To show that the functions U(x, y) = e^{u(x, y)} \cos v(x, y) and V(x, y) = e^{u(x, y)} \sin v(x,
y) are
harmonic in D , and that V(x, y) is the harmonic conjugate of U(x, y) , we proceed as follows:
1. Harmonicity of U(x, y) and V(x, y) :
A function is harmonic if it satisfies Laplace’s equation:
\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = 0
and similarly for V(x, y) .
Since f(z) = u(x, y) + iv(x, y) is analytic in D , the real part u(x, y) and the imaginary part v(x, y) satisfy
Laplace’s equation individually. That is:
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \quad
\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0.
Now consider U(x, y) = e^{u(x, y)} \cos v(x, y) . To show that U is harmonic, compute its second
derivatives:
\frac{\partial U}{\partial x} = e^{u} \left( \frac{\partial u}{\partial x} \cos v - \sin v \frac{\partial v}{\partial x} \right),
\frac{\partial^2 U}{\partial x^2} = e^{u} \left[ \left(\frac{\partial^2 u}{\partial x^2} + \left(\frac{\partial u}{\partial x}\right)^2
\right)\cos v - 2 \frac{\partial u}{\partial x} \frac{\partial v}{\partial x} \sin v - \cos v \frac{\partial^2 v}{\partial x^2} +
\left(\frac{\partial v}{\partial x}\right)^2 \cos v \right].
Perform similar computations for \frac{\partial^2 U}{\partial y^2} . Combine these terms, and use the
Cauchy-Riemann equations:
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},
and the fact that u and v satisfy Laplace’s equation. This results in:
\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = 0.
Thus, U(x, y) is harmonic. A similar calculation shows V(x, y) = e^{u(x, y)} \sin v(x, y) is also harmonic.
2. V(x, y) as the harmonic conjugate of U(x, y) :
Two functions U(x, y) and V(x, y) are harmonic conjugates if they satisfy the Cauchy-Riemann
equations:
\frac{\partial U}{\partial x} = \frac{\partial V}{\partial y}, \quad \frac{\partial U}{\partial y} = -\frac{\partial V}{\partial x}.
Compute the derivatives:
\frac{\partial U}{\partial x} = e^{u} \left( \frac{\partial u}{\partial x} \cos v - \sin v \frac{\partial v}{\partial x} \right),
\frac{\partial V}{\partial y} = e^{u} \left( \frac{\partial u}{\partial y} \sin v + \cos v \frac{\partial v}{\partial y} \right).
Using the Cauchy-Riemann equations for u and v , we find that:
\frac{\partial U}{\partial x} = \frac{\partial V}{\partial y}.
Similarly, check:
\frac{\partial U}{\partial y} = -\frac{\partial V}{\partial x}.
Thus, V(x, y) is the harmonic conjugate of U(x, y) .
