Solve (3+2i)(4-8i)

$$( 3 + 2 i ) ( 4 - 8 i )$$

Evaluate

$$28-16i$$

Solution Steps

Multiply complex numbers 3+2i and 4-8i like you multiply binomials.

$$3\times 4+3\times \left(-8i\right)+2i\times 4+2\left(-8\right)i^{2}$$

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

$$3\times 4+3\times \left(-8i\right)+2i\times 4+2\left(-8\right)\left(-1\right)$$

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$28-16i$$

Real Part

$$28$$

Solution Steps

Multiply complex numbers 3+2i and 4-8i like you multiply binomials.

$$Re(3\times 4+3\times \left(-8i\right)+2i\times 4+2\left(-8\right)i^{2})$$

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

$$Re(3\times 4+3\times \left(-8i\right)+2i\times 4+2\left(-8\right)\left(-1\right))$$

Do the multiplications in 3\times 4+3\times \left(-8i\right)+2i\times 4+2\left(-8\right)\left(-1\right).

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

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

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

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

$$Re(28-16i)$$

The real part of 28-16i is 28.

$$28$$