Problem 30 - Advanced Engineering Mathematics

Question

In Problems 27-32, let \(z=x+i y\). Find the indicated expression. $$ \operatorname{Im}\left(\bar{z}^{2}+z^{2}\right) $$

Step-by-Step Solution

Answer

The imaginary part is 0.

Step 1: Understand the Expression

We are given the expression \( \operatorname{Im}(\bar{z}^2 + z^2) \), where \( z = x + i y \). Here, \( \bar{z} \) is the complex conjugate of \( z \), which is \( x - i y \).

Step 2: Calculate \( z^2 \)

Compute \( z^2 \):\[ z^2 = (x + iy)^2 = x^2 + 2ixy - y^2. \]

Step 3: Calculate \( \bar{z}^2 \)

Compute \( \bar{z}^2 \):\[ \bar{z}^2 = (x - iy)^2 = x^2 - 2ixy - y^2. \]

Step 4: Add \( z^2 \) and \( \bar{z}^2 \)

Add \( z^2 \) and \( \bar{z}^2 \):\[ z^2 + \bar{z}^2 = (x^2 + 2ixy - y^2) + (x^2 - 2ixy - y^2) = 2x^2 - 2y^2. \]Notice that the imaginary parts cancel each other.

Step 5: Determine the Imaginary Part

The expression \( 2x^2 - 2y^2 \) is purely real. Therefore, the imaginary part is\[ \operatorname{Im}(2x^2 - 2y^2) = 0. \]

Key Concepts

Imaginary Part

In the realm of complex numbers, each number has a real part and an imaginary part. When we express a complex number as \( z = a + bi \), \( a \) is the real part, and \( bi \) is the imaginary part, where \( b \) is a real coefficient and \( i \) is the imaginary unit. The imaginary unit, \( i \), is defined as \( \sqrt{-1} \). This means that \( i^2 = -1 \).
The imaginary part of a complex number tells us how much of it is 'imaginary'. In problems, we often find ourselves needing to extract the imaginary part from expressions involving complex numbers, especially when analyzing functions or solving physics problems.

• To find the imaginary part, look at the coefficient of \( i \) in the complex number.
• If expressions are combined, identifying the imaginary part requires aggregating and simplifying the terms associated with \( i \).

This problem asks for the imaginary part of the expression \( \operatorname{Im}(\bar{z}^2 + z^2) \). By calculating \( z^2 \) and \( \bar{z}^2 \), the imaginary components end up canceling out, leaving a purely real expression. Thus, the imaginary part is 0.

Complex Conjugate

The complex conjugate of a complex number provides a way to reflect the number across the real axis in the complex plane. If you have a complex number \( z = a + bi \), its conjugate \( \bar{z} \) is \( a - bi \). This simple sign change from \( +bi \) to \( -bi \) is quite powerful.
Complex conjugates are used primarily to eliminate the imaginary part of a fraction by multiplying the numerator and denominator by the conjugate of the denominator. They are also useful in assessing the modulus of a complex number and in simplifying expressions.

• A complex conjugate reverses the sign of the imaginary component.
• For any complex number, \( z \times \bar{z} = a^2 + b^2 \), which is always real.

In the exercise, the use of \( \bar{z} \) with \( z \) helps in attaining a real outcome when these squared terms are added together, facilitating the identification of purely real expressions without imaginary parts.

Complex Arithmetic

Complex arithmetic encompasses operations like addition, multiplication, and division of complex numbers. It involves treating the imaginary unit \( i \) just like a variable, with the caveat that \( i^2 = -1 \). This makes complex arithmetic both intriguing and uniform across operations.

• **Addition/Subtraction:** Combine like terms — add or subtract real parts and imaginary parts separately.
• **Multiplication:** Distribute terms, then apply \( i^2 = -1 \) to convert complex expressions to standard form.
• **Division:** Multiply by the conjugate to eliminate the imaginary part of the denominator, simplifying the fraction into a real divide calculation.

In this particular problem, arithmetic involves squaring \( z \) and its conjugate, \( \bar{z} \). Notice how when \( z^2 \) and \( \bar{z}^2 \) are added, the imaginary parts cancel out, illustrating the elegance of complex arithmetic where balancing conjugates can effectively simplify expressions.