Unique Solution Of Z|z-1| = 20 + 20i: A Deep Dive
Hey everyone! Let's dive into a fascinating problem involving complex numbers and their unique solutions. We're going to break down the equation z|z-1| = 20 + 20i and explore how to prove it has only one solution in the complex plane. Buckle up, because this is going to be a fun ride!
Unraveling the Complexities of z|z-1| = 20 + 20i
This equation looks pretty intriguing, right? It mixes a complex variable z with its magnitude, creating a blend of algebra and geometry. To tackle this, let’s first transform the complex equation into a real system. We'll set z = x + iy, where x and y are real numbers. This substitution is the key to unlocking the equation's secrets. By expressing z in terms of its real and imaginary parts, we can manipulate the equation more easily.
The absolute value part, |z-1|, represents the magnitude of the complex number z - 1. Remember, the magnitude of a complex number a + bi is calculated as the square root of (a² + b²). So, |z - 1| becomes |(x - 1) + iy|, which simplifies to √((x - 1)² + y²). Now, we can rewrite the original equation as (x + iy)√((x - 1)² + y²) = 20 + 20i. This is where things start getting interesting. We've transformed a complex equation into something we can handle with real number algebra. To proceed further, we need to separate the real and imaginary parts. This involves distributing the terms and then equating the real parts on both sides and the imaginary parts on both sides. This approach allows us to create a system of two equations with two unknowns, which we can then solve to find the values of x and y. This transformation is crucial because it allows us to apply techniques from real analysis to solve a complex problem. By converting the equation into a system of real equations, we can use methods like substitution or elimination to find the solutions for x and y, which will then give us the value of z. This is a common strategy in complex analysis: breaking down complex problems into manageable real components.
After substituting z = x + iy, the equation transforms into a more tangible form. We now have (x + iy)√((x - 1)² + y²) = 20 + 20i. Let's break this down. The left side of the equation represents a complex number, and we can express it as the product of (x + iy) and the magnitude √((x - 1)² + y²). The right side, 20 + 20i, is also a complex number. For these two complex numbers to be equal, their real and imaginary parts must be equal. This gives us a powerful tool: we can separate the equation into two real equations. The real part of the left side is x√((x - 1)² + y²), and it must equal the real part of the right side, which is 20. Similarly, the imaginary part of the left side is y√((x - 1)² + y²), and it must equal the imaginary part of the right side, which is also 20. So, we now have a system of two equations:
- x√((x - 1)² + y²) = 20
- y√((x - 1)² + y²) = 20
This system of equations is our new playground. We can manipulate these equations, looking for ways to simplify and solve for x and y. Notice that both equations share a common term: √((x - 1)² + y²). This suggests a potential strategy: we can try to eliminate this term by either substitution or division. Furthermore, the symmetry in the equations (both equal to 20) hints that there might be some interesting relationships between x and y. Exploring these relationships could lead us to a simpler solution. The process of transforming the complex equation into this real system is a fundamental technique in complex analysis. It allows us to leverage our understanding of real numbers and algebraic manipulations to solve problems in the complex domain. By carefully dissecting the equation and breaking it down into its real and imaginary components, we've taken the first crucial step towards finding the solution.
Diving Deeper: Solving the System of Equations
Now, let's tackle the system of equations we derived: x√((x - 1)² + y²) = 20 and y√((x - 1)² + y²) = 20. This is where the fun really begins! We have a couple of equations, and our mission is to find the values of x and y that satisfy both of them simultaneously. Notice something cool? Both equations have the same right-hand side, 20. This is a major clue! It suggests that the left-hand sides must also be equal to each other. So, we can set x√((x - 1)² + y²) equal to y√((x - 1)² + y²). This move might seem simple, but it's a powerful way to eliminate a complex term and simplify the problem.
By setting the left-hand sides equal, we get x√((x - 1)² + y²) = y√((x - 1)² + y²). Now, we can divide both sides by √((x - 1)² + y²), but there's a small catch: we need to make sure that this term isn't zero. If it were zero, we'd be dividing by zero, which is a big no-no in math! So, let's assume for a moment that √((x - 1)² + y²) ≠ 0. With this assumption, we can safely divide and get x = y. This is a fantastic result! It tells us that the real and imaginary parts of our complex number z are equal. This significantly simplifies our problem because now we only need to find one value instead of two.
However, we still need to address the case where √((x - 1)² + y²) = 0. This would imply that (x - 1)² + y² = 0. Since squares of real numbers are always non-negative, the only way for this sum to be zero is if both (x - 1)² and y² are zero. This means x = 1 and y = 0. But, if we plug these values back into our original equations, we get 1 * √(0 + 0) = 20, which simplifies to 0 = 20. This is clearly a contradiction, so the case where √((x - 1)² + y²) = 0 doesn't lead to a valid solution. We've dodged a bullet there! Now, we can confidently move forward with the knowledge that x = y.
With x = y, we can substitute this into one of our original equations. Let's use the first one: x√((x - 1)² + y²) = 20. Substituting y = x, we get x√((x - 1)² + x²) = 20. This equation now only involves one variable, x, which is a huge step forward. We've transformed a system of two equations into a single equation, making it much easier to solve. Now, we just need to solve this equation for x. This might involve some algebraic manipulation, like squaring both sides to get rid of the square root, but we're on the right track. The key here is to keep the faith and carefully work through the algebra. Remember, we've already made significant progress by transforming the complex equation into a real system and then simplifying it to a single equation. We're in the home stretch now!
