Solve (3+2i)(1-i)

$$( 3 + 2 i ) ( 1 - i )$$

Evaluate

$$5-i$$

Solution Steps

Multiply complex numbers 3+2i and 1-i like you multiply binomials.

$$3\times 1+3\left(-i\right)+2i\times 1+2\left(-1\right)i^{2}$$

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

$$3\times 1+3\left(-i\right)+2i\times 1+2\left(-1\right)\left(-1\right)$$

Do the multiplications.

$$3-3i+2i+2$$

Combine the real and imaginary parts.

$$3+2+\left(-3+2\right)i$$

Do the additions.

$$5-i$$

Real Part

$$5$$

Solution Steps

Multiply complex numbers 3+2i and 1-i like you multiply binomials.

$$Re(3\times 1+3\left(-i\right)+2i\times 1+2\left(-1\right)i^{2})$$

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

$$Re(3\times 1+3\left(-i\right)+2i\times 1+2\left(-1\right)\left(-1\right))$$

Do the multiplications in 3\times 1+3\left(-i\right)+2i\times 1+2\left(-1\right)\left(-1\right).

$$Re(3-3i+2i+2)$$

Combine the real and imaginary parts in 3-3i+2i+2.

$$Re(3+2+\left(-3+2\right)i)$$

Do the additions in 3+2+\left(-3+2\right)i.

$$Re(5-i)$$

The real part of 5-i is 5.

$$5$$