written 7.3 years ago by | • modified 4.0 years ago |
written 7.3 years ago by |
Let $\tan^{-1}(\dfrac {x+iy}{x-iy})=a+ib-------- (1) $
$$∴\dfrac {+iy}{x-iy}=\tan a+ib---------(2)$$
Taking complex conjugate of Eq. (2)
We get
$\dfrac {x-iy}{x+iy}= \tan (a-ib) \\ \text {Consider } \tan 2a = \tan (a+a) $
(Adjusting ib we get)
$= \tan[(a+ib)+(a-ib)] \\ = \dfrac {\tan(a+ib) + \tan(a-ib)}{1-\tan(a+ib)\tan(a-ib)}…………………(\text {By formula }) $
Substituting values of $\tan(a+ib)$ & $\tan(a-ib) \\ =[\dfrac {[\dfrac {x+iy}{x-iy}+\dfrac {x-iy}{x+iy}]}{[1-\dfrac {(x+iy)}{(x-iy)} ×\dfrac {(x-iy)}{(x+iy)}]}] \\ = \dfrac {[\dfrac {x+iy}{x-iy} +\dfrac {x-iy}{x-iy}]}0 \\ ∴\tan 2a= \text { Not Defined } \\ ∴2a=\tan^{-1} [N.D] \\ ∴2a= \dfrac π2 \\ ∴a= \dfrac π4 $
Similarly let $2ib=\tan (ib+ib) $
Adjusting a we get
$∴\tan[(a+ib)-(a-ib)] \\ = \dfrac {\tan(a+ib) - \tan(a-ib)}{1+\tan(a+ib)\tan(a-ib)} $
(Substituting values of $\tan(a+ib)$ and $\tan(a-ib)) \\ [\dfrac {[\dfrac {x+iy}{x-iy}-\dfrac {x-iy}{x+iy}]}{1+[\dfrac {(x+iy)}{(x-iy)} ×\dfrac {(x-iy)}{(x+iy)}]}] \\ =\dfrac {(\dfrac {(x+iy)^2-(x-iy)^2}{x^2+y^2 })}{1+1} \\ =\dfrac {(x+iy+x-iy)(x+iy-x+iy)}{2(x^2+y^2 ) } \\ =\dfrac {(2x)(2iy)}{2(x^2+y^2 ) } \\ ∴\tan 2ib=\dfrac {2ixy}{x^2+y^2} \\ ∴i \tan h \space 2b = \dfrac {2 ixy}{x^2+y^2} \\ ∴\tan h 2b = \dfrac {2xy}{x^2+y^2 } \\ ∴2b=\tan h^{-1} [\dfrac {2xy}{x^2+y^2}] \\ ∵\tan h^{-1} θ= \dfrac 12 \log|\dfrac {1+θ}{1-θ}| \\ ∴2b = \dfrac 12 \log|\dfrac {1+ \dfrac {2xy}{x^2+y^2}}{1-\dfrac {2xy}{x^2+y^2 }}| \\ ∴2b=\dfrac 12 \log|\dfrac {(\dfrac {x^2+y^2+2xy}{x^2+y^2 })}{(\dfrac {x^2+y^2-2xy}{x^2+y^2 })}| \\ =\dfrac 12 \log|\dfrac {x^2+2xy+y^2}{x^2-2xy+y^2 }| \\ ∴2b = \log|\dfrac {(x+y)^2}{(x-y)^2} |^{1\2} \\ 2b= \log \dfrac {x+y}{x-y}\\ ∴b= \dfrac 12 \log \dfrac {x+y}{x-y} $
Substituting values of a & b in Wq.(1) we get
$\tan^{-1} \dfrac {(x+iy)}{(x-iy)} = \dfrac π4 + \dfrac i2 \log\dfrac {(x+y)}{(x-y)}$
Hence Proved
wrong tag. You mistakely put applied chem instead of math