An analytic function of a complex variable z = x+iy (i = √-1) is defined as f(z)= x² − y²+i ψ ( x,y) where ψ(x, y) is a real function. The value of the imaginary part of f(z) at z = (1 + i) is ___________ (round off to 2 decimal places).

The correct answer is 2.

An analytic function of a complex variable $z=x+iy$ can be written in terms of its real and imaginary parts as:
$f\left(z\right)=\varphi (x,y)+i\psi (x,y)$

Here, the real part is given as:
$\varphi (x,y)=x^2-y^2$
and the imaginary part is $\psi (x,y)$.

Since the function $f\left(z\right)$ is analytic, its real and imaginary parts must satisfy the Cauchy-Riemann equations:
$\frac{\partial \varphi }{\partial x}=\frac{\partial \psi }{\partial y}$
and
$\frac{\partial \varphi }{\partial y}=-\frac{\partial \psi }{\partial x}$

Let us compute the partial derivatives of $\varphi$ with respect to $x$ and $y$:
$\frac{\partial \varphi }{\partial x}=2x$
$\frac{\partial \varphi }{\partial y}=-2y$

Using the Cauchy-Riemann relations, we obtain:
$\frac{\partial \psi }{\partial x}=-\frac{\partial \varphi }{\partial y}=2y$
$\frac{\partial \psi }{\partial y}=\frac{\partial \varphi }{\partial x}=2x$

The total differential of $\psi (x,y)$ is given by:
$d\psi =\frac{\partial \psi }{\partial x}dx+\frac{\partial \psi }{\partial y}dy$

Substituting the partial derivatives of $\psi$ into the total differential expression:
$d\psi =2ydx+2xdy$

This expression can be rewritten as:
$d\psi =2(ydx+xdy)=2d\left(xy\right)$

Integrating both sides:
$\int d\psi =2\int d\left(xy\right)$
$\psi (x,y)=2xy+C$
where $C$ is a constant of integration. Taking $C=0$, we have:
$\psi (x,y)=2xy$

We need to evaluate the value of the imaginary part of $f\left(z\right)$ at $z=1+i$.
Comparing $z=1+i$ with $z=x+iy$, we get:
$x=1,y=1$

Substituting these values into the expression for $\psi (x,y)$:
$\psi (1,1)=2\times 1\times 1=2$

Therefore, the value of the imaginary part of the function at $z=1+i$ is 2.