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

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

Evaluate

$$16+2i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$12+8i-6i+4$$

Combine the real and imaginary parts.

$$12+4+\left(8-6\right)i$$

Do the additions.

$$16+2i$$

Real Part

$$16$$

Solution Steps

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

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

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

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

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

$$Re(12+8i-6i+4)$$

Combine the real and imaginary parts in 12+8i-6i+4.

$$Re(12+4+\left(8-6\right)i)$$

Do the additions in 12+4+\left(8-6\right)i.

$$Re(16+2i)$$

The real part of 16+2i is 16.

$$16$$