Unique Root Of Z|z-1| = 20 + 20i A Detailed Solution

Hey guys! Today, we're diving deep into a fascinating problem involving complex numbers. We're going to dissect the equation z|z-1| = 20 + 20i and demonstrate why it possesses exactly one solution in the realm of complex numbers. This journey will involve transforming the complex equation into a real system, leveraging the power of polar coordinates, and ultimately proving the uniqueness of the root. So, buckle up, and let's get started!

Transforming the Complex Equation into a Real System

To kick things off, we'll embark on a journey to translate the complex equation into a more manageable real system. Our trusty tool for this transformation is the substitution z = x + iy, where x and y are real numbers representing the real and imaginary components of z, respectively. This substitution allows us to break down the complex equation into its constituent real and imaginary parts, paving the way for further analysis. Plugging this into our original equation, we get:

(x + iy)|(x - 1) + iy| = 20 + 20i

Now, let's unravel the magnitude of the complex number (x - 1) + iy. Recall that the magnitude of a complex number a + bi is given by √(a² + b²). Applying this to our case, we have:

|(x - 1) + iy| = √((x - 1)² + y²)

Substituting this back into our equation, we arrive at:

(x + iy)√( (x - 1)² + y² ) = 20 + 20i

This equation, while still a bit formidable, is a crucial stepping stone. To further dissect it, we'll equate the real and imaginary parts on both sides. This is a valid operation since two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This gives us two equations:

Equation 1 (Real Part): x√((x - 1)² + y²) = 20

Equation 2 (Imaginary Part): y√((x - 1)² + y²) = 20

We've successfully transformed our complex equation into a system of two real equations. This is a significant achievement, as it allows us to leverage our knowledge of real analysis to tackle the problem. However, the presence of the square root still makes the system a bit unwieldy. Our next move will be to eliminate this square root, making the equations more amenable to manipulation.

To get rid of the square root, let's consider squaring both sides of both equations. This is a valid algebraic manipulation, but we need to be mindful of introducing extraneous solutions. Squaring Equation 1, we get:

x²((x - 1)² + y²) = 400

Similarly, squaring Equation 2 yields:

y²((x - 1)² + y²) = 400

Now we have a system of two equations without square roots. Notice a fascinating similarity between these equations! The right-hand sides are identical, which suggests a potential avenue for simplification. If we subtract the second equation from the first, a lot of terms will cancel out, potentially leading us to a simpler relationship between x and y. Let's perform this subtraction:

x²((x - 1)² + y²) - y²((x - 1)² + y²) = 400 - 400

This simplifies to:

(x² - y²)((x - 1)² + y²) = 0

This equation is a product of two factors equaling zero. This means that either the first factor (x² - y²) is zero, or the second factor ((x - 1)² + y²) is zero (or both). Let's analyze each case separately.

Case 1: x² - y² = 0

If x² - y² = 0, then x² = y². Taking the square root of both sides, we get two possibilities: x = y or x = -y. These represent two lines in the xy-plane, and they will play a crucial role in visualizing our solutions.

Case 2: (x - 1)² + y² = 0

Now, let's consider the second possibility: (x - 1)² + y² = 0. This equation represents the sum of two squares equaling zero. Since squares of real numbers are non-negative, the only way for this sum to be zero is if both terms are individually zero. This means (x - 1)² = 0 and y² = 0. From these, we deduce that x = 1 and y = 0. This corresponds to a single point in the xy-plane, namely (1, 0).

However, we need to be cautious here. Recall that we squared our original equations, which might have introduced extraneous solutions. We need to check if this solution (x = 1, y = 0) actually satisfies our original equations. Plugging these values into Equations 1 and 2, we get:

Equation 1: 1√((1 - 1)² + 0²) = 20 => 0 = 20 (False)

Equation 2: 0√((1 - 1)² + 0²) = 20 => 0 = 20 (False)

As we can see, the solution (x = 1, y = 0) does not satisfy our original equations. Therefore, it's an extraneous solution, and we can discard it. This leaves us with the cases x = y and x = -y to investigate further.

Leveraging Polar Coordinates for Deeper Insights

Now that we've narrowed down the possibilities to x = y and x = -y, let's employ a powerful technique: polar coordinates. Polar coordinates provide a different perspective on the problem, often simplifying equations and revealing hidden structures. Recall that polar coordinates represent a point in the plane using a distance r from the origin and an angle θ from the positive x-axis. The relationships between Cartesian coordinates (x, y) and polar coordinates (r, θ) are given by:

x = r cos θ

y = r sin θ

Substituting these into our original real equations (Equation 1 and Equation 2), we can rewrite the system in terms of r and θ. Let's start with Equation 1:

x√((x - 1)² + y²) = 20

Replacing x and y with their polar equivalents, we get:

(r cos θ)√((r cos θ - 1)² + (r sin θ)²) = 20

Now, let's simplify the expression under the square root. Expanding the terms, we have:

(r cos θ)√(r²cos²θ - 2r cos θ + 1 + r²sin²θ) = 20

We can factor out an r² from the first two terms under the square root and use the trigonometric identity cos²θ + sin²θ = 1:

(r cos θ)√(r²(cos²θ + sin²θ) - 2r cos θ + 1) = 20

This simplifies to:

(r cos θ)√(r² - 2r cos θ + 1) = 20

Now, let's apply the same transformation to Equation 2:

y√((x - 1)² + y²) = 20

Substituting x and y with their polar equivalents, we get:

