written 5.5 years ago by | • modified 5.5 years ago |
$1) e^{2φ} = \frac{cotα}{22} 2θ = nπ + \frac{π}{2} +α. $
Subject:- Applied Mathematics
Marks:- 3
Mumbai Unversity>FE>Sem1>Applied Maths1
written 5.5 years ago by | • modified 5.5 years ago |
$1) e^{2φ} = \frac{cotα}{22} 2θ = nπ + \frac{π}{2} +α. $
Subject:- Applied Mathematics
Marks:- 3
Mumbai Unversity>FE>Sem1>Applied Maths1
written 5.5 years ago by |
$tan(θ+iφ) = tanα + isecα∴ tan(θ-iφ) = tanα - isecα$ $∴ tan2θ = tan[(θ+iφiφ+ (θ-iφ)] = \frac{(tan(θ+iφ) +tan(θ-iφ))}{(1-tan(θ+iφ) tan(θ-iφ) )}$ $=\frac{(tan(θ+isecα) +tan(θ-isecα))}{(1-tan(θ+isecα) tan(θ-isecα) )}$=$\frac{ 2tanα}{(1-(tan)^2 α+(sec)^2 α )}$=$\frac{2tanα}{(-2(tan)^2 α)-cotα}$ =$tan(\frac{π}{2}+ α)$
$∴ 2θ= nπ + \frac{π}{2}+ α. (general value)$.
$Again tan(2iφ) = tan[(θ+iφ) - (θ-iφ)]$
$ = \frac{(tan(θ+isecα)-tan(θ-isecα))}{(1+tan(θ+isecα) tan(θ-isecα) )}$
$∴ itanh2φ = \frac{2isecα}{(2(sec)^2 α)}= icosα∴ tanh2φ = cosα$
$∴ 2φ = tanh^{-1} (cosα) (cosα\frac{1}{2}log[\frac{(1+cosα)}{(1-cosα)}] = 1/2log[\frac{(2(cos)^2 (α/2))}{(2(sin)^2 (α/2) )}]ogcotα/2$