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For any complex number \(z=x+i y,\) the real and imaginary parts are defined as the real numbers \(\operatorname{Re}(z)=x\) and \(\operatorname{Im}(z)=y .\) The modulus or absolute value is \(|z|=\sqrt{x^{2}+y^{2}}\) and the phase or angle is the value of \(\theta\) when \(z\) is expressed as \(z=r e^{i \theta} .\) The complex conjugate is \(z^{*}=x-i y\) (This last is the notation used by most physicists; most mathematicians use \(\bar{z}\).) For each of the following complex numbers, find the real and imaginary parts, the modulus and phase, and the complex conjugate, and sketch \(z\) and \(z^{*}\) in the complex plane: (a) \(z=1+i \quad\) (b) \(z=1-i \sqrt{3}\) (d) \(z=5 e^{i \omega t}\) (c) \(z=\sqrt{2} e^{-i \pi / 4}\) In part (d), \(\omega\) is a constant and \(t\) is the time.

Short Answer

Expert verified
Real and imaginary parts computed, moduli derived, phases found, and complex conjugates determined for each case.

Step by step solution

01

Real and Imaginary Parts for (a)

Given the complex number \( z = 1 + i \), the real part is \( \operatorname{Re}(z) = 1 \) and the imaginary part is \( \operatorname{Im}(z) = 1 \).
02

Modulus and Phase for (a)

The modulus is calculated as follows: \[|z| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}\] For the phase \( \theta \), since the real part is positive and imaginary part is positive, \( \theta = \frac{\pi}{4} \).
03

Complex Conjugate for (a)

The complex conjugate of \( z = 1 + i \) is \( z^* = 1 - i \).
04

Real and Imaginary Parts for (b)

Given \( z = 1 - i \sqrt{3} \), the real part is \( \operatorname{Re}(z) = 1 \) and the imaginary part is \( \operatorname{Im}(z) = -\sqrt{3} \).
05

Modulus and Phase for (b)

The modulus is: \[|z| = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{4} = 2\] The phase \( \theta \) can be calculated as \( \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3} \).
06

Complex Conjugate for (b)

The complex conjugate of \( z = 1 - i \sqrt{3} \) is \( z^* = 1 + i \sqrt{3} \).
07

Real and Imaginary Parts for (c)

Given \( z = \sqrt{2} e^{-i \pi / 4} \), the exponential form reveals the phase directly. The real part can be found using Euler's formula: \( z = \sqrt{2}(\cos(-\pi/4) + i \sin(-\pi/4)) = 1 - i \).Thus, \( \operatorname{Re}(z) = 1 \) and \( \operatorname{Im}(z) = -1 \).
08

Modulus and Phase for (c)

The modulus \(|z|\) is given by the coefficient of the exponential, \(\sqrt{2}\), and the phase \(\theta\) is \(-\frac{\pi}{4}\).
09

Complex Conjugate for (c)

The complex conjugate of \( z = \sqrt{2} e^{-i \pi / 4} \) is \( z^* = \sqrt{2} e^{i \pi / 4}\).
10

Real and Imaginary Parts for (d)

Given \( z = 5 e^{i \omega t} \), it is in polar form. Using Euler's formula, we write it as \( z = 5 (\cos(\omega t) + i \sin(\omega t)) \). Thus, the real part is \( 5 \cos(\omega t) \) and the imaginary part is \( 5 \sin(\omega t) \).
11

Modulus and Phase for (d)

The modulus is \(|z| = 5\), and the phase is \(\theta = \omega t \).
12

Complex Conjugate for (d)

The complex conjugate is \( z^* = 5 e^{-i \omega t} \).

