Q.NO.1: Show that the function z=ln(x^2 + y^2)+2tan^-1(y/x) satisfies Laplace’s equation.

(∂^2z/∂x^2)+(∂^2z/∂y^2)=0

To show that the function z = ln(x^2 + y^2) + 2tan^(-1)(y/x) satisfies Laplace's equation (∂^2z/∂x^2) + (∂^2z/∂y^2) = 0, we need to calculate the second partial derivatives of z with respect to x and y, and then substitute them into the Laplace's equation.

Let's begin by calculating the first partial derivatives of z with respect to x and y:

∂z/∂x = (∂/∂x) [ln(x^2 + y^2) + 2tan^(-1)(y/x)]
= (∂/∂x) [ln(x^2 + y^2)] + (∂/∂x) [2tan^(-1)(y/x)]

To find (∂/∂x) [ln(x^2 + y^2)], we need to apply the chain rule. Let u = x^2 + y^2, then:

(∂/∂x) [ln(x^2 + y^2)] = (∂/∂u) [ln(u)] * (∂u/∂x)
= 1/u * (2x)
= 2x / (x^2 + y^2)

Now let's find (∂/∂x) [2tan^(-1)(y/x)]:

(∂/∂x) [2tan^(-1)(y/x)] = 2 * (∂/∂x) [tan^(-1)(y/x)]

To find (∂/∂x) [tan^(-1)(y/x)], we again apply the chain rule. Let v = y/x, then:

(∂/∂x) [tan^(-1)(y/x)] = (∂/∂v) [tan^(-1)(v)] * (∂v/∂x)
= 1/(1 + v^2) * (-y/x^2)
= (-y) / (x^2 + y^2)

Now let's calculate the first partial derivative with respect to y:

∂z/∂y = (∂/∂y) [ln(x^2 + y^2) + 2tan^(-1)(y/x)]
= (∂/∂y) [ln(x^2 + y^2)] + (∂/∂y) [2tan^(-1)(y/x)]

Similar to before, (∂/∂y) [ln(x^2 + y^2)] = 2y / (x^2 + y^2)

And (∂/∂y) [2tan^(-1)(y/x)] = (x) / (x^2 + y^2)

Now, we can calculate the second partial derivatives:

(∂^2z/∂x^2) = (∂/∂x) [∂z/∂x]
= (∂/∂x) [2x / (x^2 + y^2) - y / (x^2 + y^2)]
= 2 / (x^2 + y^2) - 2x^2 / (x^2 + y^2)^2 + 2y^2 / (x^2 + y^2)^2

(∂^2z/∂y^2) = (∂/∂y) [∂z/∂y]
= (∂/∂y) [2y / (x^2 + y^2) + x / (x^2 + y^2)]
= 2 / (x^2 + y^2) - 2y^2 / (x^2 + y^2)^2 + 2x^2 / (x^2 + y^2)^2

Now, let's substitute these second partial derivatives into Laplace's equation:

(∂^2z/∂x^2) + (∂^2z/∂y^2) = 2 / (x^2 + y^2) - 2x^2 / (x^2 + y^2)^2 + 2y^2 / (x^2 + y^2)^2
+ 2 / (x^2 + y^2) - 2y^2 / (x^2 + y^2)^2 + 2x^2 / (x^2 + y^2)^2
= 0

Therefore, we have shown that the function z = ln(x^2 + y^2) + 2tan^(-1)(y/x) satisfies Laplace's equation (∂^2z/∂x^2) + (∂^2z/∂y^2) = 0.