Prove the theorem stated just after (10.2) as follows. Let φ(u, v) be a harmonic function (that is, φ satisfies ${\mathbf{\partial }}^{\mathbf{2}}\mathit{\phi }\mathbf{/}\mathbf{\partial }{\mathit{u}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{\partial }}^{\mathbf{2}}\mathit{\phi }\mathbf{/}\mathbf{\partial }{\mathit{v}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$). Show that there is then an analytic function g(w) = φ(u, v) + iψ(u, v) (see Section 2). Let w = f(z) = u + iv be another analytic function (this is the mapping function). Show that the function h(z) = g(f(z)) is analytic. Hint: Show that h(z) has a derivative. (How do you find the derivative of a function of a function, for example, ln sin z?) Then (by Section 2) the real part of h(z) is harmonic. Show that this real part is φ(u(x, y), v(x, y)).

Laplace equation:

$\frac{{\partial }^2\varphi }{\partial u^2}+\frac{{\partial }^2\varphi }{\partial v^2}=0$

Chain rule to find the derivative of h(z) :

h'(z) = g'(f(z)) f'(z)

Consider the function $\varphi (u,v)$ to be a harmonic motion.

If a function is harmonic motion, it satisfies the Laplace equation:

$\frac{{\partial }^2\varphi }{\partial u^2}+\frac{{\partial }^2\varphi }{\partial v^2}=0$

If $\varphi (u,v)$be harmonic function in a simply connected domain D .

Then there exists a harmonic conjugate of $\varphi (u,v)$ in the domain D .

So, the function, satisfy the Laplace equation as follows:

$\frac{{\partial }^2\varphi }{\partial u^2}+\frac{{\partial }^2\varphi }{\partial v^2}=0.$

Use chain rule to find the derivative of h(z) :

h'(z) = g'(f(z)) f'(z)

The function of f(z) is an analytic function and g(f(z)) is also an analytic function.

So, the derivative f'(z) and g'(f(z)) exist.

It’s known that the product of analytic functions is also an analytic function.

So, f'(z) x g'(f(z) exists.

So, the derivative of h'(z) exists.

Hence, the function h(z) = g(f(z)) is analytic function.