Evaluate: (x-1-i) * (x-1+i)

Expression: $\left( x-1-i \right) \times \left( x-1+i \right)$

Distribute $x-1+i$ onto $x-1-i$

$x \times \left( x-1+i \right)-1\left( x-1+i \right)-i \times \left( x-1+i \right)$

Distribute $x$ through the parentheses

${x}^{2}-x+xi-1\left( x-1+i \right)-i \times \left( x-1+i \right)$

Distribute $-1$ through the parentheses

${x}^{2}-x+xi-x+1-i-i \times \left( x-1+i \right)$

Distribute $-i$ through the parentheses

${x}^{2}-x+xi-x+1-i-xi+i-{i}^{2}$

Since two opposites add up to $0$, remove them from the expression

${x}^{2}-x-x+1-i+i-{i}^{2}$

Since two opposites add up to $0$, remove them from the expression

${x}^{2}-x-x+1-{i}^{2}$

By definition ${i}^{2}=-1$

${x}^{2}-x-x+1-\left( -1 \right)$

When there is a $-$ in front of an expression in parentheses, change the sign of each term of the expression and remove the parentheses

${x}^{2}-x-x+1+1$

Collect like terms

${x}^{2}-2x+1+1$

Add the numbers

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