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

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

Evaluate

$$-16-28i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$8-12i-16i-24$$

Combine the real and imaginary parts.

$$8-24+\left(-12-16\right)i$$

Do the additions.

$$-16-28i$$

Real Part

$$-16$$

Solution Steps

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

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

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

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

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

$$Re(8-12i-16i-24)$$

Combine the real and imaginary parts in 8-12i-16i-24.

$$Re(8-24+\left(-12-16\right)i)$$

Do the additions in 8-24+\left(-12-16\right)i.

$$Re(-16-28i)$$

The real part of -16-28i is -16.

$$-16$$