Question

Solve for all possible values of the real numbersand in the following equations ${\left(x+iy\right)}^{\mathbf{2}}\mathbf{=}{\left(x-iy\right)}^{\mathbf{2}}$.

Short answer

When x is 0 , y is a real number, and when y is 0 ,x is any real number.

Step-by-step solution

Given information

It is given that${\left(x+iy\right)}^2={\left(x-iy\right)}^2$

Definition of a complex number

Every complex number can be described as:

z=a+bi

Whereaandbare both real numbers, z is the complex number, and i is known as iota, which makes z a complex number.

Begin by stating the given information

Evaluate the left- and right-hand sides of the equation:

$$\begin{gathered} {\left(x+iy\right)}^2=x^2+2xyi-y^2 \\ {\left(x-iy\right)}^2=x^2-2xyi-y^2 \end{gathered}$$

The equation can be written as:

$$\begin{gathered} x^2+2xyi-y^2=x^2-2xyi-y^2 \\ 2xyi+2xyi=0 \end{gathered}$$

Equate like terms

Equate like terms from both sides:

$$\begin{gathered} 4xyi=0 \\ 4xy=0 \end{gathered}$$

Thus, when x is 0, y is a real number, and when y is 0, x is any real number.