Solve (x+iy)^2=x^2-y^2+2ixy

$$( x + i y ) ^ { 2 } = x ^ { 2 } - y ^ { 2 } + 2 i x y$$

Solve for x

$$x\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+iy\right)^{2}.

$$x^{2}+2ixy-y^{2}=x^{2}-y^{2}+2ixy$$

Subtract x^{2} from both sides.

$$x^{2}+2ixy-y^{2}-x^{2}=-y^{2}+2ixy$$

Combine x^{2} and -x^{2} to get 0.

$$2ixy-y^{2}=-y^{2}+2ixy$$

Subtract 2ixy from both sides.

$$2ixy-y^{2}-2ixy=-y^{2}$$

Combine 2ixy and -2ixy to get 0.

$$-y^{2}=-y^{2}$$

Cancel out -1 on both sides.

$$y^{2}=y^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for y

$$y\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+iy\right)^{2}.

$$x^{2}+2ixy-y^{2}=x^{2}-y^{2}+2ixy$$

Add y^{2} to both sides.

$$x^{2}+2ixy-y^{2}+y^{2}=x^{2}+2ixy$$

Combine -y^{2} and y^{2} to get 0.

$$x^{2}+2ixy=x^{2}+2ixy$$

Subtract 2ixy from both sides.

$$x^{2}+2ixy-2ixy=x^{2}$$

Combine 2ixy and -2ixy to get 0.

$$x^{2}=x^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any y.

$$y\in \mathrm{C}$$