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

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

Evaluate

$$16-32i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$24-24i-8i-8$$

Combine the real and imaginary parts.

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

Do the additions.

$$16-32i$$

Real Part

$$16$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(16-32i)$$

The real part of 16-32i is 16.

$$16$$