Question

In Problems 5 to 8, find the residues of the given function at all poles. Take \(z=r e^{i \theta}\), \(0 \leq \theta<2 \pi\). $$ \frac{\ln z}{1+z^{2}} $$

Short answer

The residues at the poles \( z = i \) and \( z = -i \) are both \( \frac{\pi}{4} \).

Step-by-step solution

Identify the poles

The poles occur where the denominator is zero. For the given function \( \frac{\ln z}{1+z^2} \), set \( 1+z^2 = 0 \). Solving for \( z \), we get:\[ z^2 = -1 \rightarrow z = \pm i \] Hence, the poles are at \( z = i \) and \( z = -i \).

Compute the residue at \( z = i \)

The residue of a function \( f(z) \) at a simple pole \( z_0 \) is given by \( \lim_{z \to z_0} (z - z_0) f(z) \). For \( z = i \):\[ \text{Residue at } z = i = \lim_{z \to i} (z - i) \left( \frac{\ln z}{1+z^2} \right) = \lim_{z \to i} \frac{(z - i) \ln z}{1 + z^2} \] Since \( 1 + z^2 = (z - i)(z + i) \), we get:\[ = \lim_{z \to i} \frac{(z - i) \ln z}{(z - i)(z + i)} = \lim_{z \to i} \frac{\ln z}{z + i} \] Plug in \( z = i \):\[ \text{Residue at } z = i = \frac{\ln i}{2i} \]

Compute the residue at \( z = -i \)

Similarly, for \( z = -i \):\[ \text{Residue at } z = -i = \lim_{z \to -i} (z + i) \left( \frac{\ln z}{1+z^2} \right) = \lim_{z \to -i} \frac{(z + i) \ln z}{1 + z^2} \] Simplifying using \( 1 + z^2 = (z - i)(z + i) \), we get:\[ = \lim_{z \to -i} \frac{(z + i) \ln z}{(z - i)(z + i)} = \lim_{z \to -i} \frac{\ln z}{z - i} \] Plug in \( z = -i \):\[ \text{Residue at } z = -i = \frac{\ln (-i)}{-2i} \]

Simplify the residues

To simplify:\[ \text{Residue at } z = i = \frac{\ln i}{2i} \] Recall that \( \ln i = i \frac{\pi}{2} \), so:\[ \text{Residue at } z = i = \frac{i \frac{\pi}{2}}{2i} = \frac{\pi}{4} \] Similarly, for \( z = -i \):\[ \text{Residue at } z = -i = \frac{\ln (-i)}{-2i} \] and \( \ln (-i) = -i \frac{\pi}{2} \) so:\[ \text{Residue at } z = -i = \frac{-i \frac{\pi}{2}}{-2i} = \frac{\pi}{4} \]

Key concepts

Residue Theorem

The Residue Theorem is a significant result in complex analysis that allows us to evaluate complex integrals easily. It relates integrals around closed curves to the sum of residues within that curve. A residue of a function at a point is essentially the coefficient of \(\frac{1}{z - a}\) in its Laurent series expansion around that point. For example, in the given problem, the function \(\frac{\text{ln z}}{1+z^2}\)\has poles at \(\text{z = i and z = -i}\). By calculating the residues at these poles, we can apply the residue theorem to evaluate integrals involving this function.

Poles of a Function

Poles are specific types of singularities in complex functions where the function heads towards infinity. A pole at \(z = z_0\) occurs when the function can be written as \(\frac{h(z)}{(z - z_0)^n}\), with \(h(z)\) holomorphic (complex differentiable) and \(n\) being a positive integer. In the given exercise, the poles occur where the denominator \(\text{1 + z}^2= 0\)\. This simplifies to \(\text{z}^2 = -1 \rightarrow z = \text{±i}\). Steps to identify poles:

• Set the denominator of the function to zero.
• Simplify to solve for \(z\).

In this example, solving \(1 + z^2 = 0\) gives the poles at \(\text{z = i and z = -i}\).

Logarithmic Function

A logarithmic function in the complex plane, denoted as \(\text{ln} z\), is a multi-valued function because of the periodic nature of the complex exponential function. The principal branch of \(\text{ln} z\) is usually defined on the complex plane cut along the negative real axis:

• For any complex number \(z = re^{i\theta}\), where \(\text{0 ≤ θ < 2π}\), \(\text{ln} z = \text{ln} r + iθ\).
• Recall the Euler's formula: \(\text{e}^{i\theta} = \text{cos}θ + i \text{sin}θ\).

In the given exercise, to compute the residues at poles, we use the logarithm of \(i\) and \(-i\), with \(\text{ln} i = i \frac{\text{π}}{2}\) and \(\text{ln}(-i) = -i \frac{\text{π}}{2}\).

Complex Numbers

A complex number is written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(\text{i}\) is the imaginary unit satisfying \(\text{i}^2 = -1\). Complex numbers extend the idea of one-dimensional real numbers to the two-dimensional complex plane by including \(i\). They are essential in various fields of mathematics and engineering because they simplify calculations involving oscillations.

• When working with functions like \(\frac{\text{ln} z}{1 + z^2}\), it's crucial to understand both real and imaginary parts.
• Powers and roots of complex numbers can be computed using Euler's formula, which links complex exponentiation with trigonometric functions.
• In polar form, a complex number \(z = re^{iθ}\), where \(r\) is the magnitude and \(θ\) is the argument (angle).

For instance, the poles \(z = i\) and \(-i\) in the given exercise are points on the complex plane where the function heads towards infinity.