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Chapter 1: Problem 2

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

Short Answer

Expert verified
Calculate moduli and arguments using modulus formula and tan inverse for angles. Consider quadrants for accurate angles.

Step by step solution

01

Calculate the modulus of -3 + i

For a complex number in the form \(a + bi\), the modulus is calculated as: \(\sqrt{a^2 + b^2}\). Here, \(a = -3\) and \(b = 1\), so the modulus is \(\sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}\).
02

Determine the argument of -3 + i

The argument, \(\theta\), is found using \(\tan \theta = \frac{b}{a}\). Here, \(\tan \theta = \frac{1}{-3}=-\frac{1}{3}\). Thus, \(\theta = \tan^{-1}(-\frac{1}{3})\). Since the complex number is in the second quadrant (negative real part, positive imaginary part), the argument is \(\pi - \tan^{-1}(\frac{1}{3})\).
03

Calculate the modulus of -13

For a real number \(-13\), the modulus is simply the absolute value, which is \(13\).
04

Determine the argument of -13

Since \(-13\) is a negative real number, its argument is \(\pi\).
05

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

Let \(z_1 = 1 + i\) and \(z_2 = 1 - i\). Since \(|z_1| = \sqrt{2}\) and \(\arg(z_1) = \frac{\pi}{4}\), applying De Moivre's Theorem gives \(z_1^{17} = (\sqrt{2})^{17}(\cos(\frac{17\pi}{4}) + i\sin(\frac{17\pi}{4}))\). Similarly for \(z_2 = 1 - i\), \(z_2^{17} = (\sqrt{2})^{17}(\cos(-\frac{17\pi}{4}) + i\sin(-\frac{17\pi}{4}))\). Subtracting gives us that the imaginay parts cancel out and only the real parts add up.
06

Calculate the modulus and argument of i^{4711}

First simplify \(i^{4711}\). Using \(i^4 = 1\), we know \(i^{4711} = (i^4)^{1177} \times i^3 = 1^{1177} \times i^3 = i^3 = -i\). The modulus of \(-i\) is \(1\) and the argument is \(-\frac{\pi}{2}\).
07

Calculate the modulus and argument of \(\frac{3+4i}{1-2i}\)

Multiply the numerator and denominator by the conjugate of the denominator: \( \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)} = \frac{3 + 6i + 4i - 8}{1 + 4} = \frac{-5 + 10i}{5} = -1 + 2i\). The modulus is \(\sqrt{1^2 + 2^2} = \sqrt{5}\) and the argument is \(\tan^{-1}(2)\).
08

Simplify \(\frac{1+i a}{1-i a}, a\in\mathbb{R}\)

Multiply the numerator and the denominator by the conjugate of the denominator: \( \frac{(1 + ia)(1 + ia)}{1^2 + a^2} = \frac{(1-a^2) + 2ia}{1 + a^2}\). The modulus is \(1\), since it simplifies to a form similar to euler's formula and the argument is \(2\tan^{-1}(a)\).
09

Calculate the modulus and argument of \(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\)

Multiply the numerator and the denominator by the conjugate of the denominator. The result is \( \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}} = \frac{(1-i\sqrt{3})(1-i\sqrt{3})}{4} = \frac{-2 + 2i\sqrt{3}}{4} = -\frac{1}{2} + \frac{i\sqrt{3}}{2}\). This is \(\frac{1}{2}\left(-1 + i\sqrt{3}\right)\) which lies on the unit circle and its argument is \(\frac{2\pi}{3}\).
10

Calculate the modulus of \((1-i)^n\), n in Z

The modulus of \(1 - i\) is \(\sqrt{(1)^2 + (-1)^2} = \sqrt{2}\). Therefore, \((1-i)^n\) will have a modulus of \(2^{n/2}\). The argument of \(1-i\) is \(-\frac{\pi}{4}\), so the argument of \((1-i)^n\) is \(n\cdot(-\frac{\pi}{4}) = -\frac{n\pi}{4}\).

