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

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

Evaluate

$$16-32i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$16-32i$$

Real Part

$$16$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(16-32i)$$

The real part of 16-32i is 16.

$$16$$