Unique Root: Solving Z|z-1| = 20 + 20i In Complex Numbers

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of complex numbers to tackle a rather intriguing problem. We're going to explore the equation z|z - 1| = 20 + 20i and prove that it has exactly one solution in the complex plane. Buckle up, because this is going to be a fun ride!

The Problem at Hand

Let's get straight to the heart of the matter. Our mission, should we choose to accept it (and we definitely do!), is to demonstrate that the equation:

z|z - 1| = 20 + 20i

has a single, unique solution within the realm of complex numbers. This might seem a bit daunting at first, but don't worry, we'll break it down step by step and conquer it together.

Laying the Groundwork: Transforming to a Real System

The initial step in our journey involves a clever transformation. We're going to express the complex number z in terms of its real and imaginary components. Let's represent z as:

z = x + iy

where x and y are real numbers. This substitution is a cornerstone of complex number manipulation, allowing us to bridge the gap between the abstract world of complex numbers and the more familiar territory of real numbers.

Now, let's substitute this representation of z into our original equation. This gives us:

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

This might look a bit messy, but it's progress! We've successfully replaced the single complex variable z with two real variables, x and y. This sets the stage for us to work with the magnitude of the complex number and eventually separate the real and imaginary parts of the equation.

Delving into Magnitudes: A Crucial Simplification

The next step in our adventure involves dealing with the absolute value, or magnitude, in the equation. Recall that the magnitude of a complex number a + bi is given by √(a² + b²). Applying this to our equation, we get:

(x + iy)√((x - 1)² + y²) = 20 + 20i

This is a significant simplification! We've replaced the absolute value notation with a square root expression, which is much easier to handle algebraically. Now, we have an equation that explicitly involves the real and imaginary components x and y, paving the way for further manipulation.

At this point, it's crucial to pause and appreciate what we've accomplished. We've successfully transformed the original complex equation into a form that is more amenable to analysis using real number techniques. This is a common strategy in complex number problems – breaking down complex expressions into their real and imaginary parts to leverage the power of real algebra and calculus.

Separating Reality from Imagination: Equating Real and Imaginary Parts

Now comes the pivotal moment where we separate the real and imaginary parts of our equation. This is a fundamental technique when dealing with complex equations, as it allows us to convert a single complex equation into a system of two real equations. Let's distribute the (x + iy) term:

x√((x - 1)² + y²) + iy√((x - 1)² + y²) = 20 + 20i

For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. Therefore, we can split the above equation into two separate real equations:

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

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

Voila! We've successfully transformed our single complex equation into a system of two real equations. This is a major breakthrough, as we can now employ the tools of real algebra to solve for x and y. The problem has become significantly more tractable, and we're well on our way to finding the unique solution.

Unveiling the Relationship: A Key Deduction

Observing Equations 1 and 2, a crucial relationship emerges. Notice that both equations share the same square root term. Dividing Equation 2 by Equation 1 (assuming x ≠ 0), we get:

y / x = 1

This elegantly simple equation tells us that y = x. This is a powerful deduction, as it allows us to reduce the number of variables we need to solve for. Instead of dealing with two independent variables, x and y, we now know that they are equal, effectively reducing our system to a single-variable problem.

However, we need to be cautious about the assumption we made: x ≠ 0. Let's consider the case where x = 0 separately. If x = 0, then from Equation 1, we have:

0 = 20

This is clearly a contradiction, meaning that x cannot be 0. Therefore, our division was valid, and the relationship y = x holds true.

Solving the Puzzle: A Single Equation to Conquer

Now that we know y = x, we can substitute this into either Equation 1 or Equation 2 to obtain an equation in a single variable. Let's substitute y = x into Equation 1:

x√((x - 1)² + x²) = 20

This equation, while still a bit intimidating, is a significant simplification compared to our original problem. We've gone from a complex equation to a single real equation in one variable. This is the home stretch! Let's simplify the equation further:

x√(x² - 2x + 1 + x²) = 20

x√(2x² - 2x + 1) = 20

Squaring Both Sides: Taming the Square Root

To get rid of the square root, we'll square both sides of the equation. This is a common technique, but we need to be mindful that squaring can sometimes introduce extraneous solutions. We'll need to check our final answer to make sure it satisfies the original equation.

