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Math Help - proof

  1. #1
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    proof

    prove this : [(1+itg \alpha ) /(1-itg\alpha)]^n=(1+itg (n\alpha))/(1-itg(n\alpha))
    i is for complex numbers
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  2. #2
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    Quote Originally Posted by beq!x View Post
    prove this : [(1+itg \alpha ) /(1-itg\alpha)]^n=(1+itg (n\alpha))/(1-itg(n\alpha))
    i is for complex numbers
    Hi

    \left[\frac{1+itg \alpha}{1-itg\alpha}\right]^n=\left[\frac{1+i\frac{\sin\alpha}{\cos\alpha}}{1-i\frac{\sin\alpha}{\cos\alpha}}\right]^n = \left[\frac{\cos\alpha+i\sin\alpha}{\cos\alpha-i\sin\alpha}\right]^n = \left[\frac{e^{i\alpha}}{e^{-i\alpha}}\right]^n = \frac{e^{in\alpha}}{e^{-in\alpha}}

    \left[\frac{1+itg \alpha}{1-itg\alpha}\right]^n=\frac{\cos n\alpha+i\sin n\alpha}{\cos n\alpha-i\sin n\alpha}=\frac{1+itg (n\alpha)}{1-itg (n\alpha)}
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  3. #3
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    Ahh! "tg" is "tan". I was wondering what g(\alpha) was!
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  4. #4
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    Hello, beq!x!

    Prove: . \left[\frac{1+i\tan z}{1-i\tan x}\right]^n \;=\;\frac{1+i\tan(nz)} {1-i\tan(nz)}

    Left side: . \left[\frac{1 + i\dfrac{\sin z}{\cos z}}{1 - i\dfrac{\sin z}{\cos z}}\right]^n \;=\; \left[\frac{\dfrac{\cos z + i\sin z}{\cos z}} {\dfrac{\cos x - i\sin x}{\cos z}} \right]^n \;=\; . \left[\frac{\cos z + i\sin z}{\cos z - i\sin z}\right]^n \;=\;\frac{(\cos z + i\sin z)^n}{(\cos z - i\sin z)^n}


    Applying DeMoivre's Theorem, we have: . \frac{\cos(nz) + i\sin(nz)}{\cos(nz) - i\sin(nz)}


    Divide numerator and denominator by \cos(nz)

    . . =\;\frac{\;\dfrac{\cos(nz)}{\cos(nz)} + i\dfrac{\sin(nz)}{\cos(nz)}\;} {\dfrac{\cos(nz)}{\cos(nz)} - i\dfrac{\sin(nz)}{\cos(nz)}} \;=\;\frac{1 + i\tan(nz)}{1 - i\tan(nz)} \quad\hdots\quad \text{ta-}DAA!

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