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Chapter 8: Problem 2

The modulus is always easy to compute, using e.g. \(|z|=\sqrt{z \overline{2}} .\) The argument is often harder to isolate, since inverse trigonometric functions are involved. \(\mathrm{A}\) general closed formula will be given in Exercise 21, Sect. 1.2. For instance, for real positive values of \(a\), we have $$ \operatorname{Arg} \frac{1+i a}{1-i a}=\arccos \frac{1-a^{2}}{1+a^{2}}=2 \arctan a $$

Short Answer

Expert verified
The argument is \(2 \arctan a\).

Step by step solution

01

Understand the Modulus

The modulus of a complex number is a measure of its magnitude. You're provided the formula \(|z|=\sqrt{z \overline{2}}\), though this seems incorrect for calculating a modulus directly. Remember, the modulus is actually calculated as \(|z| = \sqrt{x^2 + y^2}\) for a complex number \(z = x + yi\). However, for this task, modulus computation isn't necessary directly.
02

Simplify Given Expression

We are provided an expression for the argument: \(\operatorname{Arg} \frac{1+i a}{1-i a}\). This involves the division of two complex numbers, and such expression represents a rotation in the complex plane.
03

Apply Argument Properties

There is a formula to find the argument of a ratio of two complex numbers. For a general complex number \(z_1\) divided by \(z_2\), \(\operatorname{Arg} \left(\frac{z_1}{z_2}\right) = \operatorname{Arg}(z_1) - \operatorname{Arg}(z_2)\). This simplifies to finding, for the numerator \((1+ia)\), the argument is \(\arctan(a/1) = \arctan(a)\), and for the denominator \((1-ia)\), it is \(\arctan(-a/1) = \arctan(-a)\).
04

Use Trigonometric Identities

From step 3, applying the formula leads to: \(\operatorname{Arg} \frac{1+i a}{1-i a} = \arctan(a) - \arctan(-a)\). Using the property of tangent, this further simplifies to \(2 \arctan a\).
05

Verify with Given Relation

The given solution relates the expression to an identity involving \(\arccos\). We verify this by checking: \(\arccos \frac{1-a^2}{1+a^2} = 2 \arctan a\). Both expressions correctly equate to the same result, confirming the solution.

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Key Concepts

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

Modulus of Complex Numbers
The modulus of a complex number gives you its size or magnitude, much like the distance between a point and the origin in the coordinate plane. For a complex number expressed as \( z = x + yi \), the modulus is found using the formula \( |z| = \sqrt{x^2 + y^2} \). This formula is similar to the Pythagorean theorem, where \( x \) is the real part and \( y \) is the imaginary part. The modulus provides a crucial tool when working with complex numbers on the complex plane, as it gives the absolute value regardless of the number's location.
Argument of a Complex Number
The argument of a complex number tells us the direction angle of the number relative to the positive real axis on the complex plane. For a complex number \( z = x + yi \), the argument is calculated using \( \tan^{-1}(y/x) \) or \( \arctan(y/x) \). It measures the angle in radians from the positive x-axis to the line joining the origin to the point \((x, y)\). Finding the argument helps in transitioning to polar forms of complex numbers, which are extremely handy in advanced mathematics. Understanding the concept of argument can sometimes be challenging due to restrictions in inverse trigonometric functions, but it is invaluable in calculations involving complex divisions and transformations.
Trigonometric Identities
Trigonometric identities simplify complex calculations by reducing expressions into easier forms. In the context of complex numbers, these identities can connect angles and trigonometric functions in a way that aids in problem-solving. For example, applying \( \tan^{-1}(a) - \tan^{-1}(-a) = 2 \arctan(a) \) is a neat way of simplifying expressions during argument calculations. These identities provide shortcuts and insights during mathematical manipulations, proving essential in both basic and advanced mathematics.
Complex Division
The operation of dividing complex numbers might appear daunting, but it鈥檚 merely an extension of division using real numbers. To divide complex numbers, you multiply the numerator and the denominator by the conjugate of the denominator. For instance, dividing \( \frac{1 + ia}{1 - ia} \) involves multiplying by \(\frac{1 + ia}{1 + ia} \) to simplify the denominator to a real number. This technique transforms complex division into a problem of algebraic manipulation, providing a straightforward path to solutions, including finding the argument of the quotient as demonstrated in various mathematical problems.
Complex Plane
The complex plane, or Argand plane, is a two-dimensional plane used to represent complex numbers graphically. It maps each complex number \( z = x + yi \) to a unique point \((x, y)\). This plane is similar to the Cartesian coordinate system, with the x-axis representing the real part and the y-axis representing the imaginary part. By using the complex plane, one can visually interpret operations like addition, subtraction, and even rotation of complex numbers, which are often represented as vector transformations. Familiarity with the complex plane is essential for understanding deeper concepts in complex number theory and applications in physics and engineering.

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Most popular questions from this chapter

We have \(\left|e^{i z^{2}}\right|=e^{-R^{2} \sin 2 t}\), and use then the estimate $$ \sin (2 t) \geq \frac{4}{\pi} t \quad \text { for } \quad 0 \leq t \leq \frac{\pi}{4} $$

The oriented angle between two "vectors" \(z, w \in \mathbb{C}^{*}\) is nothing else but the argument \(\operatorname{Arg}(w / z) .\) Let now \(z=\alpha^{\prime}(0), w=\beta^{\prime}(0) .\) The chain rule gives \((f \circ\) \(\alpha)^{\prime}(0)=f^{\prime}(a) \alpha^{\prime}(0),(f \circ \beta)^{\prime}(0)=f^{\prime}(a) \beta^{\prime}(0)\), which implies $$ \operatorname{Arg} \frac{(f \circ \beta)^{\prime}(0)}{(f \circ \alpha)^{\prime}(0)}=\operatorname{Arg} \frac{\beta^{\prime}(0)}{\alpha^{\prime}(0)} $$

Check first that \(\mathcal{C}\) is stable with respect to addition and multiplication of matrices. So \(\mathcal{C}\) is a ring. The map $$ \mathbb{C} \longrightarrow \mathcal{C}, \quad a+\mathrm{i} b \longmapsto\left(\begin{array}{rr} a & -b \\ b & a \end{array}\right) $$ is an isomorphism. One can independently of the knowledge of \(\mathbb{C}\) compute, that \(\mathcal{C}\) satisfies the axioms for "the" complex number field.

The factor ring \(K:=\mathbb{R}[X] /\left(X^{2}+1\right)\) is a field, since \(\mathbb{R}\) is a field and \(X^{2}+1\) is a prime prime element in \(\mathbb{R}[X]\). This must be proved. Let us denote by \(1_{K}\) the image of 1 , and by \(\mathrm{i}_{K}\) the image of \(X\) in \(K\). Then \(K=\mathbb{R} 1_{K}+\mathbb{R i}_{K}\). The axioms for a "field of complex numbers" are now easily checked, e.g. \(\mathrm{i}_{K}^{2}=-1_{K}\).

We write \(\omega_{1}=x_{1}+\mathrm{i} y_{1}, \omega_{2}=x_{2}+\mathrm{iy}_{2}\). The fundamental parallelogram \(\mathcal{F}\) is the image of the unit square under a linear map. The volume of \(\mathcal{F}\) is then the modulus of the determinant of the transformation matrix, namely \(\left|x_{1} y_{2}-x_{2} y_{1}\right|\), in concordance to the claimed formula. The invariance follows from Exercise \(6 .\)
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