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

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

Evaluate

$$12-34i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$12-34i$$

Real Part

$$12$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(12-34i)$$

The real part of 12-34i is 12.

$$12$$