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Consider the complex number \(z=e^{i \theta}=\cos \theta+i \sin \theta .\) (a) By evaluating \(z^{2}\) two different ways, prove the trig identities \(\cos 2 \theta=\cos ^{2} \thetaQEDQED-\sin ^{2} \theta\) and \(\sin 2 \theta=2 \sin \theta \cos \theta .\) (b) Use the same technique to find corresponding identities for \(\cos 3 \theta\) and \(\sin 3 \theta\).

Short Answer

Expert verified
Use Euler's formula and equate real/imaginary components.

Step by step solution

01

Evaluate \( z^2 \) directly from exponential form

Since \( z = e^{i\theta} \), we raise it to the power of 2: \( z^2 = (e^{i\theta})^2 = e^{2i\theta} \). By Euler's formula, we know \( e^{2i\theta} = \cos(2\theta) + i\sin(2\theta) \).
02

Evaluate \( z^2 \) using trigonometric form

Given that \( z = \cos \theta + i \sin \theta \), we can calculate \( z^2 = (\cos \theta + i \sin \theta)^2 \). Using the identity \((a+b)^2 = a^2 + 2ab + b^2\), we expand it: \[ z^2 = \cos^2 \theta + 2i \cos \theta \sin \theta - \sin^2 \theta \].
03

Equate real parts to find \( \cos 2\theta \)

The real part of \( z^2 = \cos^2 \theta - \sin^2 \theta \) from the second method should equal the real part from Step 1, \( \cos 2\theta \). Thus, \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \).
04

Equate imaginary parts to find \( \sin 2\theta \)

The imaginary part of \( z^2 \) is \( 2i \cos \theta \sin \theta \) from the trigonometric form, which equals the imaginary part from Step 1, \( i \sin 2\theta \). Therefore, \( \sin 2\theta = 2 \sin \theta \cos \theta \).
05

Find \( z^3 \) using exponential form

For part (b), consider \( z^3 = (e^{i\theta})^3 = e^{3i\theta} \). By Euler’s formula, \( e^{3i\theta} = \cos(3\theta) + i\sin(3\theta) \).
06

Find \( z^3 \) using trigonometric expansion

We can expand \( z^3 = (\cos \theta + i \sin \theta)^3 \) using the binomial theorem. Calculate the terms: \[ z^3 = \cos^3 \theta + 3i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta \].
07

Simplify and equate real parts for \( \cos 3\theta \)

Group the real parts: \( \cos^3 \theta - 3\cos \theta \sin^2 \theta \). Equating it to the real part from Step 5, \( \cos 3\theta \): \( \cos 3\theta = \cos^3 \theta - 3\cos \theta \sin^2 \theta \).
08

Simplify and equate imaginary parts for \( \sin 3\theta \)

Group the imaginary parts: \( 3\cos^2 \theta \sin \theta - \sin^3 \theta \) and equate it to the imaginary part from Step 5, \( \sin 3\theta \): \( \sin 3\theta = 3\cos^2 \theta \sin \theta - \sin^3 \theta \).

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Key Concepts

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

Complex Numbers
Complex numbers are numbers of the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit, defined by \( i^2 = -1 \). The real part is \( a \) and the imaginary part is \( b \). Complex numbers extend the idea of one-dimensional real numbers to two-dimensional numbers by representing them on a plane.

The complex plane, also known as the Argand plane, is a two-dimensional plane where the x-axis represents the real part and the y-axis represents the imaginary part of complex numbers. Each point on this plane corresponds to a complex number. The formula \( z = e^{i\theta} = \cos \theta + i \sin \theta \), used in the exercise above, is a well-known representation of complex numbers that connects them to trigonometric functions.

A crucial aspect of complex numbers is their modulus and argument. The modulus is the distance from the origin to the point in the complex plane and is calculated via \( |z| = \sqrt{a^2 + b^2} \). The argument is the angle formed with the positive real axis, often denoted as \( \theta \), calculated using trigonometric functions. Complex numbers provide a powerful toolset for solving equations that do not have real solutions and play a critical role in advanced mathematics and engineering.
Euler's Formula
Euler's formula is one of the most remarkable equations in mathematics. It states that for any real number \( \theta \), \( e^{i\theta} = \cos \theta + i \sin \theta \). This formula elegantly bridges the fields of trigonometry and complex analysis.

The formula provides a way to express complex numbers in polar form, making calculations involving complex numbers more manageable. Specifically, through Euler's formula, the exponential function \( e^{i\theta} \) links exponential growth to trigonometric oscillation, describing how complex numbers rotate around the origin in the complex plane.

Using Euler's formula, we can derive several trigonometric identities, as demonstrated in the exercise with \( z^{2} = e^{2i\theta} \), leading to the identities \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \) and \( \sin 2\theta = 2\sin \theta \cos \theta \). This not only shows how trigonometric identities can be derived from complex exponentials but highlights the utility of Euler’s formula in simplifying complex calculations.

This blending of trigonometry and complex numbers simplifies many mathematical problems, offering insights into the behavior of functions and phenomena that are inherently periodic or oscillatory.
Binomial Theorem
The binomial theorem provides a formula for expanding powers of sums. Specifically, it states that \((a + b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k\), where \( \binom{n}{k} \) is a binomial coefficient. In our exercise, this theorem allows us to expand \((\cos \theta + i \sin \theta)^n\) to find expressions for higher powers of complex numbers.

When applying the binomial theorem to trigonometric forms of complex numbers, each term in the expansion represents a combination of real and imaginary parts raised to different powers. The theorem incorporates the idea of permutations of terms and coefficients, allowing a structured way to expand polynomials.

In the context of this exercise, using the binomial theorem helps derive formulas for \( \cos 3\theta \) and \( \sin 3\theta \). By expanding \( (\cos \theta + i \sin \theta)^3 \) and separating real and imaginary components, we obtain new trigonometric identities.

The binomial theorem is a powerful computational tool, not only facilitating the multiplication of expressions but also giving rise to identities across various branches of mathematics.

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Most popular questions from this chapter

2.16 \(\star\) A golfer hits his ball with speed \(v_{\mathrm{o}}\) at an angle \(\theta\) above the horizontal ground. Assuming that the angle \(\theta\) is fixed and that air resistance can be neglected, what is the minimum speed \(v_{\mathrm{o}}(\min )\) for which the ball will clear a wall of height \(h\), a distance \(d\) away? Your solution should get into trouble if the angle \(\tan \theta < h / d .\) Explain. What is \(v_{\mathrm{o}}(\min )\) if \(\theta=25^{\circ}, d=50 \mathrm{m},\) and \(h=2 \mathrm{m} ?\)

The hyperbolic functions cosh \(z\) and \(\sinh z\) are defined as follows: $$\cosh z=\frac{e^{z}+e^{-z}}{2} \quad \text { and } \quad \sinh z=\frac{e^{z}-e^{-z}}{2}$$ for any \(z,\) real or complex. (a) Sketch the behavior of both functions over a suitable range of real values of \(z\). (b) Show that \(\cosh z=\cos (i z)\). What is the corresponding relation for \(\sinh z ?\) (c) What are the derivatives of cosh \(z\) and \(\sinh z ?\) What about their integrals? ( \(\mathbf{d}\) ) Show that \(\cosh ^{2} z-\sinh ^{2} z=1\) (e) Show that \(\int d x / \sqrt{1+x^{2}}=\operatorname{arcsinh} x\). [Hint: One way to do this is to make the substitution \(x=\sinh z .]\)

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.)

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 \(2.9 .\)

Prove that \(|z|=\sqrt{z^{*} z}\) for any complex number \(z\).

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