Solve (z+1)^2=z^2+2z+1

$$( z + 1 ) ^ { 2 } = z ^ { 2 } + 2 z + 1$$

Solve for z (complex solution)

$$z\in \mathrm{C}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z+1\right)^{2}.

$$z^{2}+2z+1=z^{2}+2z+1$$

Subtract z^{2} from both sides.

$$z^{2}+2z+1-z^{2}=2z+1$$

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

$$2z+1=2z+1$$

Subtract 2z from both sides.

$$2z+1-2z=1$$

Combine 2z and -2z to get 0.

$$1=1$$

Compare 1 and 1.

$$\text{true}$$

This is true for any z.

$$z\in \mathrm{C}$$

Solve for z

$$z\in \mathrm{R}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z+1\right)^{2}.

$$z^{2}+2z+1=z^{2}+2z+1$$

Subtract z^{2} from both sides.

$$z^{2}+2z+1-z^{2}=2z+1$$

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

$$2z+1=2z+1$$

Subtract 2z from both sides.

$$2z+1-2z=1$$

Combine 2z and -2z to get 0.

$$1=1$$

Compare 1 and 1.

$$\text{true}$$

This is true for any z.

$$z\in \mathrm{R}$$