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Chapter 1: Problem 2

Calculate the absolute value and an argument for each of the following complex numbers: $$ \begin{gathered} -3+\mathrm{i} ; \quad-13 ; \quad(1+\mathrm{i})^{17}-(1-\mathrm{i})^{17} ; \quad \mathrm{i}^{4711} ; \quad \frac{3+4 \mathrm{i}}{1-2 \mathrm{i}} \\ \frac{1+\mathrm{i} a}{1-\mathrm{i} a}, a \in \mathbb{R} ; \quad \frac{1-\mathrm{i} \sqrt{3}}{1+\mathrm{i} \sqrt{3}} ; \quad(1-\mathrm{i})^{n}, n \in \mathbb{Z} \end{gathered} $$

Short Answer

Expert verified
Calculate the modulus and argument for each complex number using the specified steps for each case.

Step by step solution

01

Understanding Complex Numbers

Complex numbers are in the form of \( a + bi \), where \( a \) is the real part, and \( b \) is the imaginary part. The absolute value (or modulus) of a complex number \( z = a + bi \) is given by \( |z| = \sqrt{a^2 + b^2} \). The argument, \( \arg(z) \), is the angle the line representing the number makes with the positive real axis, typically measured in radians.
02

Calculate Absolute Value and Argument of \(-3 + i\)

1. For \( z = -3 + i \): - Real part \( a = -3 \), Imaginary part \( b = 1 \). - Absolute value: \( |-3 + i| = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \). - Argument: \( \arg(z) = \tan^{-1}\left(\frac{1}{-3}\right) \). Since the point \((-3, 1)\) is in the second quadrant, the angle is \( \pi - \tan^{-1}(1/3) \).
03

Calculate Absolute Value and Argument of \(-13\)

2. For \( z = -13 \) (purely real): - Real part \( a = -13 \), Imaginary part \( b = 0 \). - Absolute value: \( |-13| = | -13 | = 13 \). - Argument: As it lies on the negative real axis, \( \arg(z) = \pi \).
04

Calculate (\(1+i\))^{17} - (\(1-i\))^{17}

3. For \( (1+i)^{17} - (1-i)^{17} \): - Use De Moivre's Theorem. Express \( 1+i \) and \( 1-i \) in polar form. - \( 1+i = \sqrt{2}(\cos(\pi/4) + i\sin(\pi/4)) \). - \( 1-i = \sqrt{2}(\cos(-\pi/4) + i\sin(-\pi/4)) \). - \( (1+i)^{17} = (\sqrt{2})^{17}\left(\cos(17\pi/4) + i\sin(17\pi/4)\right) \). - \( (1-i)^{17} = (\sqrt{2})^{17}\left(\cos(-17\pi/4) + i\sin(-17\pi/4)\right) \). - Simplify expressions and subtract these results. This gives a purely imaginary final result based on symmetry and periodicity characteristics of trigonometric functions.
05

Calculate i^{4711}

4. For \( i^{4711} \): - Know that powers of \( i \) cycle every 4: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \). - Divide 4711 by 4: \( 4711 \div 4 = 1177 \text{ remainder } 3 \). - Thus, \( i^{4711} = i^3 = -i \). - Absolute value: 1, Argument: \( -\frac{\pi}{2} \) or equivalently \( \frac{3\pi}{2} \).
06

Calculate \(\frac{3+4i}{1-2i}\)

5. For \( \frac{3+4i}{1-2i} \): - Multiply numerator and denominator by the conjugate of the denominator: - \( \frac{3+4i}{1-2i} \times \frac{1+2i}{1+2i} = \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)} \). - Simplify: Numerator \( = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i \). - Denominator \( = 1 + 4 = 5 \). - Result: \( -1 + 2i \). - Absolute value: \( |-1 + 2i| = \sqrt{(-1)^2 + 2^2} = \sqrt{5} \). - Argument: \( \arg(z) = \tan^{-1}(2) \). Since \((-1, 2)\) is in the second quadrant, \( \pi - \tan^{-1}(2) \).
07

General Case: \(\frac{1+ia}{1-ia}\), \(a \in \mathbb{R}\)