(r sin θ)√((r cos θ - 1)² + (r sin θ)²) = 20

Following the same simplification steps as before, we arrive at:

(r sin θ)√(r² - 2r cos θ + 1) = 20

We now have our system of equations in polar coordinates:

Equation 3: (r cos θ)√(r² - 2r cos θ + 1) = 20

Equation 4: (r sin θ)√(r² - 2r cos θ + 1) = 20

Notice that the square root term is identical in both equations. This is a crucial observation! It allows us to relate r and θ directly. Since both equations equal 20, we can equate their left-hand sides:

(r cos θ)√(r² - 2r cos θ + 1) = (r sin θ)√(r² - 2r cos θ + 1)

Now, we have a single equation relating r and θ. To simplify this further, we can divide both sides by the square root term (assuming it's not zero). We'll address the case where the square root term is zero later. Dividing both sides, we get:

r cos θ = r sin θ

We can divide both sides by r (assuming r is not zero). If r were zero, then z would be zero, which doesn't satisfy the original equation. So, we have:

cos θ = sin θ

This equation tells us that the cosine and sine of θ are equal. This occurs at specific angles in the unit circle. Specifically, it happens when θ = π/4 + kπ, where k is an integer. However, since we are working with complex numbers, we only need to consider the principal values of θ within the range [0, 2π). This gives us two possible values for θ: π/4 and 5π/4.

Now, let's delve into the geometric interpretation of these angles. The angle θ = π/4 corresponds to the line y = x in the xy-plane, while the angle θ = 5π/4 corresponds to the line y = -x. These are precisely the cases we identified earlier when we analyzed x² - y² = 0. This confirms the consistency of our approach.

Proving the Uniqueness of the Root

We've made significant progress in understanding the possible solutions. We've transformed the complex equation into a real system, utilized polar coordinates, and identified potential angles for our solution. Now, the final piece of the puzzle is to demonstrate that there exists exactly one solution. To do this, we'll substitute our values of θ back into our polar equations and solve for r. Let's start with θ = π/4.

Substituting θ = π/4 into Equation 3, we get:

(r cos(π/4))√(r² - 2r cos(π/4) + 1) = 20

Recall that cos(π/4) = √2 / 2. Substituting this, we have:

(r(√2 / 2))√(r² - 2r(√2 / 2) + 1) = 20

Simplifying, we get:

(r√2 / 2)√(r² - r√2 + 1) = 20

To make this equation easier to handle, let's multiply both sides by 2/√2, which simplifies to √2:

r√(r² - r√2 + 1) = 20√2

Now, let's square both sides to eliminate the square root:

r²(r² - r√2 + 1) = 800

Expanding this, we obtain a quartic equation in r:

r⁴ - r³√2 + r² - 800 = 0

This quartic equation might seem daunting, but let's analyze it carefully. We're looking for positive real roots for r, as r represents the magnitude of a complex number. Let's define a function f(r) = r⁴ - r³√2 + r² - 800. To show that there's exactly one positive real root, we can analyze the derivative of f(r).

Let's find the derivative, f'(r):

f'(r) = 4r³ - 3r²√2 + 2r

We can factor out an r from this expression:

f'(r) = r(4r² - 3r√2 + 2)

Now, let's analyze the quadratic term 4r² - 3r√2 + 2. We can calculate its discriminant (Δ) to determine the nature of its roots:

Δ = b² - 4ac = (-3√2)² - 4(4)(2) = 18 - 32 = -14

Since the discriminant is negative, the quadratic term has no real roots. This means that the quadratic term is always positive. Therefore, the sign of f'(r) is determined by the sign of r. For positive r, f'(r) is positive, which means that f(r) is strictly increasing for positive r. This is a crucial observation!

Since f(r) is strictly increasing for positive r, it can cross the x-axis (i.e., have a root) at most once. Also, notice that f(0) = -800, which is negative. As r approaches infinity, f(r) approaches infinity, which is positive. By the Intermediate Value Theorem, there must be at least one positive real root. Since f(r) is strictly increasing, there can be only one such root. This proves that there exists a unique positive real value for r when θ = π/4.

Now, let's consider the case when θ = 5π/4. Substituting this into Equation 3, we get:

(r cos(5π/4))√(r² - 2r cos(5π/4) + 1) = 20

Recall that cos(5π/4) = -√2 / 2. Substituting this, we have:

(r(-√2 / 2))√(r² - 2r(-√2 / 2) + 1) = 20

Simplifying, we get:

(-r√2 / 2)√(r² + r√2 + 1) = 20

The left-hand side of this equation is negative for positive r, while the right-hand side is positive. This means there are no positive real solutions for r when θ = 5π/4. Therefore, this case does not yield any valid solutions.

We've exhaustively analyzed all possible cases and demonstrated that there exists exactly one solution for r when θ = π/4. This, in turn, implies that there is exactly one solution for z in the complex plane. We've successfully proven the uniqueness of the root!

Conclusion

Wow, guys! We've navigated a complex landscape of equations and concepts to arrive at a compelling conclusion. We've shown that the equation z|z-1| = 20 + 20i possesses a single, unique solution in the complex plane. This journey involved transforming the equation into a real system, leveraging the power of polar coordinates, analyzing the behavior of functions, and ultimately proving the uniqueness of the root. This problem serves as a testament to the beauty and interconnectedness of mathematical ideas, highlighting how different techniques can be combined to unravel complex challenges. Keep exploring, keep questioning, and keep the mathematical spirit alive!