Unique Solution Of Z|z-1| = 20 + 20i: A Deep Dive

Hey guys! Today, we're diving deep into a fascinating problem involving complex numbers. We're going to explore the equation z|z-1| = 20 + 20i and unravel why it has exactly one solution in the complex plane. Buckle up, because this journey will take us through the realms of complex numbers, algebraic manipulation, and a touch of geometric intuition!

Cracking the Code: Transforming the Complex Equation

Let's start by laying the groundwork. The core of our mission is to show that the equation z|z-1| = 20 + 20i possesses a lone solution within the vast expanse of complex numbers. To tackle this, our initial strategy involves transforming the complex equation into a more manageable real system. How do we do this? By expressing the complex number z in its familiar form: z = x + iy, where x and y are real numbers. This substitution is the key to unlocking the problem.

By representing z as x + iy, we can rewrite the equation in terms of real and imaginary components. This is a crucial step because it allows us to work with real numbers, which are often easier to manipulate algebraically. So, let's break it down:

• First, we substitute z with x + iy in the given equation:

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

• Next, we need to deal with the absolute value term. Remember that the magnitude (or absolute value) 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

• Now, we have a complex number on the left-hand side equal to another complex number on the right-hand side. For two complex numbers to be equal, both their real and imaginary parts must be equal. This principle allows us to separate the equation into two real equations.

  Let's express the square root term as r, where r = √[ (x-1)² + y² ]. Our equation now becomes:

  ( x + iy ) r = 20 + 20i

  Expanding the left side, we have:

  xr + iyr = 20 + 20i

• Equating the real and imaginary parts, we arrive at the following system of real equations:

  • xr = 20
  • yr = 20

Why is this transformation so important? Because now we've cleverly shifted the problem from the complex domain to the real domain. We now have a system of equations that we can solve using familiar algebraic techniques. It's like translating a sentence from a foreign language into our native tongue – suddenly, the meaning becomes clear! The transformation into a real system allows us to leverage our knowledge of real number manipulations and algebraic solutions, making the problem much more approachable. So, now that we have these two equations, we are closer to finding the unique solution of our original complex equation. This step is a classic example of how breaking down complex problems into simpler, manageable parts can lead us to the solution.

Unmasking the Solution: Diving into the Real System

Alright, we've successfully transformed our complex equation into a system of two real equations:

• xr = 20
• yr = 20

where r = √[ (x-1)² + y² ]. Now comes the exciting part – solving this system! Our aim here is to find the values of x and y that satisfy both equations simultaneously. Let's put on our detective hats and see how we can crack this case.

Notice something crucial: both xr and yr are equal to 20. This immediately tells us something significant about x and y. If two products are equal, and they share a common factor (r in this case), then the other factors must be related in a specific way. In our scenario, it implies that x must equal y. This is a key insight that simplifies our problem considerably.

So, let's set x = y. This substitution allows us to reduce the number of variables and makes the equations easier to handle. Now, let's rewrite r using this new information:

r = √[ (x-1)² + x² ]

We've replaced y with x in the expression for r. This is a powerful move because it brings us closer to having an equation with just one unknown variable. Now, we can substitute y with x in either of our original equations. Let's use xr = 20. This gives us:

x√[ (x-1)² + x² ] = 20

We now have a single equation with just x as the unknown! This is a major breakthrough. However, it's still a bit messy with that square root hanging around. The next logical step is to get rid of it. We can do this by squaring both sides of the equation. Squaring both sides is a common technique in algebra to eliminate square roots, but we need to be cautious: squaring can sometimes introduce extraneous solutions, so we'll need to check our final answers later.

Squaring both sides, we get:

x² [ (x-1)² + x² ] = 400

Now, we have a polynomial equation. Let's expand and simplify it:

x² [ x² - 2x + 1 + x² ] = 400

x² ( 2x² - 2x + 1 ) = 400

2x⁴ - 2x³ + x² = 400

2x⁴ - 2x³ + x² - 400 = 0

We've arrived at a quartic equation! Quartic equations can be tricky to solve, but don't worry, we're not going to tackle this directly just yet. We've made significant progress by reducing the original complex equation to a single quartic equation in a real variable. This is a testament to the power of algebraic manipulation. In the next section, we'll explore how to solve or, more importantly, show that this equation has only one relevant real root.

The Grand Finale: Proving Uniqueness and Finding the Root

We've journeyed through complex numbers, real systems, and now we stand before a quartic equation: 2x⁴ - 2x³ + x² - 400 = 0. Our mission is to demonstrate that this equation has only one positive real root, which will ultimately prove that our original complex equation has a unique solution. How do we achieve this? Let's bring in some powerful tools from calculus and analysis.

Consider the function f(x) = 2x⁴ - 2x³ + x² - 400. To understand the behavior of this function, especially the number and nature of its roots, we can analyze its derivative. The derivative will tell us about the function's increasing and decreasing intervals, which in turn helps us locate the roots.

Let's find the derivative of f(x):

f '(x) = 8x³ - 6x² + 2x = 2x(4x² - 3x + 1)

Now, let's analyze the critical points by setting f '(x) = 0:

2x(4x² - 3x + 1) = 0

This gives us x = 0 as one solution. The quadratic factor (4x² - 3x + 1) has a discriminant of (-3)² - 4(4)(1) = 9 - 16 = -7, which is negative. This means the quadratic has no real roots. Therefore, x = 0 is the only real root of the derivative. This tells us something crucial about the behavior of f(x). It implies that the function has only one critical point.

Now, let's investigate the sign of f '(x) for x > 0. Since 4x² - 3x + 1 has no real roots and its leading coefficient is positive, it is always positive. Thus, for x > 0, f '(x) = 2x(positive expression) > 0. This means that f(x) is strictly increasing for x > 0. This is a game-changer because a strictly increasing function can cross the x-axis (i.e., have a root) at most once.

Now let's evaluate f(0) = -400, which is negative. As f(x) is a quartic polynomial with a positive leading coefficient, it tends to +∞ as x tends to +∞. Therefore, there must be at least one positive real root since the function goes from negative to positive. Combining this with the fact that f(x) is strictly increasing for x > 0, we can definitively conclude that there is exactly one positive real root.

Let's call this unique positive real root x₀. Since x = y, we also have y₀ = x₀. Thus, we have a unique solution for x and y, and consequently, a unique solution for z in the complex plane. The corresponding unique solution for z is z = x₀ + ix₀.

In summary: We've shown that the quartic equation has only one positive real root by analyzing the derivative of the function. This, in turn, proves that the original complex equation z|z-1| = 20 + 20i has exactly one solution. We started by transforming the complex equation into a real system, then reduced it to a single quartic equation, and finally used calculus to analyze the roots of the quartic. What a journey!

Visualizing the Solution (Optional)

While we've rigorously proven the uniqueness of the solution, it's always nice to have a visual understanding. We could plot the function f(x) to see its behavior, or even attempt to plot the original equation in the complex plane using software like GeoGebra or Mathematica. This visual confirmation can provide an extra layer of confidence in our result.

Conclusion: A Triumph of Complex Analysis

Guys, we've conquered a challenging problem today! We've demonstrated that the equation z|z-1| = 20 + 20i has exactly one solution in the complex plane. This journey has showcased the power of complex number manipulation, algebraic techniques, and calculus in solving seemingly intricate problems. Remember, breaking down complex problems into smaller, manageable steps, and leveraging different mathematical tools, is often the key to success. Keep exploring, keep questioning, and keep solving!