Problem 46 Solve for all possible values of... [FREE SOLUTION]

Chapter 2: Problem 46

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

Expert verified

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 \[x = - y\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 $i^2 = 鈭�1$.
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,
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.

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91影视

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