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

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

Evaluate

$$36-2i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$24+16i-18i+12$$

Combine the real and imaginary parts.

$$24+12+\left(16-18\right)i$$

Do the additions.

$$36-2i$$

Real Part

$$36$$

Solution Steps

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

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

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

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

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

$$Re(24+16i-18i+12)$$

Combine the real and imaginary parts in 24+16i-18i+12.

$$Re(24+12+\left(16-18\right)i)$$

Do the additions in 24+12+\left(16-18\right)i.

$$Re(36-2i)$$

The real part of 36-2i is 36.

$$36$$