Answer :

Correct answer is : 2

Given, f(z) = x2 – y2 + i ψ (x, y)

$\varphi =x^2-y^2$

∵ f(z) is analytic function

$d\psi =\frac{-\partial \varphi }{\partial y}dx+\frac{\partial \varphi }{\partial x}\cdot dy$

$\frac{\partial \varphi }{\partial y}=-2y,\frac{\partial \varphi }{\partial x}=2x$

$d\psi =2ydx+2xdy$

$\int d\psi =2\int d\left(xy\right)$

ψ = 2 xy

Given, z = 1 + i

Comparing it with z = x + iy, we get :

∴ x = 1, y = 1

$\therefore {\left(\psi \right)}_{\left(1,1\right)}=2\times 1\times 1=2$

ψ = 2 when z = 1 + i