Math Problem Statement

If x + iy = (a - bi) / (a + bi) and x - iy = (a + bi) / (a - bi), show that x² + y² = 1 in both cases.

Solution

It looks like you've uploaded an image with a mathematical problem. The problem appears to involve showing that some conditions are satisfied in both cases.

Let me summarize what I can see in the image:

You are given the following conditions:

  • x+iy=abia+bix + iy = \frac{a - bi}{a + bi}
  • xiy=a+biabix - iy = \frac{a + bi}{a - bi}

The goal is to show that x2+y2=1x^2 + y^2 = 1 in both cases.

To begin, we will need to simplify the complex fractions and use algebra to prove that x2+y2=1x^2 + y^2 = 1.

Let's break this down:

Step 1: Simplify the expression for x+iyx + iy

Starting with x+iy=abia+bix + iy = \frac{a - bi}{a + bi}, we multiply both the numerator and denominator by the conjugate of the denominator abia - bi to simplify:

x+iy=(abi)(abi)(a+bi)(abi)x + iy = \frac{(a - bi)(a - bi)}{(a + bi)(a - bi)}

We know that (a+bi)(abi)=a2+b2(a + bi)(a - bi) = a^2 + b^2. Now, expand the numerator:

(abi)(abi)=a22abi+b2i2=a22abib2(a - bi)(a - bi) = a^2 - 2abi + b^2i^2 = a^2 - 2abi - b^2

This gives:

x+iy=a2b22abia2+b2x + iy = \frac{a^2 - b^2 - 2abi}{a^2 + b^2}

Now, separate the real and imaginary parts:

x=a2b2a2+b2,y=2aba2+b2x = \frac{a^2 - b^2}{a^2 + b^2}, \quad y = \frac{-2ab}{a^2 + b^2}

Step 2: Simplify the expression for xiyx - iy

Next, simplify the second equation xiy=a+biabix - iy = \frac{a + bi}{a - bi} by following a similar process:

xiy=(a+bi)(a+bi)(abi)(a+bi)=a2b2+2abia2+b2x - iy = \frac{(a + bi)(a + bi)}{(a - bi)(a + bi)} = \frac{a^2 - b^2 + 2abi}{a^2 + b^2}

Again, separating the real and imaginary parts:

x=a2b2a2+b2,y=2aba2+b2x = \frac{a^2 - b^2}{a^2 + b^2}, \quad y = \frac{2ab}{a^2 + b^2}

Step 3: Show that x2+y2=1x^2 + y^2 = 1

Now, calculate x2+y2x^2 + y^2:

For both cases, we have the same expressions for xx and yy, except for the sign of yy. In either case:

x2+y2=(a2b2a2+b2)2+(2aba2+b2)2x^2 + y^2 = \left( \frac{a^2 - b^2}{a^2 + b^2} \right)^2 + \left( \frac{2ab}{a^2 + b^2} \right)^2

Simplifying:

x2+y2=(a2b2)2(a2+b2)2+4a2b2(a2+b2)2x^2 + y^2 = \frac{(a^2 - b^2)^2}{(a^2 + b^2)^2} + \frac{4a^2b^2}{(a^2 + b^2)^2}

x2+y2=(a2b2)2+4a2b2(a2+b2)2x^2 + y^2 = \frac{(a^2 - b^2)^2 + 4a^2b^2}{(a^2 + b^2)^2}

Expanding the numerator:

(a2b2)2+4a2b2=a42a2b2+b4+4a2b2=a4+2a2b2+b4(a^2 - b^2)^2 + 4a^2b^2 = a^4 - 2a^2b^2 + b^4 + 4a^2b^2 = a^4 + 2a^2b^2 + b^4

x2+y2=(a2+b2)2(a2+b2)2=1x^2 + y^2 = \frac{(a^2 + b^2)^2}{(a^2 + b^2)^2} = 1

Thus, we have shown that x2+y2=1x^2 + y^2 = 1 in both cases.

Let me know if you need any further explanation or clarification!


Would you like more details on any specific part? Here are some related questions:

  1. What is the geometric interpretation of the equation x2+y2=1x^2 + y^2 = 1?
  2. How do complex conjugates help simplify complex expressions?
  3. What other algebraic techniques can be used to simplify complex fractions?
  4. How do the real and imaginary parts of complex numbers relate to their polar form?
  5. What are some other common problems where complex conjugates play a role?

Tip: Remember that multiplying a complex number by its conjugate eliminates the imaginary part, which is useful when simplifying complex fractions!

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Algebra
Conjugates
Fractions

Formulas

(a + bi)(a - bi) = a² + b²
x² + y² = 1
Multiplication of complex conjugates

Theorems

Properties of complex conjugates
Modulus of a complex number

Suitable Grade Level

Grades 10-12