Problem 4
Question
Verify directly that \(z w=0\) if and only if \(z=0\) or \(w=0\).
Step-by-Step Solution
Verified Answer
The statement is true: \( z w = 0 \) if and only if \( z = 0 \) or \( w = 0 \).
1Step 1: Understand the statement
The statement to verify is that for two complex numbers \( z \) and \( w \), the product \( z w = 0 \) if and only if either \( z = 0 \) or \( w = 0 \). This means we need to show two directions: if \( z w = 0 \), then \( z = 0 \) or \( w = 0 \), and conversely, if \( z = 0 \) or \( w = 0 \), then \( z w = 0 \).
2Step 2: Assume \( z w = 0 \) and show \( z = 0 \) or \( w = 0 \)
Assume \( z w = 0 \). Express \( z = a + bi \) and \( w = c + di \), where \( a, b, c, \) and \( d \) are real numbers. The product \( z w = (a + bi)(c + di) = ac - bd + (ad + bc)i = 0 \), which implies both the real part \( ac - bd = 0 \) and the imaginary part \( ad + bc = 0 \). We need to solve for \( z = 0 \) or \( w = 0 \). If either \( a = 0 \) and \( b = 0 \), then \( z = 0 \). Similarly, if \( c = 0 \) and \( d = 0 \), then \( w = 0 \). Thus, \( z = 0 \) or \( w = 0 \).
3Step 3: Assume \( z = 0 \) or \( w = 0 \) and show \( z w = 0 \)
Now assume either \( z = 0 \) or \( w = 0 \). If \( z = 0 \), then \( z = 0 + 0i \), and for any \( w \), the product \( z w = 0 \times w = 0 \). Similarly, if \( w = 0 \), then \( w = 0 + 0i \), and for any \( z \), the product \( z w = z \times 0 = 0 \). In both cases, \( z w = 0 \).
4Step 4: Conclusion
We have shown both directions: if \( z w = 0 \) then \( z = 0 \) or \( w = 0 \), and if \( z = 0 \) or \( w = 0 \), then \( z w = 0 \). Therefore, we have directly verified the statement.
Key Concepts
AlgebraComplex MultiplicationZero Product PropertyImaginary Numbers
Algebra
Algebra is a branch of mathematics that involves symbols and rules for manipulating these symbols. It allows us to represent numbers in equations and formulas.
In algebra, we often work with variables that stand for unknown values. In the context of complex numbers, these variables can take the form of expressions like \( z = a + bi \), where \( a \) and \( b \) are real numbers.
Complex numbers follow similar rules to algebraic equations, which makes it easy to work with them algebraically.
In algebra, we often work with variables that stand for unknown values. In the context of complex numbers, these variables can take the form of expressions like \( z = a + bi \), where \( a \) and \( b \) are real numbers.
Complex numbers follow similar rules to algebraic equations, which makes it easy to work with them algebraically.
Complex Multiplication
Complex multiplication involves the process of multiplying two complex numbers. Suppose you have two complex numbers \( z = a + bi \) and \( w = c + di \). When you multiply them, the result is derived by using the distributive property of multiplication over addition. This means:
\((a + bi)(c + di) = ac - bd + (ad + bc)i \).
This process follows the rules of distribution and allows us to express the product of complex numbers as another complex number, showing both real and imaginary parts.
- Calculate the product of the real parts: \( ac \)
- Calculate the product of the imaginary parts: \( bd \)
- Multiply the real part of one with the imaginary part of the other and vice versa, and then sum these products to get the imaginary part.
\((a + bi)(c + di) = ac - bd + (ad + bc)i \).
This process follows the rules of distribution and allows us to express the product of complex numbers as another complex number, showing both real and imaginary parts.
Zero Product Property
The Zero Product Property is a fundamental principle in algebra that states if the product of two numbers is zero, then at least one of the numbers must be zero. This property extends to complex numbers as well.
If \( zw = 0 \) for complex numbers \( z \) and \( w \), then either \( z \) or \( w \) must be zero. This property is crucial for solving polynomial equations, as it allows us to break down complex problems into simpler parts.
Understanding and applying this property helps students verify solutions and uncover roots of equations.
If \( zw = 0 \) for complex numbers \( z \) and \( w \), then either \( z \) or \( w \) must be zero. This property is crucial for solving polynomial equations, as it allows us to break down complex problems into simpler parts.
Understanding and applying this property helps students verify solutions and uncover roots of equations.
Imaginary Numbers
Imaginary numbers are used to extend the concept of number systems in mathematics. An imaginary number is defined as a number that gives a negative result when squared.
The standard imaginary unit is denoted as \( i \), where \( i^2 = -1 \). In complex numbers, imaginary numbers are expressed as \( bi \), where \( b \) is a real number.
By combining real numbers with imaginary numbers, we can construct complex numbers, which are essential for solving equations that do not have solutions within the real number realm, such as quadratic equations with negative discriminants.
The standard imaginary unit is denoted as \( i \), where \( i^2 = -1 \). In complex numbers, imaginary numbers are expressed as \( bi \), where \( b \) is a real number.
By combining real numbers with imaginary numbers, we can construct complex numbers, which are essential for solving equations that do not have solutions within the real number realm, such as quadratic equations with negative discriminants.
Other exercises in this chapter
Problem 4
Let \(f(z)=a z+b\) and \(g(z)=\alpha z+\beta\), where neither is the identity. Show that \(f g f^{-1} g^{-1}\) is a translation. Show also that \(f\) commutes w
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Show that the set of roots of unity for all \(n\) (that is, the set \(z\) for which \(z^{n}=1\) for some \(n\) ) is a group with respect to multiplication.
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Suppose that \(f\) is a reflection in the line \(L\), and that \(f(z)\) that \(f(z)=a(\bar{z}-\bar{b})\). As \(|a|=1\) we can write \(a=e^{i \theta} ;\) let \(c
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