Solving Complex Number Equations: A Step-by-Step Guide

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Solving Complex Number Equations: A Step-by-Step Guide

Hey there, math enthusiasts! Let's dive into the fascinating world of complex numbers and tackle some intriguing equations. We'll break down the problems step by step, making sure everyone can follow along. This guide is all about helping you understand how to solve equations involving complex numbers, ensuring you feel confident in your abilities. So, buckle up, and let's get started!

Finding Complex Numbers: Unveiling Solutions to Quadratic Equations

Part a) z² = (1 - i) / (1 + i)

Alright, guys, let's start with this equation. We need to find the complex number z that satisfies z² = (1 - i) / (1 + i). The first step here is to simplify the right-hand side of the equation. To do that, we'll multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of 1 + i is 1 - i. So, let's do it:

  • (1 - i) / (1 + i) = ((1 - i) * (1 - i)) / ((1 + i) * (1 - i))

Now, let's multiply out the numerator and the denominator separately.

  • Numerator: (1 - i) * (1 - i) = 1 - i - i + i² = 1 - 2i - 1 = -2i (since i² = -1).
  • Denominator: (1 + i) * (1 - i) = 1 - i + i - i² = 1 - (-1) = 2.

So, our equation simplifies to:

  • z² = -2i / 2 = -i

Now we have z² = -i. To find z, we can express -i in polar form. The magnitude of -i is 1, and the argument is -π/2 (or 3π/2). Therefore, -i = cos(-π/2) + i sin(-π/2). To find z, we take the square root. Remember, there will be two solutions, as squaring a complex number can result in two possibilities.

  • z = √1 * (cos((-π/2 + 2kπ) / 2) + i sin((-π/2 + 2kπ) / 2)), where k = 0, 1.

For k = 0: z₁ = cos(-π/4) + i sin(-π/4) = √2/2 - i√2/2 For k = 1: z₂ = cos(3π/4) + i sin(3π/4) = -√2/2 + i√2/2.

So, the solutions for part a) are z₁ = √2/2 - i√2/2 and z₂ = -√2/2 + i√2/2. We've successfully navigated our first challenge!

Part b) z² = (-2 + 4i) / (2 + i)

Let's get cracking on this one. We need to solve z² = (-2 + 4i) / (2 + i). Similar to the previous problem, we'll start by simplifying the right-hand side by multiplying the numerator and denominator by the conjugate of the denominator, which is 2 - i.

  • (-2 + 4i) / (2 + i) = ((-2 + 4i) * (2 - i)) / ((2 + i) * (2 - i))

Now, let's multiply out the numerator and the denominator separately.

  • Numerator: (-2 + 4i) * (2 - i) = -4 + 2i + 8i - 4i² = -4 + 10i + 4 = 10i (since i² = -1).
  • Denominator: (2 + i) * (2 - i) = 4 - 2i + 2i - i² = 4 - (-1) = 5.

So, our equation becomes:

  • z² = 10i / 5 = 2i

Now we have z² = 2i. Expressing 2i in polar form: The magnitude is 2, and the argument is π/2. Therefore, 2i = 2(cos(π/2) + i sin(π/2)). To find z, we take the square root. Again, we'll have two solutions.

  • z = √2 * (cos((π/2 + 2kπ) / 2) + i sin((π/2 + 2kπ) / 2)), where k = 0, 1.

For k = 0: z₁ = √2 * (cos(π/4) + i sin(π/4)) = √2 * (√2/2 + i√2/2) = 1 + i. For k = 1: z₂ = √2 * (cos(5π/4) + i sin(5π/4)) = √2 * (-√2/2 - i√2/2) = -1 - i.

Hence, the solutions for part b) are z₁ = 1 + i and z₂ = -1 - i. You're doing great; keep it up!

