Question

Solve for all possible values of the real numbers \(x\) and \(y\) in the following equations. $$(x+i y)^{2}=2 i x$$

Short answer

The solutions are \( x = 0, y = 0 \), \( x = 1, y = 1 \), \( x = -1, y = 1 \).

Step-by-step solution

Expand the left side

First, expand the left side of the equation \ \ (x + i y)^{2}. \ Use the distributive property to get \ \[ (x + i y)^{2} = x^{2} + 2 i x y - y^{2} \] \

Compare real and imaginary parts

Write the expanded form next to the right-hand side \ \[ x^{2} + 2ixy - y^{2} = 2ix \] Equate the real parts and the imaginary parts separately. \ Real part: \ \( x^2 - y^2 = 0 \) \ Imaginary part: \ \( 2xy = 2x \).

Solve the real part

Solve the equation for the real parts \ \( x^2 - y^2 = 0 \). \ This can be rewritten as \( x^2 = y^2 \). \ Taking the square root of both sides, we get \ \( x = y \) or \( x = -y \).

Solve the imaginary part

Now solve the equation for the imaginary parts \ \ \( 2xy = 2x \). \ Divide both sides by 2: \ \ \( xy = x \). \ This can be written as \ \ \( x(y - 1) = 0 \). \ Thus, \( x = 0 \) or \( y = 1 \).

Combine solutions

Combine the solutions from Steps 3 and 4. If \( x = 0 \), we substitute into \(x = y\) or \(x = -y\): \ \ \( x = 0, y = 0 \) \ If \(y = 1\), substitute into \( x = y \) or \( x = -y \): \ \( x = 1, y = 1 \) or \( x = -1, y = 1 \).

Key concepts

Complex Numbers

Complex numbers extend the concept of one-dimensional real numbers to two-dimensional numbers. A complex number is written in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part multiplied by \(i\) (the imaginary unit, which is defined as \(i^2 = -1\)). Complex numbers allow us to solve equations that have no real solutions and are crucial in fields such as engineering and physics.
In our exercise, \(x + i y\) is a complex number where \[x\] is the real part and \[y\] is the imaginary part.

Real and Imaginary Parts

In a complex number \(a + bi\), the term \(a\) is the real part and \(bi\) is the imaginary part. To solve complex number equations, we often need to separate these parts.
In our exercise, after expanding \( (x + i y)^{2} \), we compare the real and imaginary parts separately:
\(x^2 - y^2 = 0\) is the equation for the real part and \(2ixy = 2ix\) is for the imaginary part. By treating these separately, we can solve for the individual variables \(x\) and \(y\).

Algebraic Solutions

Algebraic solutions involve manipulating algebraic expressions to solve for unknown variables. For the real part, \(x^2 - y^2 = 0\) signifies that either \(x = y\) or \(x = -y\). For the imaginary part, we have \(2xy = 2x\). By dividing both sides by 2, we arrive at \(xy = x\), which simplifies to \[x(y-1) = 0\]. This implies either \(x=0\) or \(y=1\). Combining these provides the complete set of solutions.

Quadratic Equations

Quadratic equations take the form \[ax^2 + bx + c = 0\]. The quadratic equation in our exercise is hidden in the form \((x + i y)^2 = 2ix\). Upon expanding and isolating real and imaginary parts, we solve the resulting simpler quadratic equations.
In particular, solving \(x^2 - y^2 = 0\) gives \(x = y\) or \(x = -y\) as it forces us to find specific values of \(x\) and \(y\) that satisfy the original complex equation.