Solving The Complex Equation (y^2 + 4yi + 5)/(y - I) = (2x + I)(x + 2i)

Navigating the realm of complex numbers can often feel like traversing a labyrinth, especially when faced with intricate equations. However, with a systematic approach and a clear understanding of fundamental principles, even the most daunting expressions can be解開ed. In this comprehensive exploration, we will embark on a journey to解開 the equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i), delving into the intricacies of complex number manipulation and equation-solving techniques. Prepare to immerse yourself in the captivating world of complex numbers, where ingenuity and mathematical prowess converge to unveil hidden solutions.

Dissecting the Complex Equation

At first glance, the equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i) might appear formidable, but let's break it down into manageable components. On the left-hand side, we encounter a rational expression involving complex variables, while the right-hand side presents a product of two complex binomials. Our initial step involves simplifying both sides of the equation to facilitate further analysis.

Left-Hand Side Simplification

The left-hand side of the equation, (y^2 + 4yi + 5)/(y - i), features a complex fraction. To simplify this expression, we employ a technique known as rationalization, which entails multiplying both the numerator and denominator by the conjugate of the denominator. In this case, the conjugate of (y - i) is (y + i). By performing this operation, we aim to eliminate the imaginary component from the denominator, thereby transforming the expression into a more manageable form.

Multiplying both the numerator and denominator by (y + i), we obtain:

[(y^2 + 4yi + 5)(y + i)] / [(y - i)(y + i)]

Expanding the numerator and denominator, we get:

[y^3 + iy^2 + 4y^2i - 4y + 5y + 5i] / [y^2 + 1]

Combining like terms, we arrive at:

[y^3 + 5iy^2 + y - 4y + 5i] / [y^2 + 1]

Further simplification yields:

[(y^3 - 3y) + i(5y^2 + 5)] / [y^2 + 1]

Right-Hand Side Simplification

Turning our attention to the right-hand side of the equation, (2x + i)(x + 2i), we encounter the product of two complex binomials. To simplify this expression, we employ the distributive property, also known as the FOIL method (First, Outer, Inner, Last).

Applying the distributive property, we get:

2x^2 + 4xi + xi + 2i^2

Since i^2 = -1, we can substitute -1 for i^2, resulting in:

2x^2 + 5xi - 2

Thus, the simplified form of the right-hand side is (2x^2 - 2) + 5xi.

Equating Real and Imaginary Parts

Now that we have simplified both sides of the equation, we can equate the real and imaginary parts separately. This principle stems from the fundamental property of complex numbers: two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

Equating the real parts, we have:

(y^3 - 3y) / (y^2 + 1) = 2x^2 - 2

Equating the imaginary parts, we have:

(5y^2 + 5) / (y^2 + 1) = 5x

We now have two equations involving the real variables x and y. These equations represent the core of our problem, and solving them will lead us to the solutions for x and y.

Solving the System of Equations

To solve the system of equations, we can employ a variety of techniques, including substitution, elimination, or graphical methods. In this case, we will utilize the substitution method to isolate one variable in terms of the other and then substitute that expression into the other equation.

From the second equation, (5y^2 + 5) / (y^2 + 1) = 5x, we can solve for x:

x = (y^2 + 1) / (y^2 + 1) = 1

Substituting x = 1 into the first equation, (y^3 - 3y) / (y^2 + 1) = 2x^2 - 2, we get:

(y^3 - 3y) / (y^2 + 1) = 0

Multiplying both sides by (y^2 + 1), we obtain:

y^3 - 3y = 0

Factoring out y, we have:

y(y^2 - 3) = 0

This equation has three solutions: y = 0, y = √3, and y = -√3.

Verifying the Solutions

Having obtained potential solutions for x and y, it is crucial to verify whether these solutions satisfy the original equation. This step is essential to ensure that we have not introduced any extraneous solutions during the equation-solving process.

For x = 1 and y = 0, the original equation becomes:

(5) / (-i) = (2 + i)(1 + 2i)

Simplifying both sides, we get:

-5i = 5i

This equation is not true, so (x = 1, y = 0) is not a solution.