Part c) z̄ = z²

For part c), we have z̄ = z². This one is a bit different. Let's express z in its general form: z = x + yi, where x and y are real numbers. Then, z̄ = x - yi. Substituting these into the equation, we get:

  • x - yi = (x + yi)²

Expanding the right side:

  • x - yi = x² + 2xyi + (yi)² = x² + 2xyi - y²

Now, we can equate the real and imaginary parts:

  • Real parts: x = x² - y²
  • Imaginary parts: -y = 2xy

From the imaginary part equation, we have two possibilities:

  1. y = 0: If y = 0, then substituting into the real part equation, we get x = x², which gives us x² - x = 0, or x(x - 1) = 0. So, x = 0 or x = 1. This gives us two solutions: z = 0 and z = 1.
  2. x = -1/2: If y ≠ 0, we can divide both sides of -y = 2xy by 2y (since y ≠ 0), resulting in x = -1/2. Substituting x = -1/2 into the real part equation: -1/2 = (-1/2)² - y², -1/2 = 1/4 - y², y² = 1/4 + 1/2 = 3/4. Therefore, y = ±√3/2. This gives us two additional solutions: z = -1/2 + i√3/2 and z = -1/2 - i√3/2.

So, the solutions for part c) are z = 0, z = 1, z = -1/2 + i√3/2, and z = -1/2 - i√3/2. Fantastic work!

Solving Complex Number Equations: Unraveling More Complexities

Part a) z / (1 + z) + z̄ / (1 + z̄) = 1

Let's get stuck into this equation, shall we? This one looks a little more involved, but don't worry, we'll break it down. We'll again use z = x + yi and z̄ = x - yi. Substitute these into the equation:

  • (x + yi) / (1 + x + yi) + (x - yi) / (1 + x - yi) = 1

Now, this gets a little messy, but the key here is to rationalize the denominators. That means multiplying the numerator and denominator of each fraction by the conjugate of the denominator. It's a bit of a process, so let's focus on one fraction at a time.

For the first fraction, we multiply the numerator and denominator by (1 + x - yi).

  • ((x + yi) * (1 + x - yi)) / ((1 + x + yi) * (1 + x - yi))

  • Numerator: (x + yi)(1 + x - yi) = x + x² - xyi + yi + xyi - y²i² = x + x² + y² + yi.

  • Denominator: (1 + x + yi)(1 + x - yi) = (1 + x)² - (yi)² = (1 + x)² + y² = 1 + 2x + x² + y².

So the first fraction becomes: (x + x² + y² + yi) / (1 + 2x + x² + y²).

For the second fraction, multiply the numerator and denominator by (1 + x + yi).

  • ((x - yi) * (1 + x + yi)) / ((1 + x - yi) * (1 + x + yi))

  • Numerator: (x - yi)(1 + x + yi) = x + x² + xyi - yi - xyi - y²i² = x + x² + y² - yi.

  • Denominator: (1 + x - yi)(1 + x + yi) = (1 + x)² - (yi)² = (1 + x)² + y² = 1 + 2x + x² + y².

So the second fraction becomes: (x + x² + y² - yi) / (1 + 2x + x² + y²).

Now the equation is:

  • (x + x² + y² + yi) / (1 + 2x + x² + y²) + (x + x² + y² - yi) / (1 + 2x + x² + y²) = 1

Combine the fractions since they have the same denominator:

  • (2x + 2x² + 2y²) / (1 + 2x + x² + y²) = 1

Multiply both sides by the denominator:

  • 2x + 2x² + 2y² = 1 + 2x + x² + y²

Simplify the equation:

  • x² + y² = 1/2

This represents a circle centered at the origin with a radius of √(1/2) or √2/2. Therefore, the solution is all complex numbers z that lie on this circle, excluding z = -1 (since it makes the original equation undefined). You're demonstrating great problem-solving skills, keep it up!

Part b) (2 + i - z) / (1 - z) ∈ R

This one is a real head-scratcher, but we'll get through it. The condition (2 + i - z) / (1 - z) ∈ R means that the result of the division is a real number. Let's again use z = x + yi and substitute into the equation.

  • (2 + i - (x + yi)) / (1 - (x + yi))

Simplify:

  • ((2 - x) + i(1 - y)) / ((1 - x) - yi)

For this to be a real number, the imaginary part must be equal to 0, which means the fraction, when simplified, must not have an imaginary component. To simplify, we multiply the numerator and denominator by the conjugate of the denominator, which is (1 - x) + yi.

