If fx-iy-x3-3xy2-i- x-y where i-1 and fx-iy is an analytic fucntion then - x-y is

F(x+iy)=(x^3 3xy^2)+iϕ (x,y) =Ψ (x,y)+iϕ (x,y) Where =Ψ (x,y)=x^3 3xy^2 then Ψ _x=3x^2 3y^2 amp; Ψ _y= 6xy Now using total derivative concept, dϕ= ( dϕdx )dx ...

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Engineering Mathematics

Complex Function ( Limit, Analytic Function)

If fx+iy=x3-3...

Question

If $f (x + i y) = x^{3} - 3 x y^{2} + i ϕ (x, y)$ where $i = \sqrt{- 1}$ and $f (x + i y)$ is an analytic fucntion then $ϕ (x, y)$ is

y^{3} - 3 x^{2} y

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3 x^{2} y - y^{3}

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x^{4} - 4 x^{2} y

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x y - y^{2}

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Solution

The correct option is B $3 x^{2} y - y^{3}$
$f (x + i y) = (x^{3} - 3 x y^{2}) + i ϕ (x, y)$
$= Ψ (x, y) + i ϕ (x, y)$
Where $= Ψ (x, y) = x^{3} - 3 x y^{2}$
then $Ψ_{x} = 3 x^{2} - 3 y^{2}$ & $Ψ_{y} = - 6 x y$
Now using total derivative concept,
$d ϕ = (\frac{d ϕ}{d x}) d x + (\frac{d ϕ}{d y}) d y$
$= - Ψ_{y} d x + Ψ_{x} d y$ (Using C-R equations)
$= - (- 6 x y) d x + (3 x^{2} - 3 y^{2}) d y$
$d ϕ = d (3 x^{2} y - y^{3})$
$\Rightarrow ϕ = 3 x^{2} y - y^{3} + C$

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If f(x+iy)=x3−3xy2+iϕ(x,y) where i=√−1 and f(x+iy) is an analytic fucntion then ϕ(x,y) is

The correct option is B 3x2y−y3f(x+iy)=(x3−3xy2)+iϕ(x,y) =Ψ(x,y)+iϕ(x,y) Where =Ψ(x,y)=x3−3xy2 then Ψx=3x2−3y2 & Ψy=−6xy Now using total derivative concept, dϕ=(dϕdx)dx+(dϕdy)dy =−Ψydx+Ψxdy (Using C-R equations) =−(−6xy)dx+(3x2−3y2)dy dϕ=d(3x2y−y3) ⇒ϕ=3x2y−y3+C

If $f (x + i y) = x^{3} - 3 x y^{2} + i ϕ (x, y)$ where $i = \sqrt{- 1}$ and $f (x + i y)$ is an analytic fucntion then $ϕ (x, y)$ is