Problem 68
Question
Recall that the symbol \(\overline{z}\) represents the complex con- jugate of \(z .\) If \(z=a+b i\) and \(w=c+d i,\) prove each statement. $$ \overline{z w}=\overline{z} \cdot \overline{w} $$
Step-by-Step Solution
Verified Answer
Thus, \(\overline{z w} = \overline{z} \cdot \overline{w}\) holds true.
1Step 1: Write out the expressions for zw and conjugates
Given two complex numbers \(z = a + bi\) and \(w = c + di\), the product \(zw\) is computed as follows: \(zw = (a + bi)(c + di) = ac + adi + bci + bdi^2\). Since \(i^2 = -1\), this simplifies to \(zw = (ac - bd) + (ad + bc)i\). Next, we need the conjugate of \(zw\), which is \(\overline{zw} = (ac - bd) - (ad + bc)i\).
2Step 2: Find the conjugates of z and w individually
Start with \(z = a + bi\), its conjugate is \(\overline{z} = a - bi\). For \(w = c + di\), its conjugate is \(\overline{w} = c - di\).
3Step 3: Compute the product of the conjugates
Now calculate the product \(\overline{z} \cdot \overline{w} = (a - bi)(c - di)\). Use the distributive property to expand: \((a - bi)(c - di) = ac - adi - bci + bdi^2\). Simplify using \(i^2 = -1\) to get \(ac - bd - (ad + bc)i\).
4Step 4: Compare the expressions
Both expressions \(\overline{zw} = (ac - bd) - (ad + bc)i\) and \(\overline{z} \cdot \overline{w} = (ac - bd) - (ad + bc)i\) are identical. Thus, \(\overline{z w} = \overline{z} \cdot \overline{w}\).
Key Concepts
Complex NumbersConjugate MultiplicationImaginary Unit Properties
Complex Numbers
Complex numbers are an extension of the real numbers and are typically denoted by the letter \(z\). They have the form \(z = a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit. The part \(a\) is known as the real part, and \(bi\) as the imaginary part.
- The real part aids in positioning the number along the horizontal axis of a complex plane.
- The imaginary part determines the number's position along the vertical axis.
Conjugate Multiplication
Conjugate multiplication involves multiplying the conjugates of two complex numbers. For a given complex number \(z = a + bi\), its conjugate is \(\overline{z} = a - bi\). When dealing with two complex numbers, \(z = a + bi\) and \(w = c + di\), their conjugate multiplication is expressed as \(\overline{z} \cdot \overline{w}\).
- First, calculate \(z \cdot w\) and take the conjugate of the result: \(\overline{zw}\).
- Then compute \(\overline{z} \cdot \overline{w}\) directly.
Imaginary Unit Properties
The imaginary unit, denoted by \(i\), is defined by the property \(i^2 = -1\). This property is crucial when working with complex numbers, especially in calculations involving conjugates and multiplication.
- In the multiplication of complex numbers, like \((a + bi)(c + di)\), the term \(bdi^2\) simplifies to \(-bd\).
- This transformation stems directly from the fundamental property of \(i\), facilitating the simplification of expressions involving \(i\).
Other exercises in this chapter
Problem 67
\(47-70\) The given equation involves a power of the variable. Find all real solutions of the equation. $$ \sqrt[3]{x}=5 $$
View solution Problem 68
\(67-70=\) Determine the values of the variable for which the expression is defined as a real number. $$ \sqrt{3 x^{2}-5 x+2} $$
View solution Problem 68
65–68 ? Solve the equation for the variable x. The constants a and b represent positive real numbers. $$ \sqrt{x}+a \sqrt[3]{x}+b \sqrt[6]{x}+a b=0 $$
View solution Problem 68
Use the discriminant to determine the number of real solutions of the equation. Do not solve the equation. \(x^{2}-r x+s=0 \quad(s>0, r>2 \sqrt{s})\)
View solution