Solve (6-3i)(4-2i)

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

Evaluate

$$18-24i$$

Solution Steps

Multiply complex numbers 6-3i and 4-2i like you multiply binomials.

$$6\times 4+6\times \left(-2i\right)-3i\times 4-3\left(-2\right)i^{2}$$

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

$$6\times 4+6\times \left(-2i\right)-3i\times 4-3\left(-2\right)\left(-1\right)$$

Do the multiplications.

$$24-12i-12i-6$$

Combine the real and imaginary parts.

$$24-6+\left(-12-12\right)i$$

Do the additions.

$$18-24i$$

Real Part

$$18$$

Solution Steps

Multiply complex numbers 6-3i and 4-2i like you multiply binomials.

$$Re(6\times 4+6\times \left(-2i\right)-3i\times 4-3\left(-2\right)i^{2})$$

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

$$Re(6\times 4+6\times \left(-2i\right)-3i\times 4-3\left(-2\right)\left(-1\right))$$

Do the multiplications in 6\times 4+6\times \left(-2i\right)-3i\times 4-3\left(-2\right)\left(-1\right).

$$Re(24-12i-12i-6)$$

Combine the real and imaginary parts in 24-12i-12i-6.

$$Re(24-6+\left(-12-12\right)i)$$

Do the additions in 24-6+\left(-12-12\right)i.

$$Re(18-24i)$$

The real part of 18-24i is 18.

$$18$$