Solve (-1+i)(2+i)/-1

$$\frac { ( - 1 + i ) ( 2 + i ) } { - 1 }$$

Evaluate

$$3-i$$

Solution Steps

Multiply complex numbers -1+i and 2+i like you multiply binomials.

$$\frac{-2-i+2i+i^{2}}{-1}$$

By definition, i^{2} is -1.

$$\frac{-2-i+2i-1}{-1}$$

Combine the real and imaginary parts in -2-i+2i-1.

$$\frac{-2-1+\left(-1+2\right)i}{-1}$$

Do the additions in -2-1+\left(-1+2\right)i.

$$\frac{-3+i}{-1}$$

Divide -3+i by -1 to get 3-i.

$$3-i$$

Real Part

$$3$$

Solution Steps

Multiply complex numbers -1+i and 2+i like you multiply binomials.

$$Re(\frac{-2-i+2i+i^{2}}{-1})$$

By definition, i^{2} is -1.

$$Re(\frac{-2-i+2i-1}{-1})$$

Combine the real and imaginary parts in -2-i+2i-1.

$$Re(\frac{-2-1+\left(-1+2\right)i}{-1})$$

Do the additions in -2-1+\left(-1+2\right)i.

$$Re(\frac{-3+i}{-1})$$

Divide -3+i by -1 to get 3-i.

$$Re(3-i)$$

The real part of 3-i is 3.

$$3$$