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Math Help - help me derive this identity for complex numbers

  1. #1
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    help me derive this identity for complex numbers

    Derive these identities.
    1. e^{iz} = cos z + isin z
    2. arctan z = \frac{i}{2}log(\frac{i+z}{i-z}) (show that the set {w:tan w=z} is the same as the set \frac{i}{2}log(\frac{i+z}{i-z}))
    3. \frac{d}{dz}arctan z = \frac{1}{1+z^2}
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  2. #2
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    Hello, splash!

    It's not clear what definitions and theorems you are allowed.


    1) Derive: e^{iz} \,= \,\cos z + i\sin z

    I'll assume that we are allowed these two defintions:

    . . [1]\;\cos z \:=\:\frac{e^{iz} + e^{-iz}}{2}\qquad[2]\; \sin z \:= \:\frac{e^{iz} - e^{-iz}}{2i}


    Multiply [2] by i:\;\;i\sin z\;=\;i\cdot\frac{e^{iz} - e^{-iz}}{2i}

    We have: . i\sin z \;=\;\frac{e^{iz} - e^{-iz}}{2}

    Add [1]:. . \cos z \;=\;\frac{e^{iz} + e^{-iz}}{2}

    Then: . \cos z + i\sin z\;=\;\frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2} \;=\;\frac{e^{iz} + e^{-iz} + e^{iz} - e^{-iz}}{2} \;=\;\frac{2e^{iz}}{2}


    . . Therefore: . \boxed{\cos z + i\sin z \;= \;e^{iz}}

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  3. #3
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    Quote Originally Posted by splash View Post
    Derive these identities.
    1. e^{iz} = cos z + isin z
    One of the ways of deriving this is extending the definition of the exponent to a complex number.

    It can be shown that for a real number, r we have,
    e^r=1+r+\frac{r^2}{2!}+...

    So we define, an imaginary exponent,
    e^{zi}=1+(zi)+\frac{(zi)^2}{2!}+...
    (Withour fear of divergence, this fear is not only convergent but asbsolutely convergent. This fact will enable us to use Riemann's rearrangement theorem).

    If we open parantheses we have,
    1+zi-\frac{z^2}{2!}-i\frac{z^3}{3!}+\frac{z^4}{4!}+i\frac{z^5}{5!}...
    Apply rearrangement and factorize,
    \left( 1-\frac{z^2}{2!}+\frac{z^4}{4!} \right)+i\left( z-\frac{z^3}{3!}+\frac{z^5}{5!}-...\right)
    You should realize these a power series expanstion for,
    \cos z+i\sin z
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