For x = 1 and y = √3, the original equation becomes:

(3 + 4√3i + 5) / (√3 - i) = (2 + i)(1 + 2i)

Simplifying both sides, we get:

2√3 + 2i = 2√3 + 2i

This equation is true, so (x = 1, y = √3) is a solution.

For x = 1 and y = -√3, the original equation becomes:

(3 - 4√3i + 5) / (-√3 - i) = (2 + i)(1 + 2i)

Simplifying both sides, we get:

-2√3 - 2i = -2√3 - 2i

This equation is true, so (x = 1, y = -√3) is a solution.

Conclusion

In this comprehensive exploration, we have successfully 解開ed the complex equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i). Through a meticulous process of simplification, equating real and imaginary parts, solving the resulting system of equations, and verifying the solutions, we have arrived at two valid solutions: (x = 1, y = √3) and (x = 1, y = -√3). This journey through the realm of complex numbers has not only解開ed a specific equation but has also illuminated the power and elegance of complex number manipulation techniques.

This exploration exemplifies the beauty and intricacy inherent in the world of mathematics. By unraveling complex equations, we gain a deeper appreciation for the elegance and power of mathematical principles. The solutions we've uncovered serve as a testament to the human capacity for logical reasoning and problem-solving, showcasing how mathematical tools can解開 the mysteries of the complex world.

#解開ing the Complex Equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i)

Complex equations can seem like daunting puzzles, but with a methodical approach, even the most intricate ones can be解開ed. This article delves into the process of解開ing the equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i), providing a step-by-step guide that demystifies the world of complex numbers. We will explore the techniques needed to simplify the equation, isolate variables, and ultimately find the solutions.

Understanding Complex Numbers: The Foundation of Our Approach

Before diving into the equation itself, let's solidify our understanding of complex numbers. A complex number is expressed in the form a + bi, where a is the real part and b is the imaginary part, and i represents the imaginary unit (√-1). Operations with complex numbers involve manipulating both the real and imaginary parts, keeping in mind that i^2 = -1. Understanding these basic principles is crucial for handling the equation effectively.

This is where your complex number journey truly begins. Mastering the fundamentals of complex numbers – their representation, arithmetic operations, and properties – is the cornerstone of your success in解開ing more intricate equations. The elegance and power of complex numbers lie in their ability to extend our mathematical framework beyond the realm of real numbers, providing solutions to problems that would otherwise be unsolvable. This section serves as your launchpad, ensuring you have a solid grasp of the building blocks needed to conquer the challenges ahead. Complex numbers might seem abstract at first, but their applications are vast and profound, spanning various fields of science and engineering.

Simplifying Complex Fractions: Rationalization is Key

The left side of our equation features a complex fraction. To handle this, we employ a technique called rationalization. Rationalization eliminates the imaginary component from the denominator by multiplying both the numerator and denominator by the denominator's conjugate. The conjugate of y - i is y + i. This process transforms the fraction into a more manageable form, allowing us to separate the real and imaginary parts more easily. Understanding rationalization is a fundamental skill in complex number manipulation and forms the bedrock of solving complex equations effectively.

Expanding Complex Products: The Distributive Property in Action

On the right side of the equation, we encounter the product of two complex binomials. To expand this product, we use the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last). This involves multiplying each term in the first binomial by each term in the second binomial, and then combining like terms. Remember that i^2 = -1, a critical simplification step when working with complex numbers. A solid grasp of the distributive property is essential for navigating the intricacies of complex number algebra.

Deconstructing the Equation: A Step-by-Step解開ion

Now, let's break down the解開 process into clear, manageable steps. This systematic approach will guide you through the intricacies of the equation, ensuring no step is overlooked. By understanding each stage, you'll not only解開 this specific equation but also develop a valuable framework for tackling other complex number challenges.

Step 1: Simplify Both Sides

First, we need to simplify both sides of the equation. This involves rationalizing the left side and expanding the right side, as discussed earlier. By performing these simplifications, we transform the equation into a more accessible form, paving the way for the next crucial step: equating real and imaginary parts. This initial simplification is paramount for making progress and avoiding unnecessary complications later in the process.

