Question: Show that $\tan^{-1}(\dfrac {x+iy}{x-iy})=\dfrac {\pi}4 +\dfrac i2 \log (\dfrac {x+y}{x-y})$
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FE > Semester 1 > Applied Mathematics 1

Marks : 03

Years : DEC 2015

 modified 2.1 years ago  • written 2.1 years ago by Aksh_31 • 600
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wrong tag. You mistakely put applied chem instead of math

 written 2.1 years ago by Krrish ♦ 190
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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

$\dfrac {x-iy}{x+iy}= \tan⁡ (a-ib) \\ \text {Consider } \tan⁡ 2a = \tan⁡ (a+a)$

$= \tan⁡[(a+ib)+(a-ib)] \\ = \dfrac {\tan⁡(a+ib) + \tan⁡(a-ib)}{1-\tan⁡(a+ib)\tan⁡(a-ib)}…………………(\text {By formula })$

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 $2ib=\tan (ib+ib)$

$∴\tan⁡[(a+ib)-(a-ib)] \\ = \dfrac {\tan⁡(a+ib) - \tan⁡(a-ib)}{1+\tan⁡(a+ib)\tan⁡(a-ib)}$

(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