Problem 3
Question
Show that \(z \bar{z}\) is always a real number.
Step-by-Step Solution
Verified Answer
\( z \bar{z} = a^2 + b^2 \) is a real number.
1Step 1: Define a Complex Number
A complex number \( z \) can be expressed in the form \( z = a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit with the property \( i^2 = -1 \).
2Step 2: Identify the Conjugate
The conjugate of a complex number \( z = a + bi \) is \( \bar{z} = a - bi \). It changes the sign of the imaginary part of the complex number.
3Step 3: Multiply the Complex Number and Its Conjugate
Compute \( z \bar{z} \) using the previously defined \( z \) and \( \bar{z} \): \[ z \bar{z} = (a + bi)(a - bi). \]
4Step 4: Use Algebraic Expansion
Expand the expression \((a + bi)(a - bi)\) using the formula \((x + y)(x - y) = x^2 - y^2\):\[ z \bar{z} = a^2 - (bi)^2. \]
5Step 5: Simplify the Expression
Since \( (bi)^2 = b^2i^2 \), and \( i^2 = -1 \), we have \( (bi)^2 = -b^2 \). Substitute back into the expression: \[ z \bar{z} = a^2 + b^2. \]
6Step 6: Conclude with Real Number Property
The expression \( a^2 + b^2 \) is the sum of squares of real numbers \( a \) and \( b \), which is always a non-negative real number.
Key Concepts
Conjugate of a Complex NumberMultiplication of Complex NumbersReal NumbersImaginary Unit
Conjugate of a Complex Number
In the fascinating world of complex numbers, every complex number has a mirror version called its conjugate. For a complex number expressed as \( z = a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit, the conjugate \( \bar{z} \) is simply \( a - bi \). This operation involves flipping the sign of the imaginary part, leaving the real part unchanged.
- Complex number example: \( z = 3 + 4i \) with conjugate \( \bar{z} = 3 - 4i \)
- More broadly, for any complex number \( z = a + bi \), conjugate \( \bar{z} \) is \( a - bi \).
Multiplication of Complex Numbers
Multiplying complex numbers might seem tricky at first, but it's a lot like expanding expressions when we first learned about real numbers. When we multiply a complex number by its conjugate, we use a special algebraic formula:
- If \( z = a + bi \) and \( \bar{z} = a - bi \), then the product is \( z\bar{z} = (a + bi)(a - bi) \).
Real Numbers
Real numbers are the numbers we use every day - from counting apples to measuring distances. They include all the numbers along the number line: integers, fractions, and even pi or square roots.
This expression results in a real number because it's simply a sum of squares, which can never be negative. Whether we have \( a^2 \) or \( b^2 \), both are inherently non-negative, so their addition remains non-negative.
By observing the multiplication result of \( z\bar{z} \), we notice it confirms the magnitude or modulus of the complex number and is always a real number.
- Real numbers make up the set of all rational and irrational numbers.
This expression results in a real number because it's simply a sum of squares, which can never be negative. Whether we have \( a^2 \) or \( b^2 \), both are inherently non-negative, so their addition remains non-negative.
By observing the multiplication result of \( z\bar{z} \), we notice it confirms the magnitude or modulus of the complex number and is always a real number.
Imaginary Unit
The imaginary unit, denoted as \( i \), is a mathematical concept that enables us to extend the real number system. Its defining property is that when it is squared, it equals -1: \( i^2 = -1 \). This simple property allows mathematicians to expand the number system to include solutions to equations that don't have real-number solutions, such as the square root of a negative number.
When multiplying complex numbers, understanding \( i^2 = -1 \) is crucial for simplifying products like \( (bi)^2 \), since it turns into \( -b^2 \). This transformation is a cornerstone for proving that \( z\bar{z} \) results in a real number when complex numbers are involved.
- Key property: \( i^2 = -1 \)
- Example: \( \sqrt{-1} = i \)
When multiplying complex numbers, understanding \( i^2 = -1 \) is crucial for simplifying products like \( (bi)^2 \), since it turns into \( -b^2 \). This transformation is a cornerstone for proving that \( z\bar{z} \) results in a real number when complex numbers are involved.
Other exercises in this chapter
Problem 2
Sketch the curve \(z(t)=t^{2}+2 t+i(t+1)\) (a) for \(-1 \leq t \leq 0\). (b) for \(1 \leq t \leq 2\).
View solution Problem 3
Find all solutions to the following depressed cubics. (a) \(27 x^{3}-9 x-2=0 .\) Hint: Get an equivalent monic polynomial. (b) \(x^{3}-27 x+54=0\)
View solution Problem 3
Which of the following points lie inside the circle \(|z-i|=2\) ? Explain your answers. (a) \(\frac{1}{2}+i\). (b) \(\sqrt{2}+i(\sqrt{2}+1)\) (c) \(2+3 i\). (d)
View solution Problem 3
Represent the following complex numbers in polar form. (a) \(-4\). (b) \(6-6 i\). (c) \(-7 i\), (d) \(-2 \sqrt{3}-2 i\) (e) \(\frac{1}{(1-i)^{2}}\). (f) \(\frac
View solution