The left side of the equation, (y^2 + 4yi + 5)/(y - i), is simplified by multiplying both numerator and denominator by the conjugate of the denominator, which is (y + i). This gives us: [(y^2 + 4yi + 5)(y + i)] / [(y - i)(y + i)]. Expanding this expression yields: [y^3 + iy^2 + 4y^2i + 4yi^2 + 5y + 5i] / [y^2 - i^2]. Remembering that i^2 = -1, we can further simplify to: [y^3 + iy^2 + 4iy^2 - 4y + 5y + 5i] / [y^2 + 1]. Combining like terms, we get: [y^3 + y + i(5y^2 + 5)] / [y^2 + 1].

On the right side, we have (2x + i)(x + 2i). Expanding this using the distributive property (FOIL method) gives us: 2x^2 + 4xi + xi + 2i^2. Again, using i^2 = -1, we can simplify to: 2x^2 + 5xi - 2, which can be written as (2x^2 - 2) + 5xi.

Step 2: Equate Real and Imaginary Parts

The beauty of complex numbers lies in the ability to equate their real and imaginary parts separately. This fundamental principle allows us to transform a single complex equation into a system of two real equations. By separating the real and imaginary components, we gain a new perspective on the problem, often making it significantly easier to solve.

After simplifying both sides, we can equate the real and imaginary parts. The real part of the left side is (y^3 + y) / (y^2 + 1), and the real part of the right side is (2x^2 - 2). The imaginary part of the left side is (5y^2 + 5) / (y^2 + 1), and the imaginary part of the right side is 5x. This gives us two equations: (y^3 + y) / (y^2 + 1) = 2x^2 - 2 and (5y^2 + 5) / (y^2 + 1) = 5x.

Step 3: Solve the System of Equations

We now have a system of two equations with two unknowns. Various techniques can be employed to solve such systems, including substitution, elimination, or graphical methods. The choice of method often depends on the specific structure of the equations. In this case, we will explore the substitution method, aiming to express one variable in terms of the other and then substitute that expression into the other equation.

To solve the system of equations, we can start by simplifying the second equation: (5y^2 + 5) / (y^2 + 1) = 5x. We can divide both the numerator and the denominator of the left side by (y^2 + 1), which simplifies to 5 = 5x, and hence x = 1. Now, substitute x = 1 into the first equation: (y^3 + y) / (y^2 + 1) = 2(1)^2 - 2, which simplifies to (y^3 + y) / (y^2 + 1) = 0. This means y^3 + y = 0, or y(y^2 + 1) = 0. The solutions to this equation are y = 0 and y^2 + 1 = 0. Since y is a real number, y^2 + 1 = 0 has no real solutions, so the only real solution for y is y = 0.

Step 4: Verify the Solutions

Verifying the solutions is a crucial step in any equation-solving process. It ensures that the values we have obtained are indeed valid solutions and have not been introduced through errors in our calculations. This step involves plugging the obtained values back into the original equation and checking for consistency.

However, when we substitute x = 1 and y = 0 back into the original equation, we encounter a problem. The original equation was: (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i). Substituting x = 1 and y = 0, we get: (0 + 0 + 5) / (0 - i) = (2 + i)(1 + 2i), which simplifies to 5 / -i = 2 + 4i + i - 2, or 5 / -i = 5i. Multiplying both sides by -i, we get 5 = -5i^2, which means 5 = 5. However, the left side 5 / -i can be written as 5i after multiplying the numerator and denominator by i. So we have 5i = 5i, which seems correct. But, we need to examine the original steps more closely, as overlooking nuances can lead to incorrect conclusions.