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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 provides a measure of its size or magnitude. For a complex number given by \( z = x + iy \), the modulus is calculated using the formula: \[ |z| = \sqrt{x^2 + y^2} \] This formula is derived from applying the Pythagorean theorem to the complex plane. Imagine plotting \( z \) on the plane with the real part \( x \) on the horizontal axis and the imaginary part \( y \) on the vertical axis.
  • The modulus \(|z|\) represents the distance from the origin \((0,0)\) to the point \((x, y)\).
  • This distance directly corresponds to the hypotenuse of a right triangle where \(x\) and \(y\) are the perpendicular sides.
For example, consider \( z = 1 + i \). Here, \( |z| = \sqrt{1^2 + 1^2} = \sqrt{2} \). The modulus helps in understanding how far the number is from the origin, irrespective of its direction.
Complex Conjugate
The complex conjugate of a complex number is obtained simply by changing the sign of the imaginary part. If you start with a complex number \( z = x + iy \), its complex conjugate is represented as \( z^* = x - iy \).
  • In many contexts, the conjugate is denoted as \( \bar{z} \), especially in mathematics, but physicists often use \( z^* \).
  • The real part of the number remains unchanged, but the imaginary part takes the opposite sign.
The complex conjugate is crucial in various mathematical calculations because when a complex number is multiplied by its conjugate, it results in a real number: \[ z \cdot z^* = (x + iy)(x - iy) = x^2 + y^2 \] This product is particularly useful for simplifying division of complex numbers and for deriving their modulus.
Polar Form of Complex Numbers
Polar form is a brilliant way to represent complex numbers using a radius and an angle rather than using Cartesian coordinates. Any complex number \( z = x + iy \) can be expressed in polar form as \( z = r e^{i\theta} \). Here's the breakdown:
  • \( r \), the modulus, represents the distance from the origin to the point \((x, y)\).
  • \( \theta \), the phase or argument, is the angle made with the positive real axis.
The transformation from Cartesian to polar involves: \[ r = \sqrt{x^2 + y^2} \] \[ \theta = \tan^{-1}(y/x) \] Using Euler’s formula, \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), you can easily convert to and from this form. This form is particularly useful for multiplication and division of complex numbers, as their moduli multiply and their arguments add. For instance, \( z = \sqrt{2} e^{-i \pi/4} \) is already in polar form. Here, \( \sqrt{2} \) is the modulus and \(-\pi/4\) is the angle.
Complex Plane Sketching
Complex plane sketching provides a visual representation of complex numbers.
  • On this plane, the x-axis represents the real part, and the y-axis represents the imaginary part.
  • Each complex number corresponds to a point \((x, y)\) on this plane.
This could also be seen as a 2D plane where complex numbers live. When sketching, you plot a number like \( z = 1 + i \) by moving one unit along the real axis and one unit up the imaginary axis. The complex conjugate \( z^* = 1 - i \) would be one unit down from the real point on the plane.
  • Such sketching helps quickly visualize operations, like addition and subtraction, by considering the vector movement in the plane.
  • By drawing, you can also understand how the modulus denotes distance and the argument shows direction.
Sketching provides a useful tool both for practical calculations and for theoretical understanding.

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Most popular questions from this chapter

Use the series definition (2.72) of \(e^{z}\) to prove that \(^{12} d e^{z} / d z=e^{z}\).

There are certain simple one-dimensional problems where the equation of motion (Newton's second law) can always be solved, or at least reduced to the problem of doing an integral. One of these (which we have met a couple of times in this chapter) is the motion of a one-dimensional particle subject to a force that depends only on the velocity \(v\), that is, \(F=F(v)\). Write down Newton's second law and separate the variables by rewriting it as \(m d v / F(v)=d t .\) Now integrate both sides of this equation and show that $$t=m \int_{v_{0}}^{v} \frac{d v^{\prime}}{F\left(v^{\prime}\right)}$$ Provided you can do the integral, this gives \(t\) as a function of \(v\). You can then solve to give \(v\) as a function of \(t .\) Use this method to solve the special case that \(F(v)=F_{\mathrm{o}},\) a constant, and comment on your result. This method of separation of variables is used again in Problems 2.8 and \(2.9 .\)

A charged particle of mass \(m\) and positive charge \(q\) moves in uniform electric and magnetic fields, \(\mathbf{E}\) and \(\mathbf{B}\), both pointing in the \(z\) direction. The net force on the particle is \(\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})\). Write down the equation of motion for the particle and resolve it into its three components. Solve the equations and describe the particle's motion.

Problem 2.7 is about a class of one-dimensional problems that can always be reduced to doing an integral. Here is another. Show that if the net force on a one-dimensional particle depends only on position, \(F=F(x),\) then Newton's second law can be solved to find \(v\) as a function of \(x\) given by $$v^{2}=v_{\mathrm{o}}^{2}+\frac{2}{m} \int_{x_{0}}^{x} F\left(x^{\prime}\right) d x^{\prime}$$ [Hint: Use the chain rule to prove the following handy relation, which we could call the " \(v\) dv/d \(x\) rule": If you regard \(v\) as a function of \(x,\) then $$\dot{v}=v \frac{d v}{d x}=\frac{1}{2} \frac{d v^{2}}{d x}$$ Use this to rewrite Newton's second law in the separated form \(m d\left(v^{2}\right)=2 F(x) d x\) and then integrate from \(x_{\mathrm{o}}\) to \(x .\) ] Comment on your result for the case that \(F(x)\) is actually a constant. (You may recognise your solution as a statement about kinetic energy and work, both of which we shall be discussing in Chapter 4.)

[Computer] Use suitable graph-plotting software to plot graphs of the trajectory (2.36) of a projectile thrown at 45 ^above the horizontal and subject to linear air resistance for four different values of the drag coefficient, ranging from a significant amount of drag down to no drag at all. Put all four trajectories on the same plot. [Hint: In the absence of any given numbers, you may as well choose convenient values. For example, why not take \(v_{x \mathrm{o}}=v_{y \mathrm{o}}=1\) and \(g=1 .\) (This amounts to choosing your units of length and time so that these parameters have the value \(1 .\).) With these choices, the strength of the drag is given by the one parameter \(v_{\text {ter }}=\tau,\) and you might choose to plot the trajectories for \(v_{\text {ter }}=0.3,1,3,\) and \(\infty\) (that is, no drag at all), and for times from \(t=0\) to \(3 .\) For the case that \(v_{\text {ter }}=\infty\) you'll probably want to write out the trajectory separately.]

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