6. For \( \frac{1+ia}{1-ia} \): - Use conjugate: \( \frac{1+ia}{1-ia} \times \frac{1+ia}{1+ia} = \frac{(1+ia)^2}{1+a^2} \). - Numerator: \( (1+ia)^2 = 1 + 2ia - a^2 \). - Denominator: \( 1 + a^2 \). - Result: \( \frac{1-a^2}{1+a^2} + i\cdot\frac{2a}{1+a^2} \). - Absolute value is 1 due to symmetry: \( |z| = 1 \). - Argument: \( \tan^{-1}\left(\frac{2a}{1-a^2}\right) \).
08

Calculate \(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\)

7. For \( \frac{1-i\sqrt{3}}{1+i\sqrt{3}} \): - Multiply by conjugate: \( \frac{1-i\sqrt{3}}{1+i\sqrt{3}} \times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} = \frac{(1-i\sqrt{3})^2}{1+3}= \frac{1-2i\sqrt{3}-3}{4} \). - Simplify to: \( \frac{-2-2i\sqrt{3}}{4} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \). - Absolute value: \( |z| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = 1 \). - Argument: \( \arg(z) = \tan^{-1}\left(-\sqrt{3}/1\right) = -\frac{2\pi}{3} \) or equivalently \( \frac{4\pi}{3} \).
09

Calculate \((1-i)^{n}\), \(n \in \mathbb{Z}\)

8. For \( (1-i)^n \): - Base \( 1-i \) is expressed in polar form: \( \sqrt{2}(\cos(-\pi/4) + i\sin(-\pi/4)) \). - By De Moivre's Theorem: \( (1-i)^n = (\sqrt{2})^n \left( \cos\left(-\frac{n\pi}{4}\right) + i\sin\left(-\frac{n\pi}{4}\right) \right) \). - Evaluate based on specific \( n \) values. Generally, \( (\sqrt{2})^n \) for modulus and \( -\frac{n\pi}{4} \) for argument adjusted as needed for quadrant positioning.

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Key Concepts

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

Absolute Value of Complex Numbers
The absolute value, or modulus, of a complex number gives us an idea of its "size". For a complex number in the form of \( z = a + bi \), the absolute value is calculated using the formula \( |z| = \sqrt{a^2 + b^2} \). This formula essentially treats the complex number as a point described by Cartesian coordinates \((a, b)\) and calculates the Euclidean distance from that point to the origin \((0, 0)\) on the complex plane.
  • For a purely imaginary or real number, such as \(-13\), the absolute value simplifies to the absolute value of the real or imaginary component since the other component is zero.
  • The absolute value is always a non-negative real number.
  • For example, the absolute value of the complex number \(-3 + i\) is \(\sqrt{10}\), calculated by finding the distance from the point \((-3, 1)\) to the origin.
Understanding this concept is critical for visualizing and analyzing the position of complex numbers in the complex plane.
Argument of Complex Numbers
The argument of a complex number is the angle the line representing the complex number makes with the positive real axis in the complex plane. This angle gives us the direction of the vector representation and is usually measured in radians. For any complex number \( z = a + bi \), the argument is calculated using \( \arg(z) = \tan^{-1}\left(\frac{b}{a}\right) \).
  • The calculated angle must be adjusted based on the quadrant in which the point \((a, b)\) is located, since the tangent function is periodic.
  • For example, the argument of \(-3 + i\), residing in the second quadrant, is \(\pi - \tan^{-1}\left(\frac{1}{3}\right)\).
  • Similarly, for a purely real number like \(-13\), the argument is \(\pi\), since it lies on the negative real axis.
The argument helps in transforming the rectangular form of a complex number into its polar form, which is essential for multiplication and division operations.
De Moivre's Theorem
De Moivre's Theorem is a powerful tool for raising complex numbers to integer powers and extracting roots. It states that if a complex number is in polar form \( z = r(\cos \theta + i\sin \theta) \), then \( z^n = r^n (\cos(n\theta) + i\sin(n\theta)) \).
  • This theorem is particularly useful when dealing with expressions like \((1+i)^{17}\) or \((1-i)^{n}\).
  • To use De Moivre's Theorem, the complex number needs to be in polar form: for \(1+i\), it is \(\sqrt{2}(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}))\).
  • Applying the theorem simplifies the complex multiplications to manageable trigonometric problems.
It's crucial for solving more complex problems involving powers of complex numbers, offering an elegant method over Cartesian computations.
Complex Number Polar Form
The polar form of a complex number is an elegant way of expressing complex numbers using their magnitude (absolute value) and direction (argument). It is given as \( z = r(\cos \theta + i\sin \theta) \), where \( r = |z| \) is the modulus and \( \theta = \arg(z) \) is the argument of the complex number.
  • Converting a complex number like \(1+i\) into polar form involves calculating \( r = \sqrt{2} \) and \( \theta = \frac{\pi}{4} \).
  • This transformation allows for easier computation of powers and roots using De Moivre's Theorem.
  • Polar form is very useful in multiplying and dividing complex numbers, as it simplifies the operations to working with their moduli and arguments separately.
Understanding the polar representation enhances comprehension of the geometric interpretation of complex numbers and prepares us for complex analyses like rotation and scaling in the complex plane.

