Dumelang guys!!!

In this post, we will explore the real and imaginary parts of the complex logarithm function \( w = \log{z} \),

and show that they satisfy the Cauchy-Riemann equations when \( z \neq 0 \).  Additionally, we will compute the partial derivatives of these parts.

Expressing \( w = \log{z} \) in terms of real and imaginary parts

Let \( z = x + iy \) be a complex number, where \( x \) and \( y \) are real numbers. The complex logarithm function is defined as:

$w = \log{z}$

We can express the logarithm in polar form, where:

$z = r e^{i\theta}$

Here, \( r = |z| = \sqrt{x^2 + y^2} \) is the modulus of \( z \), and \( \theta = \arg(z) = \tan^{-1}\left( \frac{y}{x} \right) \) is the argument of \( z \). Then, the logarithm becomes:

$w = \log{z} = \log{r} + i \theta$

Thus, we can separate the real and imaginary parts of \( w \) as follows:

- The real part of \( w \) is:

$u(x, y) = \log{r} = \frac{1}{2} \log(x^2 + y^2)$

- The imaginary part of \( w \) is:

$v(x, y) = \theta = \arg{(z)} = \tan^{-1}\left( \frac{y}{x} \right)$

Cauchy-Riemann Equations

For a complex function \( f(z) = u(x, y) + iv(x, y) \), the Cauchy-Riemann equations state that:

1. \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \),

2. \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \).

We will now compute the partial derivatives of \( u(x, y) \) and \( v(x, y) \) and check if these equations are satisfied.

Computing the Partial Derivatives

1. Derivatives of the real part \( u(x, y) \)

The real part of \( w \) is given by:

$u(x, y) = \frac{1}{2} \log(x^2 + y^2)$

Now, let's compute the partial derivatives:

$\frac{\partial u}{\partial x} = \frac{1}{2} \cdot \frac{2x}{x^2 + y^2} = \frac{x}{x^2 + y^2}$

$\frac{\partial u}{\partial y} = \frac{1}{2} \cdot \frac{2y}{x^2 + y^2} = \frac{y}{x^2 + y^2}$

2. Derivatives of the imaginary part \( v(x, y) \)

The imaginary part of \( w \) is:

$v(x, y) = \tan^{-1}\left( \frac{y}{x} \right)$

Now, let's compute the partial derivatives:

$\frac{\partial v}{\partial x} = \frac{\partial}{\partial x} \left( \tan^{-1} \left( \frac{y}{x} \right) \right) = \frac{-y}{x^2 + y^2}$

$\frac{\partial v}{\partial y} = \frac{\partial}{\partial y} \left( \tan^{-1} \left( \frac{y}{x} \right) \right) = \frac{x}{x^2 + y^2}$

Verifying the Cauchy-Riemann Equations

We now verify the Cauchy-Riemann equations by checking if the following hold:

First equation:

$\frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2}, \quad \frac{\partial v}{\partial y} = \frac{x}{x^2 + y^2}$

Thus, the first Cauchy-Riemann equation is satisfied.

Second equation:

$\frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2}, \quad -\frac{\partial v}{\partial x} = -\left( \frac{-y}{x^2 + y^2} \right) = \frac{y}{x^2 + y^2}$

Thus, the second Cauchy-Riemann equatin is also satisfied.

Since both Cauchy-Riemann equations are satisfied, the real and imaginary parts of \( w = \log{z} \) satisfy the Cauchy-Riemann equations for \( z \neq 0 \).

Derivatives of the Function \( w = \log{z} \)

We can now compute the derivative of \( w \) with respect to \( z \).

Since \( w = \log{z} \), hence $w=\log{z}$ is analytic except for $x^2+y^2=0\Rightarrow{x=y=0}\Rightarrow{x+iy=0}\Rightarrow{z=0}$. 

Now, $w=u+iv$, then  the derivative of \( w \) with respect to \( z \) is:

$\frac{dw}{dz}=\frac{\partial{u}}{\partial{x}}+i\frac{\partial v}{\partial x}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}=\frac{x-iy}{x^2+y^2}=\frac{x-iy}{(x+iy)(x-iy)}=\frac{1}{x+iy}=\frac{1}{z}$.

Thus, the derivative of the complex logarithm function is:

$\frac{d}{dz} \log{z} = \frac{1}{z}$

We have shown that the real and imaginary parts of \( w = \log{z} \) satisfy the Cauchy-Riemann equations for \( z \neq 0 \), and we have also computed the derivative of the logarithm function. This function is analytic everywhere in the complex plane except at \( z = 0 \), where it has a singularity.

Thank you for reading! Feel free to leave any questions or comments below.