y)\to (0,0)}e^x\frac{2isin(y)}{y}

    if we change to polar coordinates we get

    \lim_{r\to 0}e^{r\cos(\theta)}\frac{2isin(r\sin(\theta))}{r\s  in(\theta)}

    Now as r \to 0 we get

    2i
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  3. #3
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    What about theta?

    Also, how is sin(rsin(theta)) over rsin(theta) simplifying?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by dwsmith View Post
    What about theta?

    Also, how is sin(rsin(theta)) over rsin(theta) simplifying?
    Remember that

    \lim_{u \to 0}\frac{\sin(au)}{au}=1 for any real number a
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