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Most popular questions from this chapter

Prove HEINE's theorem (E. HEINE, 1872 ): If \(K \subset \mathbb{C}\) is compact and \(f: K \rightarrow \mathbb{C}\) is continuous then \(f\) is uniformly continuous on \(K\), i.e. for each \(e>0\) there exists a \(\delta>0\) so that for all \(z, z^{\prime} \in K\) with \(\left|z-z^{\prime}\right|<\delta\) $$ \left|f(z)-f\left(z^{\prime}\right)\right|<\varepsilon $$

Determine all \(z \in \mathbb{C}\) such that \(z^{3}-\mathrm{i}=0\)

Let \(M \subset \mathbb{R}^{p}\). A point \(a \in \mathbb{R}^{p}\) is called an accumulation point of \(M\) if for each \(\varepsilon\)-ball \(U_{\varepsilon}(a)\) there holds $$ U_{c}(a) \cap(M \backslash\\{a\\}) \neq \emptyset $$ In each \(\varepsilon\)-ball for \(a\) there is therefore a point of \(M\) different from \(a\) Notation. \(M^{\prime}:=\left\\{x \in \mathbb{R}^{p} ; x\right.\) is an accumulation point of \(\left.M\right\\}\) Show: For a subset \(A \subset \mathbb{R}^{p}\) the following are equivalent: (a) \(A\) is closed, i.e. \(\mathbb{R}^{p}-A\) is open. (b) For each convergent sequence \(\left(a_{n}\right), a_{n} \in A\) we have \(\lim _{n \rightarrow \infty} a_{n} \in A\). (c) \(A\) \supset \(A^{\prime}\) Show that in addition: $$ \bar{A}:=A \cup A^{\prime} $$ is always closed, and for each closed set \(B \subset \mathbb{R}^{p}\) with \(B \supset A\) we have \(B \supset \bar{A}\). \(\bar{A}\) is called the closure (or closed hull) of \(A\).

Verify for \(z=x+\mathrm{i} y \in \mathbb{C}\) the inequalities $$ \frac{|x|+|y|}{\sqrt{2}} \leq|z|=\sqrt{x^{2}+y^{2}} \leq|x|+|y| $$ and $$ \max \\{|x|,|y|\\} \leq|z| \leq \sqrt{2} \max \\{|x|,|y|\\} $$

Find the real and imaginary parts of each of the following complex numbers: $$ \begin{gathered} \frac{\mathrm{i}-1}{\mathrm{i}+1} ; \quad \frac{3+4 \mathrm{i}}{1-2 \mathrm{i}} ; \mathrm{i}^{n}, n \in \mathbb{Z} ; \quad\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)^{n}, n \in \mathbb{Z} ; \\ \left(\frac{1+\mathrm{i} \sqrt{3}}{2}\right)^{n}, n \in \mathbb{Z} ; \quad \sum_{\nu=0}^{7}\left(\frac{1-\mathrm{i}}{\sqrt{2}}\right)^{\nu} ; \frac{(1+\mathrm{i})^{4}}{(1-\mathrm{i})^{3}}+\frac{(1-\mathrm{i})^{4}}{(1+\mathrm{i})^{3}} \end{gathered} $$
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