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Chapter 1: Problem 20

(a) For each of the following complex numbers calculate the principal value of the logarithm: $$ \text { i; } \quad-\mathrm{i} ; \quad-1 ; \quad x \in \mathbb{R}, x>0 ; \quad 1+\mathrm{i} $$ (b) Calculate the principal values of the following numbers and compare them. $$ (i(i-1))^{1} \text { and } i^{1} \cdot(i-1)^{1} $$ (c) Calculate $$ \left\\{a^{b}\right\\}:=\\{\exp (b \log |a|+i b \operatorname{Arg} a) \exp (2 \pi \mathrm{i} b k) ; k \in \mathbb{Z}\\} $$ for $$ (a, b) \in\\{(-1, \mathrm{i}),(1, \sqrt{2}),(-2, \sqrt{2})\\} $$

Short Answer

Expert verified
The principal logarithms are calculated for complex numbers, while multiplication equality holds, with complex calculations influenced by the base sign during exponentiation.

Step by step solution

01

Define Principal Value of Logarithm

The principal value of the logarithm for a complex number \( z \) is given by \( ext{Log}(z) = ext{ln}|z| + i \text{Arg}(z) \), where \( \text{Arg}(z) \) is the principal argument of \( z \), restricted to \( (-\pi, \pi] \).
02

Calculate Logarithm for \( i \)

For \( i \), the modulus \( |i| = 1 \) and the principal argument \( \text{Arg}(i) = \frac{\pi}{2} \). Thus, \( \text{Log}(i) = \text{ln}(1) + i \frac{\pi}{2} = i \frac{\pi}{2} \).
03

Calculate Logarithm for \( -i \)

For \( -i \), the modulus \( |-i| = 1 \) and the principal argument \( \text{Arg}(-i) = -\frac{\pi}{2} \). Thus \( \text{Log}(-i) = \text{ln}(1) + i(-\frac{\pi}{2}) = -i \frac{\pi}{2} \).
04

Calculate Logarithm for \( -1 \)

For \( -1 \), the modulus \( |-1| = 1 \) and the principal argument \( \text{Arg}(-1) = \pi \). Thus \( \text{Log}(-1) = \text{ln}(1) + i \pi = i \pi \).
05

Calculate Logarithm for \( x \in \mathbb{R}, x>0 \)

For a positive real number \( x \), the modulus \( |x| = x \) and the principal argument \( \text{Arg}(x) = 0 \). Thus \( \text{Log}(x) = \text{ln}(x) + i \cdot 0 = \text{ln}(x) \).
06

Calculate Logarithm for \( 1 + i \)

For \( 1+i \), the modulus \( |1+i| = \sqrt{2} \) and the principal argument \( \text{Arg}(1+i) = \frac{\pi}{4} \). Therefore, \( \text{Log}(1+i) = \text{ln}(\sqrt{2}) + i \frac{\pi}{4} = \frac{\ln(2)}{2} + i \frac{\pi}{4} \).
07

Compute \( (i(i-1))^1 \)

First calculate \( i(i-1) = i^2 - i = -1 - i \). Its modulus is \( \sqrt{2} \) and argument \( -\frac{3\pi}{4} \). Thus, \( (i(i-1)) = \sqrt{2} \exp(-i \frac{3\pi}{4}) \).
08

Compute \( i^1 \cdot (i-1)^1 \)

Calculate \( (i-1) = -1+i \). Its modulus is \( \sqrt{2} \) and argument \( \frac{3\pi}{4} \). Thus, \( i \cdot (i-1) = \exp(i \frac{\pi}{2} + \ln(\sqrt{2}) + i \frac{3\pi}{4}) \).
09

Compare Principal Values

The principal values are equivalent for both expressions as they both convert back to original complex multiplication results. Thus, both complex numbers have same magnitude and value under multiplication.
10

Compute \( a^b \) for \((-1, i)\)

For \((-1, i)\), use \( \exp(0) = 1 \) since exponent multiplies to zero due to \( \log|-1| = 0 \). No imaginary part change again resets return of initial value.
11

Compute \( a^b \) for \((1, \sqrt{2})\)

Evaluate \( 1^{\sqrt{2}} = 1 \) since exponent impacts non-existing imaginary to no change repeat since zero logarithms.
12

Compute \( a^b \) for \((-2, \sqrt{2})\)

From \( a = -2 \), \( \log|-2|, \pi \) combinations lead to calculations equivalent in procedures showing variation due to negative intricate exponent effect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Numbers
A complex number is a number that has both a real and an imaginary part, usually expressed in the form \( z = a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit satisfying \( i^2 = -1 \). Complex numbers extend the real numbers and enable the solution of equations that have no real solutions, such as \( x^2 + 1 = 0 \). Understanding complex numbers is crucial for fields such as engineering, physics, and applied mathematics.

Each complex number has a
  • Real part: denoted as \( a \) in \( a + bi \).
  • Imaginary part: denoted as \( b \) in \( a + bi \). The imaginary part includes \( i \), making it distinguishable from the real part.
Complex numbers can be graphically represented on a plane called the complex plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part. This geometric interpretation helps make operations on complex numbers such as addition, subtraction, and multiplication more intuitive.

The modulus of a complex number \( z = a + bi \) is its distance from the origin in the complex plane, calculated as \(|z| = \sqrt{a^2 + b^2} \). The modulus is always a non-negative real number and provides a measure of the size or magnitude of the complex number.
Argument of a Complex Number
The argument of a complex number is the angle formed between the positive real axis and the line representing the complex number in the complex plane. For a complex number \( z = a + bi \), the argument is denoted as \( \text{Arg}(z) \). It provides the direction of the complex number in the plane.

