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

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

Evaluate

$$8-i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$6-4i+3i+2$$

Combine the real and imaginary parts.

$$6+2+\left(-4+3\right)i$$

Do the additions.

$$8-i$$

Real Part

$$8$$

Solution Steps

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

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

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

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

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

$$Re(6-4i+3i+2)$$

Combine the real and imaginary parts in 6-4i+3i+2.

$$Re(6+2+\left(-4+3\right)i)$$

Do the additions in 6+2+\left(-4+3\right)i.

$$Re(8-i)$$

The real part of 8-i is 8.

$$8$$