Squaring both sides, we get:

x²(2x² - 2x + 1) = 400

Expanding the left side, we obtain a quartic equation:

2x⁴ - 2x³ + x² = 400

2x⁴ - 2x³ + x² - 400 = 0

The Quartic Challenge: Finding the Real Root

We've arrived at a quartic equation, which might seem daunting. Solving quartic equations analytically can be quite complex, but in this case, we're in luck! We only need to show that there is one real solution. We don't necessarily need to find the exact solution, although that would be a bonus.

Let's consider the function:

f(x) = 2x⁴ - 2x³ + x² - 400

To show that this equation has exactly one real root, we can use a combination of calculus and observation. First, let's analyze the behavior of the function as x approaches positive and negative infinity.

As x approaches positive or negative infinity, the term 2x⁴ dominates the other terms, so f(x) approaches positive infinity. This tells us that the graph of the function rises on both ends.

Now, let's evaluate the function at a few strategic points:

• f(0) = -400
• f(4) = 2(256) - 2(64) + 16 - 400 = 512 - 128 + 16 - 400 = 0

We've struck gold! We found that f(4) = 0, which means that x = 4 is a root of the equation. This is a crucial step, as it gives us a concrete solution to work with.

Proof of Uniqueness: A Calculus Perspective

To prove that x = 4 is the only real solution, we can use calculus. Let's find the derivative of f(x):

f'(x) = 8x³ - 6x² + 2x

We can factor out a 2x:

f'(x) = 2x(4x² - 3x + 1)

Now, let's analyze the quadratic term 4x² - 3x + 1. The discriminant of this quadratic is:

Δ = (-3)² - 4(4)(1) = 9 - 16 = -7

Since the discriminant is negative, the quadratic has no real roots. This means that the only real root of f'(x) = 0 is x = 0.

Now, let's analyze the sign of f'(x):

• For x < 0, f'(x) < 0 (since 2x is negative and the quadratic is always positive)
• For 0 < x, f'(x) > 0 (since 2x is positive and the quadratic is always positive)

This tells us that f(x) is decreasing for x < 0 and increasing for x > 0. Therefore, f(x) has a global minimum at x = 0. Since f(0) = -400 and f(x) approaches positive infinity as x approaches infinity, there can be only one real root for x > 0. We already found that root: x = 4.

Therefore, we've proven that the quartic equation has exactly one real root, which is x = 4.

The Grand Finale: Constructing the Complex Solution

We've found that x = 4, and since y = x, we also have y = 4. Therefore, the unique complex solution is:

z = x + iy = 4 + 4i

To be absolutely sure, let's plug this solution back into the original equation:

(4 + 4i)|(4 + 4i) - 1| = (4 + 4i)|3 + 4i| = (4 + 4i)√(3² + 4²) = (4 + 4i) * 5 = 20 + 20i

It checks out! We've successfully found the unique solution to the equation.

Conclusion: A Triumph of Mathematical Deduction

We've embarked on a mathematical journey, starting with a complex equation and, through a series of clever transformations and deductions, arriving at a single, unique solution. We've shown that the equation z|z - 1| = 20 + 20i has exactly one solution in the complex plane, which is z = 4 + 4i. This problem beautifully illustrates the power of complex number techniques, algebraic manipulation, and calculus in solving seemingly intractable problems. Give yourself a pat on the back, guys – we nailed it!

This exploration highlights the elegance and interconnectedness of mathematics. By combining techniques from algebra, calculus, and complex number theory, we were able to unravel the mystery and arrive at a satisfying conclusion. Keep exploring, keep questioning, and keep the mathematical flame burning bright!