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

Short Answer

Expert verified
\(|z|=\sqrt{z^{*} z}\) since \(z^{*}z=a^2+b^2\) and \(|z|=\sqrt{a^2+b^2}\).

Step by step solution

01

Understanding Modulus of a Complex Number

Given a complex number \(z = a + bi\), its modulus \(|z|\) is defined as \(|z| = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.
02

Conjugate of a Complex Number

The conjugate of \(z = a + bi\) is \(z^{*} = a - bi\). This flips the sign of the imaginary part of the complex number while keeping the real part the same.
03

Multiplication of z and its Conjugate

To find \(z^{*}z\), multiply \(z\) by its conjugate: \((a - bi)(a + bi) = a^2 + abi - abi - (bi)^2\). The middle terms cancel out, and since \(i^2 = -1\), you get \(a^2 + b^2\).
04

Relating Modulus to the Expression

The expression \(z^{*}z = a^2 + b^2\) matches the inside of the modulus definition \(|z| = \sqrt{a^2 + b^2}\). Therefore, taking the square root of both sides of \(z^{*}z\), we show that \(|z| = \sqrt{z^{*}z}\).

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Key Concepts

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

Conjugate of a Complex Number
Complex numbers are numbers that have both a real part and an imaginary part. A complex number is typically written as \(z = a + bi\), where \(a\) is the real part, \(b\) is the imaginary part, and \(i\) is the imaginary unit, which satisfies \(i^2 = -1\). The conjugate of a complex number, noted as \(z^{*}\), is formed by changing the sign of the imaginary component. For the complex number \(z = a + bi\), its conjugate \(z^{*}\) would be \(a - bi\). This simply involves negating the imaginary part, leaving the real part unchanged.

Understanding conjugates is crucial in complex number arithmetic, especially when simplifying expressions or solving equations. Conjugates have a special property: when a complex number is multiplied by its conjugate, it results in a real number. This is because the imaginary parts cancel each other out, showing their importance in finding the modulus and in simplifying expressions.
Multiplication of Complex Numbers
Multiplying complex numbers takes a bit more work than multiplying real numbers. If you have two complex numbers, \(z_1 = a + bi\) and \(z_2 = c + di\), you multiply them by distributing each term just like polynomials:

  • First, multiply the real parts: \(ac\).
  • Next, multiply the outer parts: \(adi\).
  • Then, multiply the inner parts: \(bci\).
  • Finally, multiply the imaginary parts: \(bdi^2\).
The resulting expression becomes \(ac + adi + bci + bdi^2\). Remember, \(i^2 = -1\), so you replace \(bdi^2\) with \(-bd\). Combine the real terms and the imaginary terms to simplify. Thus, the product will be \((ac - bd) + (ad + bc)i\).

When multiplying a complex number by its conjugate, using this method effectively reveals why it results in a real number, as in \((a + bi)(a - bi) = a^2 + b^2\), which is crucial for understanding the connection to modulus.
Square Root and Modulus
The modulus of a complex number is a measure of its "size" or distance from the origin in the complex plane. For any complex number \(z = a + bi\), the modulus is computed as \(|z| = \sqrt{a^2 + b^2}\). This definition of modulus makes use of the Pythagorean Theorem, reflecting the distance from the point \((a, b)\) to the origin \((0, 0)\) in a Cartesian plane.

When you multiply the complex number by its conjugate as shown earlier, \(z^{*}z = a^2 + b^2\), you arrive at a real number which is precisely the square of the modulus. Therefore, \(|z| = \sqrt{z^{*}z}\). This relationship allows you to connect the algebraic and geometric interpretations of complex numbers, thereby proving the initial proposition. Hence, understanding modulus not only gives a number but also an insight into the geometric nature of complex numbers by associating them to magnitudes.

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Most popular questions from this chapter

For each of the following two pairs of numbers compute \(z+w, z-w, z w,\) and \(z / w\) (a) \(z=6+8 i\) and \(w=3-4 i \quad\) (b) \(z=8 e^{i \pi / 3}\) and \(w=4 e^{i \pi / 6}\) Notice that for adding and subtracting complex numbers, the form \(x+i y\) is more convenient, but for multiplying and especially dividing, the form \(r e^{i \theta}\) is more convenient. In part (a), a clever trick for finding \(z / w\) without converting to the form \(r e^{i \theta}\) is to multiply top and bottom by \(w^{*}\); try this one both ways.

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.)

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.

I kick a puck of mass \(m\) up an incline (angle of slope \(=\theta\) ) with initial speed \(v_{\mathrm{o}}\). There is no friction between the puck and the incline, but there is air resistance with magnitude \(f(v)=c v^{2} .\) Write down and solve Newton's second law for the puck's velocity as a function of \(t\) on the upward journey. How long does the upward journey last?

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 \(x=\sinh z .]\)

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