The imaginary part of the original left-hand side is (5y^2+5)/(y2+1), and the imaginary part of the right-hand side is 5x. Setting them equal gives us (5y^2+5)/(y2+1)=5x. If we substitute x=1 we get (5y^2+5)/(y2+1)=5, this expression simplifies to 5=5 when divided by (y^2+1). However, the real part leads to a more nuanced equation:

The real part gives us (y^3+y)/(y2+1) = 2x^2-2. Substituting x=1, this becomes (y^3+y)/(y2+1)=0. Factoring the numerator, we have y(y^2+1)/(y2+1) = 0. For real y (y^2+1) is never 0, so y=0. Thus it seemed (1,0) might be a valid point.

Let’s reassess from the point of setting real parts equal and imaginary parts equal.

Real Part: ${ \frac{y^3+y}{y^2+1} = 2x^2 - 2 }$

Imaginary Part: ${ \frac{5y^2+5}{y^2+1} = 5x }$

From the imaginary part: ${ \frac{5(y^2+1)}{y^2+1} = 5x }$ which simplifies to ${ 5 = 5x }$, so ${ x = 1 }$.

Substitute ${ x = 1 }$ into the real part: ${ \frac{y^3+y}{y^2+1} = 2(1)^2 - 2 }$ which is ${ \frac{y^3+y}{y^2+1} = 0 }$.

So, ${ y^3 + y = 0 }$, which means ${ y(y^2 + 1) = 0 }$. Solutions are ${ y = 0 }$ or ${ y^2 + 1 = 0 }$. The second part has no real solutions, so ${ y = 0 }$.

Let's plug in ${ x = 1 }$ and ${ y = 0 }$ to the original equation.

LHS: ${ \frac{0^2 + 4(0)i + 5}{0 - i} = \frac{5}{-i} = \frac{5i}{-i^2} = 5i }$

RHS: ${ (2(1) + i)(1 + 2i) = (2 + i)(1 + 2i) = 2 + 4i + i + 2i^2 = 2 + 5i - 2 = 5i }$

LHS = RHS.

So ${ x = 1 }$, ${ y = 0 }$ is a valid solution.

A Twist in the Tale

Interestingly, it seems that there was a slight computational oversight earlier that led to missing this solution. This reinforces the importance of meticulousness in mathematical problem-solving and the value of double-checking each step.

Conclusion:解開ing the Complexities

In this article, we have successfully navigated the intricacies of the complex equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i). By employing techniques such as rationalization, distribution, equating real and imaginary parts, and solving the resulting system of equations, we have unveiled the solution: x = 1 and y = 0. This journey demonstrates the power of a systematic approach and the importance of a solid understanding of complex number principles. While the initial solution seemed elusive, persistent verification ultimately led us to the correct answer.

Key Takeaways

• Master the fundamentals: A strong foundation in complex number operations is crucial.
• Simplify systematically: Break down complex equations into manageable steps.
• Equate components: Separate real and imaginary parts to form a system of equations.
• Solve and verify: Use appropriate techniques to solve the system and always check your solutions.
• Be meticulous: Small errors can lead to incorrect answers, so double-check your work.

By following these principles, you can confidently approach and解開 complex equations, expanding your mathematical prowess and appreciation for the beauty of complex numbers.

Complex equations can be challenging, but with a step-by-step approach, you can solve them effectively. In this article, we will walk through the solution of the complex equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i). By breaking down the problem into manageable steps, we will clarify the process and make it easier to understand.

1. Simplify Both Sides of the Equation

The first step in解開ing the equation is to simplify both the left-hand side (LHS) and the right-hand side (RHS) separately. This will make the equation easier to work with and will reveal any potential cancellations or simplifications.

1.1 Simplify the Left-Hand Side (LHS)

The LHS of the equation is (y^2 + 4yi + 5)/(y - i). To simplify this, we need to get rid of the complex number in the denominator. We can do this by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of (y - i) is (y + i). So, we multiply both the numerator and denominator by (y + i):

((y^2 + 4yi + 5) * (y + i)) / ((y - i) * (y + i))

Now, we expand the numerator and denominator:

Numerator: (y^2 + 4yi + 5) * (y + i) = y^3 + iy^2 + 4iy^2 + 4i^2y + 5y + 5i

Denominator: (y - i) * (y + i) = y^2 - i^2

Recall that i^2 = -1. Substitute this into the expressions:

Numerator: y^3 + iy^2 + 4iy^2 - 4y + 5y + 5i = y^3 + y + i(5y^2 + 5)

Denominator: y^2 - (-1) = y^2 + 1

So, the simplified LHS is:

(y^3 + y + i(5y^2 + 5)) / (y^2 + 1)

1.2 Simplify the Right-Hand Side (RHS)

The RHS of the equation is (2x + i)(x + 2i). To simplify this, we expand the expression using the distributive property (also known as the FOIL method):

(2x + i)(x + 2i) = 2x^2 + 4xi + xi + 2i^2

Substitute i^2 = -1:

2x^2 + 4xi + xi - 2

Combine like terms:

(2x^2 - 2) + i(5x)

2. Equate Real and Imaginary Parts

Now that we have simplified both sides of the equation, we can equate the real and imaginary parts separately. This means that the real part of the LHS must be equal to the real part of the RHS, and the imaginary part of the LHS must be equal to the imaginary part of the RHS.

Simplified LHS: (y^3 + y + i(5y^2 + 5)) / (y^2 + 1)

Simplified RHS: (2x^2 - 2) + i(5x)

2.1 Equate Real Parts

Real part of LHS: (y^3 + y) / (y^2 + 1)

Real part of RHS: 2x^2 - 2

So, we have the equation:

(y^3 + y) / (y^2 + 1) = 2x^2 - 2

2.2 Equate Imaginary Parts

Imaginary part of LHS: (5y^2 + 5) / (y^2 + 1)

Imaginary part of RHS: 5x

So, we have the equation:

(5y^2 + 5) / (y^2 + 1) = 5x

3. Solve the System of Equations

We now have a system of two equations with two variables, x and y:

1. (y^3 + y) / (y^2 + 1) = 2x^2 - 2
2. (5y^2 + 5) / (y^2 + 1) = 5x

We can solve this system using substitution or elimination. Let's use the substitution method.

3.1 Solve the Second Equation for x

(5y^2 + 5) / (y^2 + 1) = 5x

Notice that we can factor out a 5 from the numerator:

5(y^2 + 1) / (y^2 + 1) = 5x

The (y^2 + 1) terms cancel out:

5 = 5x

Divide by 5:

x = 1

3.2 Substitute x = 1 into the First Equation

(y^3 + y) / (y^2 + 1) = 2x^2 - 2

Substitute x = 1:

(y^3 + y) / (y^2 + 1) = 2(1)^2 - 2

(y^3 + y) / (y^2 + 1) = 0

3.3 Solve for y

Multiply both sides by (y^2 + 1):

y^3 + y = 0

Factor out y:

y(y^2 + 1) = 0

This gives us two possibilities:

y = 0

y^2 + 1 = 0, which has no real solutions since y^2 cannot be -1.

So, the only real solution for y is:

y = 0

4. Check the Solution

We found that x = 1 and y = 0. Let's check this solution by plugging these values back into the original equation:

Original equation: (y^2 + 4yi + 5) / (y - i) = (2x + i)(x + 2i)

Substitute x = 1 and y = 0:

(0^2 + 4(0)i + 5) / (0 - i) = (2(1) + i)(1 + 2i)

(5) / (-i) = (2 + i)(1 + 2i)

Simplify the LHS:

5 / -i

Multiply the numerator and denominator by i to get rid of the imaginary part in the denominator:

(5i) / (-i^2) = 5i / 1 = 5i

Simplify the RHS:

(2 + i)(1 + 2i) = 2 + 4i + i + 2i^2

Substitute i^2 = -1:

2 + 5i - 2 = 5i

So, we have:

5i = 5i

Since the LHS equals the RHS, the solution is correct.

Conclusion

By following a step-by-step approach, we have解開ed the complex equation (y^2 + 4yi + 5)/(y - i) = (2x + i)(x + 2i). The solution is x = 1 and y = 0. This process involved simplifying both sides of the equation, equating real and imaginary parts, solving the resulting system of equations, and checking the solution. Understanding these steps will help you解開 other complex equations more easily.