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

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

Evaluate

$$20-10i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$12-16i+6i+8$$

Combine the real and imaginary parts.

$$12+8+\left(-16+6\right)i$$

Do the additions.

$$20-10i$$

Real Part

$$20$$

Solution Steps

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

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

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

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

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

$$Re(12-16i+6i+8)$$

Combine the real and imaginary parts in 12-16i+6i+8.

$$Re(12+8+\left(-16+6\right)i)$$

Do the additions in 12+8+\left(-16+6\right)i.

$$Re(20-10i)$$

The real part of 20-10i is 20.

$$20$$