This angle is generally expressed in radians and is determined using the formula:
  • \( \text{Arg}(z) = \text{atan2}(b, a) \), where \( \text{atan2} \) is a variation of the arctangent function that takes into account the signs of both \( a \) and \( b \) to determine the correct quadrant of the complex number.
An important aspect of complex numbers is the principal argument, which restricts the angle to the interval \((-\pi, \pi] \). This restriction ensures each complex number has a unique principal value.

Understanding the argument is essential for performing advanced operations like finding the principal value of the logarithm of a complex number, as it defines the imaginary component of the logarithm, ensuring that calculations involving complex exponentiation are correctly interpreted.
Complex Exponentiation
Complex exponentiation extends the concept of raising numbers to power from real numbers to complex numbers. This involves the use of Euler's formula, which states \( e^{i\theta} = \cos \theta + i \sin \theta \). Therefore, raising a complex number \( z = re^{i\phi} \) to a power involves using its polar form where \( r \) is the modulus and \( \phi \) is the argument.

Performing exponentiation on a complex number \( a^b \) often uses the identity:
  • \( a^b = \exp(b \cdot \log(a)) \), where \( \log(a) \) is the logarithm of the complex number \( a \).
The principal value of \( a^b \) involves computing
  • \( \exp(b \cdot (\ln|a| + i\cdot \text{Arg}(a))) \).
This results in a new complex number that reflects the power effects on both magnitude and direction (argument). It's crucial to pay attention to the principal argument for accurate results in these calculations.

Complex exponentiation is essential for understanding phenomena such as waveforms in physics and signal processing, where complex numbers and exponentials are used to model and analyze oscillatory phenomena.

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Most popular questions from this chapter

If \(u: \mathbb{R}^{2} \rightarrow \mathbb{R}\) is a harmonic polynomial (in two real variables), then $$ f(z)=2 u\left(\frac{z}{2}, \frac{z}{2 \mathrm{i}}\right)-u(0,0) $$ is an analytic function with real part \(u\).

Definition of the tangent and cotangent For \(z \in \mathbb{C} \backslash\\{(k+1 / 2) \pi ; k \in \mathbb{Z}\\}\) let $$ \tan z:=\frac{\sin z}{\cos z} $$ and for \(z \in \mathbb{C} \backslash\\{k \pi ; k \in \mathbb{Z}\\}\) let $$ \cot z:=\frac{\cos z}{\sin z} $$ Show: $$ \begin{gathered} \tan z=\frac{1}{\mathrm{i}} \frac{\exp (2 \mathrm{i} z)-1}{\exp (2 \mathrm{i} z)+1}, \quad \cot z=\mathrm{i} \frac{\exp (2 \mathrm{i} z)+1}{\exp (2 \mathrm{i} z)-1} \\ \tan (z+\pi / 2)=-\cot z, \quad \tan (-z)=-\tan z, \quad \tan z=\tan (z+\pi) \\\ \tan z=\cot z-2 \cot (2 z), \quad \cot (z+\pi)=\cot z \end{gathered} $$

Binomial series For \(\alpha \in \mathbb{C}\) and \(\nu \in \mathbb{N}\) let $$ \left(\begin{array}{c} \alpha \\ 0 \end{array}\right):=1 \quad \text { and } \quad\left(\begin{array}{l} \alpha \\ \nu \end{array}\right):=\prod_{j=1}^{\nu} \frac{\alpha-j+1}{j} $$ Show: \(\sum_{\nu=0}^{\infty}\left(\begin{array}{c}\alpha \\\ \nu\end{array}\right) z^{\nu}\) is absolutely convergent for all \(z \in \mathbb{C}\) with \(|z|<1\) Let \(b_{\alpha}(z):=\sum_{\nu=0}^{\infty}\left(\begin{array}{c}\alpha \\\ \nu\end{array}\right) z^{\nu} .\) Show: For all \(z \in \mathbb{C}\) with \(|z|<1\) and arbitrary \(\alpha, \beta \in \mathbb{C}\) we have $$ b_{\alpha+\beta}(z)=b_{\alpha}(z) b_{\beta}(z) $$ Remark. We shall see later that for \(z \in \mathbb{C}\) with \(|z|<1\) there holds $$ b_{\alpha}(z)=(1+z)^{\alpha}:=\exp (\alpha \log (1+z)) $$ For \(\alpha=n \in \mathbb{N}_{0}\), one obtains the binomial formula $$ (1+z)^{n}=\sum_{v=0}^{n}\left(\begin{array}{l} n \\ \nu \end{array}\right) z^{\nu} $$

Write the following functions in the form \(f=u+\mathrm{i} v\) and give explicit formulas for \(u\) and \(v\). (a) \(f(z)=\sin z\) (b) \(f(z)=\cos z\) (c) \(f(z)=\sinh z\), (d) \(f(z)=\cosh z\) (e) \(f(z)=\exp \left(z^{2}\right)\) (f) \(f(z)=z^{3}+z\). Show that in every case the CAUCHY-RIEMANN equations are satisfied (for all \(z \in \mathbb{C}\) ), and conclude that these functions are analytic in \(\mathbb{C}\).

Let us consider the upper half-plane \(\mathbb{H}=\\{z \in \mathbb{C} ; \quad \operatorname{Im} z>0\\}\) and the unit disk \(\mathbb{E}=\\{q \in \mathbb{C} ; \quad|q|<1\\}\) Show: The function $$ f(z):=\frac{z-\mathrm{i}}{z+\mathrm{i}} $$ provides a globally conformal map of H onto \(\mathbb{E}\). What is its inverse map? The map \(f\) is also called the Cayley map (A. CAYLEY, 1846 ).
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