Solve left(x+zright)^2=x^2+y^2

$${ \left(x+z \right) }^{ 2 } = { x }^{ 2 } + { y }^{ 2 }$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{y^{2}}{2z}-\frac{z}{2}\text{, }&z\neq 0\\x\in \mathrm{C}\text{, }&z=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}+2xz+z^{2}=x^{2}+y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+2xz+z^{2}-x^{2}=y^{2}$$

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

$$2xz+z^{2}=y^{2}$$

Subtract z^{2} from both sides.

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

The equation is in standard form.

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

Divide both sides by 2z.

$$\frac{2zx}{2z}=\frac{\left(y-z\right)\left(y+z\right)}{2z}$$

Dividing by 2z undoes the multiplication by 2z.

$$x=\frac{\left(y-z\right)\left(y+z\right)}{2z}$$

Divide \left(y-z\right)\left(y+z\right) by 2z.

$$x=\frac{y^{2}}{2z}-\frac{z}{2}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{y^{2}}{2z}-\frac{z}{2}\text{, }&z\neq 0\\x\in \mathrm{R}\text{, }&z=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}+2xz+z^{2}=x^{2}+y^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+2xz+z^{2}-x^{2}=y^{2}$$

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

$$2xz+z^{2}=y^{2}$$

Subtract z^{2} from both sides.

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

The equation is in standard form.

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

Divide both sides by 2z.

$$\frac{2zx}{2z}=\frac{\left(y-z\right)\left(y+z\right)}{2z}$$

Dividing by 2z undoes the multiplication by 2z.

$$x=\frac{\left(y-z\right)\left(y+z\right)}{2z}$$

Divide \left(y-z\right)\left(y+z\right) by 2z.

$$x=\frac{y^{2}}{2z}-\frac{z}{2}$$

Solve for y (complex solution)

$$y=-\sqrt{z}\sqrt{2x+z}$$

$$y=\sqrt{z}\sqrt{2x+z}$$

Solve for y

$$y=\sqrt{z\left(2x+z\right)}$$

$$y=-\sqrt{z\left(2x+z\right)}\text{, }\left(z\leq 0\text{ or }x\geq -\frac{z}{2}\right)\text{ and }\left(z\geq 0\text{ or }x\leq -\frac{z}{2}\right)$$