Problem 36 - Advanced Engineering Mathematics

Question

\(x^{2}-y^{2}-4 x+(-2 x y-4 y) i=0+0 i\) implies \(x^{2}-y^{2}-4 x=0\) and \(y(-2 x-4)=0 .\) If \(y=0\) then \(x(x-4)=0\) and so \(z=0\) and \(z=4 .\) If \(-2 x-4=0\) or \(x=-2\) then \(12-y^{2}=0\) or \(y=\pm 2 \sqrt{3} .\) This gives \(z=-2+2 \sqrt{3} i\) and \(z=-2-2 \sqrt{3} i\)

Step-by-Step Solution

Answer

Complex solutions: 0, 4, -2±2√3i.

Step 1: Separate Real and Imaginary Parts

Given the complex equation \(x^2 - y^2 - 4x + (-2xy - 4y)i = 0 + 0i\), separate it into real and imaginary components. The real part provides the equation \(x^2 - y^2 - 4x = 0\), and the imaginary part gives \(-2xy - 4y = 0\). These two will be solved separately.

Step 2: Solve Real Part for x

From the equation \(x^2 - y^2 - 4x = 0\), factor it to get \((x^2 - 4x) - y^2 = 0\). If \(y = 0\), it simplifies to \(x^2 - 4x = 0\). This can be factored further as \(x(x - 4) = 0\), indicating that \(x = 0\) or \(x = 4\).

Step 3: Check Imaginary Part When y = 0

Checking with \(y = 0\) in the imaginary component equation \(-2xy - 4y = 0\), you confirm it satisfies the equation since the whole term becomes zero.

Step 4: Solve Imaginary Part for y

If \(y eq 0\), divide the equation \(-2xy - 4y = 0\) by \(y\) to get \(-2x - 4 = 0\). Solve to find \(x = -2\).

Step 5: Calculate y when x = -2

Substituting \(x = -2\) in the real equation \(x^2 - y^2 - 4x = 0\), compute \((-2)^2 - y^2 - 4(-2) = 0\), which simplifies to \(4 - y^2 + 8 = 0\); \(12 - y^2 = 0\). Thus, \(y = \pm 2\sqrt{3}\).

Step 6: Conclude Complex Solutions

The solutions for \(z\) can now be compiled as, for \(y = 0\), \(z = 0\) and \(z = 4\). For \(x = -2\) and \(y = \pm 2\sqrt{3}\), the solutions are \(z = -2 + 2\sqrt{3}i\) and \(z = -2 - 2\sqrt{3}i\).

Key Concepts

Real and Imaginary Parts

In complex numbers, any value can be expressed in the form of a real part and an imaginary part.
Complex numbers take the format of \( a + bi \), where \( a \) and \( b \) are real numbers. Here, \( a \) is the real part, and \( b \) is the coefficient of the imaginary part represented by \( i \), which is \( \sqrt{-1} \).
When solving equations involving complex numbers, separating these components is crucial.

• The real part consists of all components without the imaginary unit \( i \).
• The imaginary part consists exclusively of components containing \( i \).

In the given equation, \( x^2 - y^2 - 4x + (-2xy - 4y)i = 0 + 0i \), notice how the equation is divided into real and imaginary components:

• Real: \( x^2 - y^2 - 4x = 0 \)
• Imaginary: \( -2xy - 4y = 0 \)

These separate equations are then solved individually, allowing us to better isolate each variable's contribution, facilitating the calculation of potential solutions.

Factoring Equations

Factoring is a vital skill when working with polynomial equations, as it greatly simplifies the solving process. Factoring involves rewriting an equation as a product of its factors.
For instance, the equation \( x^2 - y^2 - 4x = 0 \) can be tackled by initially focusing on terms related to \( x \).
If \( y = 0 \), the equation modifies to \( x^2 - 4x = 0 \).
By factoring out the common term, the expression becomes \( x(x - 4) = 0 \).
This reveals that the possible values for \( x \) are those which make each factor zero:

• If \( x = 0 \), the equation is satisfied.
• If \( x - 4 = 0 \), then \( x = 4 \).

Understanding this concept makes it easier to break down complex equations, reducing them into simpler expressions that are much more manageable to solve.

Complex Solutions

To determine the complex solutions of an equation, you must solve both the real and imaginary parts separately and then combine their solutions.
Once each component has been addressed individually, combine results to express the complete set of solutions in terms of complex numbers.
In our example:

• Solving the real component \( x^2 - y^2 - 4x = 0 \) gives possible real solutions for \( x \) depending on values of \( y \).
• The imaginary component \( -2xy - 4y = 0 \), separated and simplified, yields solutions concerning \( y \).

Combining these solutions, considering \( y = 0 \) gives us \( z = 0 \) and \( z = 4 \).
Meanwhile, solving for when \( y \) is non-zero, specifically \( x = -2 \) with \( y = \pm 2\sqrt{3} \), yields the following complex solutions:

• \( z = -2 + 2\sqrt{3}i \)
• \( z = -2 - 2\sqrt{3}i \)

These solutions showcase the interplay between real numbers and their imaginary counterparts, providing a complete picture of solutions within the complex number plane.