Question

Solve for all possible values of the real numbers \(x\) and \(y\) in the following equations. $$\frac{x+i y}{x-i y}=-i$$

Short answer

The solutions are all pairs where \(x = - y\).

Step-by-step solution

Express in Complex Form

Rewrite the equation \(\frac{x + i y}{x - i y} = -i\) using the complex number format \(a + bi\). This will help in simplifying the equation.

Multiply by the Conjugate

Multiply both the numerator and denominator by the conjugate of the denominator. Hence, multiply by \(x + i y\): \(\frac{(x+i y)(x+i y)}{(x-i y)(x+i y)}\).

Simplify the Numerator and Denominator

Simplify the expressions: the numerator will be \((x+i y)^2\) and the denominator will be \((x^2 + y^2)\) (since \(x - iy\) and \(x + iy\) are conjugates).

Substitute and Equate Real and Imaginary Parts

This results in \((x^2 - y^2 + 2ixy)/(x^2 + y^2) = -i\). Equate the real and imaginary parts separately: the real part must be 0 and imaginary part must be \(-1\).

Set Real Part to Zero

Equate the real part to zero: \(x^2 - y^2 = 0\). This gives \(x^2 = y^2\), which simplifies to \x = y\ or \-y\.

Set Imaginary Part to -1

Equate the imaginary part to \(-1\): \frac{2xy}{x^2 + y^2} = -1\. Substitute \(x = y\) and \x = -y\ into the equation.

Simplify Further for Solutions

For \(x = y\) we get \frac{2x^2}{2x^2} = -1\ which is not possible. For \(x = -y\) we get \frac{-2x^2}{2x^2} = -1\ which is valid. Thus, \[x = -y\].

Conclusion

The solutions to the given equation are all the pairs where \[x = - y\].

Key concepts

complex numbers

Let's begin by understanding what complex numbers are. A complex number is a number that can be expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit that satisfies the equation \(i^2 = −1\).
Complex numbers are an extension of the real numbers and include both a real part (\(a\)) and an imaginary part (\(bi\)). Here are some key points to remember:

• The complex number \(a + bi\) can be visualized as a point in the complex plane.
• The real part, \(a\), aligns with the x-axis,
and
• the imaginary part, \(bi\), aligns with the y-axis.

Conjugate of a complex number is another useful concept. For a complex number \(a + bi\), the conjugate is \(a - bi\). Complex conjugates are useful in complex fraction manipulations as seen in the given problem.

algebra

Algebra plays a key role in solving complex number problems. Here, basic algebraic operations help us manipulate and simplify the equation.

• First, rewrite the given equation \(\frac{x+i y}{x-i y} = -i\) in a more manageable form.
• Then, multiply the numerator and the denominator by the conjugate of the denominator, \(x + iy\). This step utilizes the property of conjugates to simplify the complex fraction.
• Next, algebraic manipulation involves expanding the numerator \((x + iy)(x + iy)\), which results in \(x^2 + 2ixy - y^2\).

The denominator becomes \(x^2 + y^2\) because the product of conjugates (\(x - iy\) and \(x + iy\)) equals \(x^2 + y^2\).
Equate the real and imaginary parts separately to solve for \(x\) and \(y\). The equation simplifies to \(\frac{x^2 - y^2 + 2ixy}{x^2 + y^2} = -i\). Setting the real part to 0 helps find one relation, and setting the imaginary part to -1 yields another condition.

mathematical methods

Solving equations with complex numbers often involves a variety of mathematical methods. In this problem, the key steps were:

• Using the concept of complex conjugation to facilitate simplification.
• Expressing the given equation in a suitable form for algebraic manipulation.
• Separating the real and imaginary parts of the resulting equation.

To solve \(\frac{x + iy}{x - iy} = -i\), multiply by the conjugate: \(\frac{(x + iy)^2}{x^2 + y^2}\). When expressed in its real and imaginary parts, it leads to \(x^2 - y^2 + 2ixy\) in the numerator and \(x^2 + y^2\) in the denominator.

• Setting the real part to zero was a crucial step, leading to \(x^2 = y^2\). This gives \(x = y\) or \(x = -y\).
• Setting the imaginary part yielded \(2xy\) over \(x^2 + y^2\) to -1.

Upon substituting the possible values, only \(x = -y\) satisfies the initial equation, demonstrating the valid approach to solving such equations.