  • (((2 - x) + i(1 - y)) * ((1 - x) + yi)) / (((1 - x) - yi) * ((1 - x) + yi))

  • Numerator: (2 - x)(1 - x) + (2 - x)yi + i(1 - y)(1 - x) + i²y(1 - y) = (2 - 2x - x + x² - y + xy) + i(2y - xy + 1 - x - y + xy) = 2 - 3x + x² - y + 1i(y - x + xy - xy) = 2 - 3x + x² - y + i(y - x).

  • Denominator: (1 - x)² - (yi)² = (1 - x)² + y².

So the expression becomes:

  • (2 - 3x + x² - y + i(y - x)) / ((1 - x)² + y²)

For the entire expression to be a real number, the imaginary part of the numerator must be 0, which means (y - x) = 0, which implies y = x. Also, the denominator cannot be 0, so (1 - x)² + y² ≠ 0. Since y = x, this translates to (1 - x)² + x² ≠ 0. Expanding this: 1 - 2x + x² + x² ≠ 0, or 2x² - 2x + 1 ≠ 0. This quadratic has no real roots, so the condition (1 - x)² + y² ≠ 0 is always true.

Therefore, the solution is z = x + xi, where x is any real number except x = 1 (because if x = 1, the denominator of the original fraction would be 0). This represents a line, y = x, with a point z = 1 (or 1 + 0i) excluded. You're doing incredible work, well done!

Part c) Re((2i - z) / (i - z)) = 1

Last one, guys! Let's conquer this problem. Here, Re((2i - z) / (i - z)) = 1 means the real part of the expression is equal to 1. Substitute z = x + yi:

  • Re((2i - (x + yi)) / (i - (x + yi))) = 1

Simplify:

  • Re((-x + i(2 - y)) / (-x + i(1 - y))) = 1

Again, let's rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is -x - i(1 - y).

  • ((-x + i(2 - y)) * (-x - i(1 - y))) / ((-x + i(1 - y)) * (-x - i(1 - y)))

  • Numerator: x² + ix(1 - y) - ix(2 - y) - i²(2 - y)(1 - y) = x² + ix - ixy - 2ix + ixy + 2 - 3y + y² = x² + 2 + y² - 3y - ix.

  • Denominator: x² - i²(1 - y)² = x² + (1 - y)² = x² + 1 - 2y + y².

So the expression becomes:

  • (x² + 2 + y² - 3y - ix) / (x² + 1 - 2y + y²)

Therefore, the real part is:

  • (x² + y² - 3y + 2) / (x² + y² - 2y + 1) = 1

Multiply both sides by the denominator:

  • x² + y² - 3y + 2 = x² + y² - 2y + 1

Simplify:

  • -y = -1

So, y = 1. This represents a horizontal line where the imaginary part of z is always 1, with the exception of z = i (because if z = i, the original equation would be undefined). Congratulations on solving all of these problems!

Showing (1 + i)ⁿ + (1 - i)ⁿ

To show (1 + i)ⁿ + (1 - i)ⁿ for any n, we can use the polar form of complex numbers or the binomial theorem. Let's try the polar form first.

We know that 1 + i = √2(cos(π/4) + i sin(π/4)) and 1 - i = √2(cos(-π/4) + i sin(-π/4)). Using De Moivre's theorem:

(1 + i)ⁿ = (√2)ⁿ(cos(nπ/4) + i sin(nπ/4)) and (1 - i)ⁿ = (√2)ⁿ(cos(-nπ/4) + i sin(-nπ/4)).

Therefore,

(1 + i)ⁿ + (1 - i)ⁿ = (√2)ⁿ(cos(nπ/4) + i sin(nπ/4)) + (√2)ⁿ(cos(-nπ/4) + i sin(-nπ/4)).

Since cos(-x) = cos(x) and sin(-x) = -sin(x), the equation simplifies to:

(√2)ⁿ(cos(nπ/4) + i sin(nπ/4) + cos(nπ/4) - i sin(nπ/4)) = (√2)ⁿ * 2 * cos(nπ/4).

So, (1 + i)ⁿ + (1 - i)ⁿ = 2^(n/2 + 1) * cos(nπ/4). This expression is always a real number, as expected. This also demonstrates that we can solve complex number problems with different approaches, enhancing our understanding and skill. Keep practicing, and you'll become a complex number whiz! Great work today! You have successfully solved complex equations and gained a deeper understanding of complex number concepts.