Problem 75
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. $$(\bar{z})^{2}=\overline{z^{2}}$$
Step-by-Step Solution
Verified Answer
The two expressions are equal; there is a proof that \((\bar{z})^2 = \overline{z^2}\).
1Step 1: Understand Complex Conjugate
The complex conjugate of a complex number \( z = a + bi \) is \( \bar{z} = a - bi \). Likewise, for \( w = c + di \), its complex conjugate is \( \bar{w} = c - di \).
2Step 2: Calculate \((\bar{z})^2\)
First, find \((\bar{z})^2\) by squaring the complex conjugate of \( z \). We have \( \bar{z} = a - bi \), so \((\bar{z})^2 = (a - bi)^2 = a^2 - 2abi + (bi)^2 = a^2 - 2abi - b^2 \). This simplifies to \( a^2 - b^2 - 2abi \).
3Step 3: Calculate \( z^2 \) and its conjugate
Now, find \( z^2 \) by squaring \( z \). We have \( z = a + bi \), so \( z^2 = (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 + 2abi - b^2 \). Its complex conjugate is \( \overline{z^2} = \overline{a^2 + 2abi - b^2} = a^2 - 2abi - b^2 \) (reversing the sign of the imaginary part).
4Step 4: Compare Results
Compare the two results: \((\bar{z})^2 = a^2 - b^2 - 2abi \) and \( \overline{z^2} = a^2 - b^2 - 2abi \). They are equal, thus proving \((\bar{z})^2 = \overline{z^2}\).
Key Concepts
Complex ConjugateSquaring Complex NumbersProperties of Complex Numbers
Complex Conjugate
The complex conjugate is a fascinating aspect of complex numbers. If you have a complex number in the form of \( z = a + bi \), its complex conjugate is represented by \( \bar{z} = a - bi \). This involves changing the sign of the imaginary part while keeping the real part the same. The complex conjugate is useful in complex number arithmetic, especially for simplifying division and multiplication of complex numbers. It helps in finding the absolute value of complex numbers by eliminating the imaginary part when multiplied by the original number.
- The notation \( \bar{z} \) is used to denote the conjugate of \( z \).
- Conjugates are crucial in forming a real number through multiplication.
- Useful for finding the modulus, \( |z| = \sqrt{z \cdot \bar{z}} \).
Squaring Complex Numbers
Squaring a complex number isn't too different from squaring real numbers, but it does involve careful handling of terms. For a complex number \( z = a + bi \), squaring it involves expanding the binomial:\[z^2 = (a + bi)^2 = a^2 + 2abi + (bi)^2\]Breaking this down, - You first square the real part: \( a^2 \).- Then, you double the product of the real and imaginary parts: \( 2abi \).- Finally, you square the imaginary unit: \( (bi)^2 = -b^2 \) due to \( i^2 = -1 \).So, it simplifies to \( a^2 + 2abi - b^2 \). This detail highlights why understanding the imaginary unit \( i \) is vital. Squaring a conjugate, \( \bar{z} = a - bi \), following a similar process produces \( a^2 - 2abi - b^2 \). It's intriguing to see how squaring affects a complex number and its conjugate, emphasizing their mathematical properties.
Properties of Complex Numbers
Complex numbers hold intriguing properties that extend the real number system to solve more complex equations. They consist of a real part and an imaginary part, represented as \( z = a + bi \). These properties are essential in complex analysis below:
- **Addition and Subtraction**: Performed by adding or subtracting the corresponding real and imaginary parts. For example, \((a + bi) + (c + di) = (a+c) + (b+d)i\).
- **Multiplication**: Follow the distributive property, ensuring \(i^2 = -1\) for imaginary components. For example, \((a + bi)(c + di) = ac + adi + bci - bd\).
- **Division**: Involves multiplying the numerator and denominator by the complex conjugate of the divisor to eliminate the imaginary part in the denominator.
- **Modulus**: Given by \( |z| = \sqrt{a^2 + b^2} \), showing the distance from the origin in the complex plane.
- **Polar Form**: Expressed as \( z = r(\cos\theta + i\sin\theta) \), where \( r \) is the modulus and \( \theta \) the argument, facilitating multiplication and division.
Other exercises in this chapter
Problem 74
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