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Chapter 1: Problem 75

Evaluate the given expression for \(z=3-4 i\) and \(w=5+2 i\) $$z \cdot \bar{z}$$

Short Answer

Expert verified
The value of \(z \cdot \bar{z}\) is 25.

Step by step solution

01

Understand the Problem

We need to evaluate the expression \(z \cdot \bar{z}\), where \(z\) is a complex number and \(\bar{z}\) is its complex conjugate. The complex conjugate of \(z = a + bi\) is \(\bar{z} = a - bi\). For \(z = 3 - 4i\), the conjugate is \(\bar{z} = 3 + 4i\).
02

Multiply z and its Conjugate

Multiply \(z = 3 - 4i\) by its conjugate \(\bar{z} = 3 + 4i\): \[(3 - 4i)(3 + 4i)\]. Use the formula for difference of squares: \(a^2 - b^2 = (a-b)(a+b)\), where \(a = 3\) and \(b = 4i\).
03

Calculations Using Difference of Squares

Compute the expression \((3)^2 - (4i)^2 = 9 - 16i^2\). Since \(i^2 = -1\), substitute to get \(9 - 16(-1) = 9 + 16 = 25\).
04

Conclusion of the Calculation

The value of the expression \(z \cdot \bar{z}\) is \(25\). This evaluates the magnitude squared of \(z\).

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Key Concepts

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

Complex Conjugate
Complex numbers are written in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part. For any complex number \(z = a + bi\), its complex conjugate \(\bar{z}\) is \(a - bi\). This swapping of the sign on the imaginary part is crucial.
  • Given a complex number \(z = 3 - 4i\).
  • The conjugate would be \(\bar{z} = 3 + 4i\).
The complex conjugate helps in various calculations, such as finding the modulus or in simplifying division involving complex numbers. It essentially reflects the complex number with respect to the real axis on the complex plane.
Magnitude of a Complex Number
The magnitude (or modulus) of a complex number \(z = a + bi\) is calculated using the formula \(|z| = \sqrt{a^2 + b^2}\). This formula is derived from the Pythagorean theorem, as complex numbers can be represented as points in a plane with real and imaginary parts as coordinates.
  • For \(z = 3 - 4i\), the magnitude is \(|3 - 4i| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5\).
When you multiply a complex number by its conjugate, the result \(z \cdot \bar{z}\) is the square of the magnitude. Thus, it's equal to \((|z|)^2\) without any imaginary component. This simplifies many mathematical operations and helps in gaining insights into the nature of the complex number.
Difference of Squares
The difference of squares is a simple algebraic shortcut. It states: for any two numbers \(a\) and \(b\), \((a-b)(a+b) = a^2 - b^2\).
  • In case of multiplying a number by its conjugate, use this rule.
  • For instance, \((3 - 4i)(3 + 4i) = 3^2 - (4i)^2\).
It's essential to apply special properties of imaginary numbers here. Recall that \(i^2 = -1\). Thus, \((4i)^2 = 16i^2 = 16(-1) = -16\), converting this to \(3^2 - (-16)\), simplifies to \(9 + 16\).
This technique not only improves efficiency in calculations but also demonstrates an interesting relationship between algebra and complex numbers.

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Most popular questions from this chapter

Gas Mileage The gas mileage \(q\) (measured in milgal) for a particular vehicle, driven at \(v\) mi/h, is given by the formula \(g=10+0.9 v-0.01 v^{2},\) as long as \(v\) is between \(10 \mathrm{mi} / \mathrm{h}\) and \(75 \mathrm{mi} / \mathrm{h}\). For what range of speeds is the vehicle's mileage 30 milgal or better?

Use scientific notation, the Laws of Exponents, and a calculator to perform the indicated operations. State your answer rounded to the number of significant digits indicated by the given data. $$\left(7.2 \times 10^{-9}\right)\left(1.806 \times 10^{-12}\right)$$

It follows from Kepler's Third Law of planetary motion that the average distance from a planet to the sun (in meters) is $$d=\left(\frac{G M}{4 \pi^{2}}\right)^{1 / 3} T^{2 / 3}$$ where \(M=1.99 \times 10^{30} \mathrm{kg}\) is the mass of the sun, \(G=6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant, and \(T\) is the period of the planet's orbit (in seconds). Use the fact that the period of the earth's orbit is about 365.25 days to find the distance from the earth to the sun.

Prove the following Laws of Exponents. (a) Law \(6:\left(\frac{a}{b}\right)^{-n}=\frac{b^{n}}{a^{n}}\) (b) Law \(7: \frac{a^{-n}}{b^{-m}}=\frac{b^{m}}{a^{n}}\)

Radicals Simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers. (a) \(\sqrt[3]{y \sqrt{y}}\) (b) \(\sqrt{\frac{16 u^{3} v}{u v^{5}}}\)
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