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Key Concepts

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

Modulus of a complex number
The modulus of a complex number, often referred to as the absolute value, is pivotal in complex number analysis. It provides a measure of the distance from the complex number to the origin in the complex plane. This distance is always non-negative. For a complex number in the form of \( a + bi \), the modulus is calculated as \( \sqrt{a^2 + b^2} \).
For example, the modulus of \(-3 + i\) is calculated as \( \sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \).
  • Real numbers have their modulus equal to their absolute value.
  • The modulus is helpful in defining the length of a vector when the complex number is represented as a vector.
Understanding modulus is crucial for grasping the size and scale of complex numbers.
Argument of a complex number
The argument of a complex number provides an angle, usually expressed in radians, which the vector from the origin to the complex number makes with the positive real axis. It can also be seen as a way to determine direction in the complex plane. For a complex number \( a + bi \), the argument \( \theta \) is typically found using the arctan function: \( \theta = \tan^{-1}(\frac{b}{a}) \).
Understanding which quadrant the number lies in is crucial, as it influences the calculation:
  • If \( a > 0 \), use \( \theta = \tan^{-1}(\frac{b}{a}) \).
  • If \( a < 0 \) and \( b \geq 0 \), use \( \theta = \tan^{-1}(\frac{b}{a}) + \pi \).
  • If \( a < 0 \) and \( b < 0 \), use \( \theta = \tan^{-1}(\frac{b}{a}) - \pi \).
  • If \( a = 0 \), the angle depends on \( b \).
For example, the argument of \(-3 + i\) is \( \pi - \tan^{-1}(\frac{1}{3}) \), reflecting its position in the second quadrant.
De Moivre's Theorem
De Moivre's Theorem provides a formula for finding powers and roots of complex numbers. The theorem states that for any complex number \( z = r(\cos \theta + i \sin \theta) \) and an integer \( n \), \( z^n = r^n(\cos(n\theta) + i\sin(n\theta)) \). This is particularly useful for computations like \((1+i)^{17}\) or \((1-i)^{n}\) in exercises.
De Moivre's Theorem connects polar and exponential forms of complex numbers, offering:
  • Simplicity in raising powers of complex numbers.
  • Support in calculating roots through its inverse operation.
By using the theorem, solving complex expressions becomes manageable. Utilizing it, we determined \((1 + i)^{17} - (1 - i)^{17}\) and verified values for different powers.
Complex Conjugate
The complex conjugate of a complex number \( a + bi \) is \( a - bi \). This operation involves changing the sign of the imaginary part. It plays a significant role in simplifying division of complex numbers, especially with expressions like \( \frac{3+4i}{1-2i} \).
Using the conjugate, we can turn division into multiplication, easing calculations:
  • The product of a number and its conjugate gives a real number: \( (a+bi)(a-bi) = a^2 + b^2 \).
  • Conjugates are used to rationalize denominators in complex fractions.
By using the conjugate in our exercise, we simplified calculations and found appropriate modulus and arguments for the expressions.
Imaginary Unit
The imaginary unit, denoted as \( i \), is fundamental in complex numbers with \( i^2 = -1 \). It acts as a basic building block for complex numbers, allowing for the representation of numbers that are not real. Operations with \( i \) follow consistent rules, and powers of \( i \) exhibit cyclic behavior:
  • \( i^1 = i \)
  • \( i^2 = -1 \)
  • \( i^3 = -i \)
  • \( i^4 = 1 \) and repeats this cycle.
This cycle helps simplify expressions like \( i^{4711} \), an operation fundamental in understanding higher powers of \( i \). Using this cycle of powers, we reduce higher powers to a manageable form, revealing the essence of imaginary numbers in complex arithmetic processes.

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Most popular questions from this chapter

The function \(f: \mathbb{C} \rightarrow \mathbb{C}\) $$ f(z)= \begin{cases}\exp \left(-1 / z^{4}\right) & \text { for } z \neq 0 \\ 0 & \text { for } z=0\end{cases} $$ satisfies the CAUCHY-RIEMANN equations for all \(z \in \mathbb{C}\) and is complex differentiable for all \(z \in \mathbb{C}^{\bullet}\), but not at the origin.

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 \(\mathbb{C}\).

For \(z, w \in \mathbb{C}^{\bullet}\) show $$ \log (z w)=\log (z)+\log (w)+2 \pi \mathrm{i} k(z, w) \quad \text { where } $$ $$ k(z, w)=\left\\{\begin{aligned} 0, & \text { if }-\pi<\operatorname{Arg} z+\operatorname{Arg} w \leq \pi, \\ +1, & \text { if }-2 \pi<\operatorname{Arg} z+\operatorname{Arg} w \leq-\pi, \\\ -1, & \text { if } \quad \pi<\operatorname{Arg} z+\operatorname{Arg} w \leq 2 \pi. \end{aligned}\right. $$

Determine all the harmonic functions $$ u: \mathbb{C}^{\bullet}=\mathbb{R}^{2} \backslash\\{(0,0)\\} \longrightarrow \mathbb{R} $$ that depend only on \(r:=\sqrt{x^{2}+y^{2}} .\)

Let \(\left(a_{n}\right)_{n \geq 0}\) and \(\left(b_{n}\right)_{n \geq 0}\) be two sequences of complex numbers such that \(a_{n}=\) \(b_{n}-b_{n+1}, n \geq 0\) Show: The series \(\sum_{n=0}^{\infty} a_{n}\) is convergent if and only if the sequence \(\left(b_{n}\right)\) is convergent, and then $$ \sum_{n=0}^{\infty} a_{n}=b_{0}-\lim _{n \rightarrow \infty} b_{n+1} $$ Example: \(\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}=1 .\)
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