Question

Let u(x,y)= 2x(1-y) for all x,y e []. Then v(x,y) such that f(z)=u(x,y)+iv(x,y) is analytic is,

Answer 1

GIVEN :-

• $\sf \: u(x,y)= 2x(1-y)$

Such that.

• $\sf \: f(z)=u(x,y)+iv(x,y) \: is \: analytic$

TO FIND :-

• $\sf \: \: v(x,y) =$

$\sf \: Let \: \: f(z) = u + iv$

$\sf \: If \: \: it \: \: is \: \: analytic \: \: then \: \: its \: \: derivative -$

$\sf \: f'(z) = \frac{ \partial u}{ \partial x} - i \frac{ \partial v}{ \partial x} \: \: \: \: \: \: \: \: .....(i)$

$\sf \: Also \: \: by \: \: Cauchy \: \: Remman \: \: Equation$

$\sf \: 1. \: \: \green{ \sf \frac{ \partial u}{ \partial x} = \frac{ \partial v}{ \partial y} } \: \: \: and$

$\sf \: 2. \: \: \green{ \sf\frac{ \partial v}{ \partial x} = - \frac{ \partial u}{ \partial y} }$

$\sf \: Put \: 2. \: \: cauchy \: eq. \: into \: eq. \:( i)$

$\sf \: f'(z) = \frac{ \partial u}{ \partial x} - \bigg(i \frac{ - \partial u}{ \partial y} \bigg)\:$

$\sf \: f'(z) = \frac{ \partial u}{ \partial x} + \bigg(i \frac{ \partial u}{ \partial y} \bigg)\: \: \: \: \: \: \: \: .....(ii)$

$\sf \: First \: \: -$

$\sf \: \frac{ \partial u }{ \partial x} = \frac{ \partial }{ \partial x}(2x(1 - y))$

$\implies \sf \: \frac{ \partial u }{ \partial x} = \frac{ \partial }{ \partial x}(2x -2x y)$

$\implies \sf \: \frac{ \partial u }{ \partial x} = (2 -2y)$

$\sf \: Second -$

$\sf \: \frac{ \partial u }{ \partial y} = \frac{ \partial }{ \partial y}(2x(1 - y))$

$\implies \sf \: \frac{ \partial u }{ \partial y} = \frac{ \partial }{ \partial y}(2x -2x y)$

$\implies \sf \: \frac{ \partial u }{ \partial y} = ( -2x)$

$\sf \: Putting \: \: the \: \: value \: \: in \: \: eq. (ii)$

$\sf \: f'(z) = \frac{ \partial u}{ \partial x} + i\bigg( \frac{ \partial u}{ \partial y} \bigg)\:$

$\sf \: f'(z) = (2 - 2y) + i( - 2x)$

$\sf \: f'(z) = 2 - 2y - 2ix$

$\sf \: NOW \: replace \: ( x,y) \: by \: (z ,0)$

$\sf \: f'(z) = 2 - 2(0) - 2i(z)$

$\sf \: where \: z = x + iy$

$\sf \: f'(z) = 2 - 2i(z)$

$\sf \: Now \: \: integrating \: \: w.r.t \: (z)$

$\sf \: f(z) = \int( 2 - 2iz)dz$

$\sf \: f(z) = 2z - 2i( \frac{ {z}^{2} }{2} ) + c$

$\sf \: f(z) = 2z - i {z}^{2} + c$

$\sf \: Put \: z=x + iy$

$\sf \: f(z) = 2(x + iy)- i {( x - iy)}^{2} + c$

$\sf \: f(z) = 2x + 2iy- i ( {x}^{2} - 2ixy + (iy) {}^{2}) + c$

$\sf \: f(z) = 2x + 2iy- i {x}^{2} - 2xy + i (y) {}^{2} + c$

$\sf \: f(z) = 2x - 2xy + 2iy- i {x}^{2} + i (y) {}^{2} + c$

$\sf \: f(z) = 2x (1 - y) + i( 2y- {x}^{2} + y{}^{2} ) + c$

$\sf \: f(z) = real \: + imaginary \: \: part$

$\sf \: By \: \: comparison -$

$\sf \: f(z)=u(x,y)+iv(x,y)$

$\green{ \boxed{ \sf \: v(x,y) = 2y - {x}^{2} + {y}^{2} }}$