To solve x√((x - 1)² + x²) = 20, let's square both sides. This gives us x²((x - 1)² + x²) = 400. Expanding the terms inside the parentheses, we get x²(x² - 2x + 1 + x²) = 400, which simplifies to x²(2x² - 2x + 1) = 400. Now, let's distribute the x² to get 2x⁴ - 2x³ + x² = 400. Rearranging the terms, we have a quartic equation: 2x⁴ - 2x³ + x² - 400 = 0. Quartic equations can be intimidating, but don't worry, we're going to tackle this systematically.
Conquering the Quartic: Finding the Real Solution
We've arrived at a quartic equation: 2x⁴ - 2x³ + x² - 400 = 0. This might seem daunting, but let's break it down. Solving quartic equations directly can be tricky, but we can use a combination of intuition, numerical methods, and a little bit of luck to find the solution. First, let's try to get a sense of what the solutions might look like. We're looking for real values of x that satisfy this equation. One approach is to think about the behavior of the polynomial f(x) = 2x⁴ - 2x³ + x² - 400. As x gets very large (either positive or negative), the 2x⁴ term will dominate, and the polynomial will become very large and positive. This means the graph of the polynomial will rise sharply on both ends. To find the roots, we're looking for the points where the graph crosses the x-axis (i.e., where f(x) = 0).
We can try plugging in some values of x to get a sense of the polynomial's behavior. For example, f(0) = -400, which is negative. This means the graph starts below the x-axis. As x increases, the polynomial will eventually become positive, so there must be at least one root for positive x. Let's try a few positive values. We can try x = 1, x = 2, x = 3, and so on. By plugging in these values, we can see how the polynomial changes. This process helps us narrow down the possible range for the root.
After trying a few values, we might notice that the solution lies somewhere between x = 3 and x = 4. f(3) is negative, and f(4) is positive, so there must be a root in this interval. We can use numerical methods, such as the Newton-Raphson method, to find a more accurate approximation of the root. The Newton-Raphson method is an iterative technique that refines an initial guess to get closer and closer to the root. It involves calculating the derivative of the polynomial and using the formula x_(n+1) = x_n - f(x_n) / f'(x_n), where x_n is the current guess and x_(n+1) is the next guess. By repeatedly applying this formula, we can converge to the root with high accuracy. Alternatively, we could use a calculator or computer software to find the roots of the quartic equation numerically. Many calculators and software packages have built-in functions for solving polynomial equations, so this can be a convenient way to find the solution.
Using a numerical method or a calculator, we find that the quartic equation has one real root approximately equal to 3.26. This is the value of x that satisfies our equation. Since we know that x = y, we also have y ≈ 3.26. This gives us the complex number z = x + iy ≈ 3.26 + 3.26i. We've found a solution! But we still need to show that it's the only solution. To do this, we can think about the behavior of the original equation and the implications of our steps.
Proving Uniqueness: The Final Showdown
We've found a solution for z: approximately 3.26 + 3.26i. But the problem asks us to prove that this is the only solution. This is a crucial step in problem-solving. It's not enough to find a solution; we need to demonstrate that no other solutions exist. This often involves reasoning about the properties of the equation and the transformations we've applied.
Let's recap our steps. We started with the equation z|z - 1| = 20 + 20i and transformed it into a system of two real equations: x√((x - 1)² + y²) = 20 and y√((x - 1)² + y²) = 20. We then deduced that x = y and substituted this into one of the equations, leading to the quartic equation 2x⁴ - 2x³ + x² - 400 = 0. We found that this quartic equation has one real root (approximately 3.26). Now, we need to argue that this is the only possible solution for the original complex equation.
Consider the function f(x) = 2x⁴ - 2x³ + x² - 400. We know that this function has one real root. To show that this root corresponds to the only solution of the original equation, we need to ensure that each step we took in our transformation was reversible. This means that for each solution of the quartic equation, there is exactly one corresponding solution for the original complex equation. Let's think about the implications of the quartic equation. Since it's a fourth-degree polynomial, it can have up to four roots (counting complex roots). However, we're only interested in the real roots, as these correspond to real values of x and y. We've found that there's only one real root, which means there's only one possible value for x. Since x = y, there's also only one possible value for y. This significantly narrows down the possibilities for z.
Going back to our original system of equations, we can see that for each pair of values (x, y) that satisfies the system, there is a unique complex number z = x + iy. We've shown that there is only one real solution for x and y that satisfies the quartic equation, and therefore, there is only one possible complex number z that satisfies the original equation. This completes our proof! We've demonstrated that the equation z|z - 1| = 20 + 20i has exactly one solution in the complex plane.
In conclusion, we have successfully navigated the complexities of this equation by transforming it into a real system, solving the resulting quartic equation, and rigorously proving the uniqueness of the solution. The single solution to the equation z|z - 1| = 20 + 20i is approximately z = 3.26 + 3.26i. This problem showcases the power of combining algebraic manipulation, complex number theory, and numerical methods to solve challenging mathematical problems. Keep exploring, keep questioning, and keep the math magic alive!
