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Chapter 8: Problem 8

If \(z=3-2 i\), find (i) \(z^{*}\) and (ii) \(z z^{*}\). (iii) Express the real and imaginary parts of \(z\) in terms of \(z\) and \(z^{*}\).

Short Answer

Expert verified
(i) \( z^* = 3 + 2i \); (ii) \( z z^* = 13 \); (iii) Real: 3, Imaginary: -2.

Step by step solution

01

Find the Complex Conjugate of z

The complex conjugate of a complex number \( z = a + bi \) is given by \( z^* = a - bi \). For the given \( z = 3 - 2i \), the complex conjugate is \( z^* = 3 + 2i \).
02

Calculate z times z*

To find \( z z^* \), multiply \( z = 3 - 2i \) by its conjugate \( z^* = 3 + 2i \): \[ z z^* = (3 - 2i)(3 + 2i) \]Use the formula for the product of conjugates, which is \( (a+bi)(a-bi) = a^2-b^2i^2 \): \[ = 3^2 - (2i)^2 = 9 - 4i^2 \]Since \( i^2 = -1 \), the expression becomes:\[ = 9 + 4 = 13 \]
03

Express Real and Imaginary Parts

The real part of \( z \), \( \text{Re}(z) \), is the average of \( z \) and \( z^* \): \[ \text{Re}(z) = \frac{z + z^*}{2} = \frac{(3-2i) + (3+2i)}{2} = \frac{6}{2} = 3 \]The imaginary part of \( z \), \( \text{Im}(z) \), can be found by taking the difference and dividing by \( 2i \): \[ \text{Im}(z) = \frac{z - z^*}{2i} = \frac{(3-2i) - (3+2i)}{2i} = \frac{-4i}{2i} = -2 \].

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Key Concepts

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

Complex Conjugate
A complex number is typically represented in the form of \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The complex conjugate of a complex number is formed by changing the sign of the imaginary part while leaving the real part unchanged. Let's see how this process works:
  • Given a complex number \( z = a + bi \).
  • The complex conjugate is \( z^* = a - bi \).
For example, in the exercise, we have \( z = 3 - 2i \). Its complex conjugate is simply \( z^* = 3 + 2i \). Finding the conjugate is crucial as it helps simplify expressions and easily calculate magnitudes for complex numbers.
Real Part of Complex Number
The real part of a complex number is straightforward to find as it is the coefficient of the real number term. For example, if you have a complex number \( z = a + bi \), \( a \) represents the real part.
However, there is a neat trick when you know both the complex number and its conjugate:
  • The real part can be calculated as the average of the complex number and its complex conjugate.
Mathematically, this is expressed as:\[ \text{Re}(z) = \frac{z + z^*}{2} \]Using the numbers from the exercise, \( z = 3 - 2i \) and \( z^* = 3 + 2i \), we find:\[ \text{Re}(z) = \frac{(3 - 2i) + (3 + 2i)}{2} = \frac{6}{2} = 3 \]
Imaginary Part of Complex Number
The imaginary part of a complex number is perhaps a bit trickier to extract than the real part. It is generally the coefficient of \( i \) in the expression \( z = a + bi \), which is \( b \).
But similar to the real part, there's a formula involving the complex conjugate that can help:
  • The imaginary part can be calculated using the difference between the complex number and its conjugate, divided by \( 2i \).
This relation can be written as:\[ \text{Im}(z) = \frac{z - z^*}{2i} \]For the example \( z = 3 - 2i \) and \( z^* = 3 + 2i \):\[ \text{Im}(z) = \frac{(3 - 2i) - (3 + 2i)}{2i} = \frac{-4i}{2i} = -2 \]
Multiplication of Complex Numbers
Multiplying complex numbers follows the distributive and associative laws, similar to real numbers. When multiplying, you generally treat \( i \) as if it's a variable, but keep in mind the special property \( i^2 = -1 \).
When multiplying a complex number by its conjugate \( (a + bi)(a - bi) \), this results in what is known as the magnitude squared of the complex number:
  • The expression simplifies significantly because the "cross terms" cancel out, which are the product terms involving \( i \).
Here's the formula applied:\[ z z^* = (a + bi)(a - bi) = a^2 - (bi)^2 \]Given \( z = 3 - 2i \), its conjugate \( z^* = 3 + 2i \), we find:\[ z z^* = (3 - 2i)(3 + 2i) = 3^2 - (2i)^2 = 9 - (2i)^2 \]Remembering \( i^2 = -1 \), we further simplify to:\[ = 9 - 4(-1) = 9 + 4 = 13 \]

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Most popular questions from this chapter

\(2 e^{\pi i / 6}\)

\((2+3 i)-(2-3 i)\)

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Solve the equations: \(z^{2}-2 z+4=0\)

(i) Express the complex function \(f(x)=3 x^{2}+(1+2 i) x+2(i-1)\) in the form \(f(x)=g(x)+i h(x)\), where \(g(x)\) and \(h(x)\) are real. (ii) solve \(g(x)=0, h(x)=0\), then \(f(x)=0 .\) (iii) find \(|f(x)|^{2}\).
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