91Ó°ÊÓ

First, we calculate the second derivatives of the function with respect to \(x\) and \(y\). For \(x^{2} - y^{2}\): $$ f_{xx} = \frac{\partial^2}{\partial x^2} (x^{2} - y^{2}) = 2 $$ $$ f_{yy} = \frac{\partial^2}{\partial y^2} (x^{2} - y^{2}) = -2 $$ Now we check if the Laplace equation holds, that is \(f_{xx} + f_{yy} = 0\). In this case, \(2 - 2 = 0\), so \(x^2 - y^2\) is a harmonic function.

Now, let's find the harmonic conjugate \(g(x, y)\) of the function \(x^{2} - y^{2}\). To find the conjugate function, we use the Cauchy-Riemann equations: \begin{align*} &f_x = g_y\\ &f_y = -g_x \end{align*} Computing the first derivatives, we have $$ f_x = 2x, \quad f_y = -2y $$ Now we can find the conjugate function \(g(x, y)\). First we integrate the partial derivatives of \(g\). From \(f_x = g_y\): $$ g_y = 2x \implies g(x, y) = \int 2x \, dy = 2xy + h(x) $$ Now from \(f_y = -g_x\): $$ -2y = g_x - h'(x) \implies g(x, y) = \int -2y \, dx = -2xy + k(y) $$ Comparing both expressions for \(g(x, y)\), we have $$ g(x, y) = 2xy + h_1(x) = -2xy + h_2(y) $$ Since any function of the form \(g(x, y) = 2xy + C\) will satisfy the Cauchy-Riemann equations, all these functions are harmonic conjugates of \(x^2 - y^2\). So, the harmonic conjugate of \(x^{2} - y^{2}\) is given by \(g(x, y) = 2xy + C\) where C is an arbitrary constant.

Following the same process as before, we first check if the function is harmonic by calculating the second derivatives: $$ f_{xx} = \frac{\partial^2}{\partial x^2} (e^{x}\cos{y}) = e^{x}\cos{y} $$ $$ f_{yy} = \frac{\partial^2}{\partial y^2} (e^{x}\cos{y}) = -e^{x}\cos{y} $$ Now we check if the Laplace equation holds, that is \(f_{xx} + f_{yy} = 0\). In this case, \(e^{x}\cos{y} - e^{x}\cos{y} = 0\), so \(e^x\cos{y}\) is a harmonic function.

For the harmonic conjugate, we first compute the first derivatives: $$ f_x = e^x\cos{y}, \quad f_y = -e^x\sin{y} $$ Using the Cauchy-Riemann equations, we find the conjugate function \(g(x, y)\) as follows. From \(f_x = g_y\): $$ g_y = e^x\cos{y} \implies g(x, y) = \int e^x\cos{y} \, dy = e^x\sin{y} + h(x) $$ Now from \(f_y = -g_x\): $$ -g_x = -e^x\sin{y} \implies g(x, y) = \int e^x\sin{y} \, dx = e^x\sin{y} + k(y) $$ Comparing both expressions for \(g(x, y)\), we have $$ g(x, y) = e^x\sin{y} + h_1(x) = e^x\sin{y} + h_2(y) $$ Thus, the harmonic conjugate of \(e^x\cos{y}\) is given by \(g(x, y) = e^x\sin{y} + C\) where C is an arbitrary constant. The verification and harmonic conjugates for the remaining functions (c), (d), and (e) follow the same steps outlined above.