f(z)=((1−z)/(1+z)) u(x,y)=.. v(x,y)=..


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 48161 by gunawan last updated on 20/Nov/18

$f (z) = \frac{1 - z}{1 + z}$ $u (x, y) = . .$ $v (x, y) = . .$
Commented by maxmathsup by imad last updated on 20/Nov/18

$l e t z = x + i y \Rightarrow f (z) = \frac{1 - x - i y}{1 + x + i y} = \frac{(1 - x - i y) (1 + x - i y)}{{(1 + x)}^{2} + y^{2}}$ $= \frac{1 - x^{2} - i y (1 - x) - i y (1 + x) - y^{2}}{x^{2} + y^{2} + 2 x + 1} = \frac{1 - x^{2} - y^{2} - i (y - x y - y - x y)}{x^{2} + y^{2} + 2 x + 1}$ $= \frac{1 - x^{2} - y^{2}}{x^{2} + y^{2} + 2 x + 1} + i \frac{2 x y}{x^{2} + y^{2} + 2 x + 1} = u + i v \Rightarrow$ $u (x, y) = \frac{1 - x^{2} - y^{2}}{x^{2} + y^{2} + 2 x + 1} a n d v (x, y) = \frac{2 x y}{x^{2} + y^{2} + 2 x + 1}$
	Answered by ajfour last updated on 21/Nov/18

	$f (z) = u + i v$ $= \frac{(1 - x) - i y}{(1 + x) + i y}$ $= \frac{[(1 - x) - i y] [(1 + x) - i y]}{{(1 + x)}^{2} + y^{2}}$ $= \frac{1 - x^{2} - y^{2}}{{(1 + x)}^{2} + y^{2}} - i \frac{2 y}{(1 + x^{2}) + y^{2}} .$
	Commented by gunawan last updated on 21/Nov/18

	$nice Sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum