Problem 74
Question
Recall that the symbol \(\bar{z}\) represents the complex conjugate of \(z .\) If \(z=a+b i\) and \(w=c+d i,\) prove each statement. $$\overline{z w}=\bar{z} \cdot \bar{w}$$
Step-by-Step Solution
Verified Answer
The complex conjugate of a product equals the product of conjugates: \( \overline{zw} = \bar{z} \cdot \bar{w} \).
1Step 1: Write down the expressions for complex conjugates
The complex conjugate of any complex number is obtained by reversing the sign of its imaginary part. For the given numbers, the complex conjugates are \( \bar{z} = a - bi \) and \( \bar{w} = c - di \).
2Step 2: Calculate the product of z and w
The product of \( z = a + bi \) and \( w = c + di \) is calculated using the distributive property: \[ zw = (a + bi)(c + di) = ac + adi + bci + bdi^2 \]By substituting \( i^2 = -1 \), it simplifies to: \[ zw = ac + adi + bci - bd = (ac - bd) + (ad + bc)i \]
3Step 3: Calculate the complex conjugate of zw
Now, we need to determine \( \overline{zw} \), which involves reversing the sign of the imaginary part of \( zw \):\[ \overline{zw} = (ac - bd) - (ad + bc)i \]
4Step 4: Calculate the product of conjugates \(\bar{z}\) and \(\bar{w}\)
Using the expressions from Step 1, compute the product \( \bar{z} \cdot \bar{w} \):\[ \bar{z} \cdot \bar{w} = (a - bi)(c - di) = ac - adi - bci + bdi^2 \]Since \( i^2 = -1 \), this simplifies to:\[ \bar{z} \cdot \bar{w} = ac + bd - (ad + bc)i \]
5Step 5: Verify both expressions are equal
Both expressions \( \overline{zw} = (ac - bd) - (ad + bc)i \) and \( \bar{z} \cdot \bar{w} = (ac - bd) - (ad + bc)i \) are identical after simplification. This proves that \( \overline{zw} = \bar{z} \cdot \bar{w} \).
Key Concepts
Complex ConjugateDistributive PropertyImaginary Unit
Complex Conjugate
A complex conjugate of a complex number is found by changing the sign of its imaginary part. For example, if a complex number is represented as \( z = a + bi \), where \( a \) and \( b \) are real numbers, its complex conjugate is \( \bar{z} = a - bi \).
The importance of complex conjugates lies in their ability to simplify the division and multiplication of complex numbers and provide a real number when multiplied by their original complex number.
The importance of complex conjugates lies in their ability to simplify the division and multiplication of complex numbers and provide a real number when multiplied by their original complex number.
- For any complex number \( z \), the product \( z \cdot \bar{z} \) is always a real number: \( z \cdot \bar{z} = (a + bi)(a - bi) = a^2 + b^2 \).
- Complex conjugates are often used to find the modulus or absolute value of a complex number. The modulus of \( z \) is given by \( |z| = \sqrt{z \cdot \bar{z}} = \sqrt{a^2 + b^2} \).
- This property is especially useful in complex number equations and transformations.
Distributive Property
The distributive property allows us to multiply a sum by another number. This property is crucial when dealing with complex numbers, especially when multiplying two complex numbers together. For the expression \( (a + bi)(c + di) \), the distributive property is applied as follows:
- First, distribute \( a \) over \( (c + di) \): \( a(c + di) = ac + adi \).
- Then, distribute \( bi \) over \( (c + di) \): \( bi(c + di) = bci + bdi^2 \). Since \( i^2 = -1 \), this becomes \( bci - bd \).
Imaginary Unit
The imaginary unit, denoted as \( i \), is a fundamental component of complex numbers. It is defined by its property \( i^2 = -1 \). This definition is the cornerstone for calculations involving complex numbers, enabling their applications in various fields of mathematics and engineering.
- When multiplying powers of \( i \), the cycle: \( i^0 = 1 \), \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \), and so forth, repeats every four terms.
- This cyclic behavior simplifies computations, reducing higher powers of \( i \) to simpler expressions.
- The imaginary unit transforms real numbers into complex numbers, as seen in \( a + bi \), where \( a \) is the real part, and \( bi \) is the imaginary part.
Other exercises in this chapter
Problem 73
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