Solve (-4+6i)(-1+4i)

$$( - 4 + 6 i ) ( - 1 + 4 i )$$

Evaluate

$$-20-22i$$

Solution Steps

Multiply complex numbers -4+6i and -1+4i like you multiply binomials.

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

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

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

Do the multiplications.

$$4-16i-6i-24$$

Combine the real and imaginary parts.

$$4-24+\left(-16-6\right)i$$

Do the additions.

$$-20-22i$$

Real Part

$$-20$$

Solution Steps

Multiply complex numbers -4+6i and -1+4i like you multiply binomials.

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

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

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

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

$$Re(4-16i-6i-24)$$

Combine the real and imaginary parts in 4-16i-6i-24.

$$Re(4-24+\left(-16-6\right)i)$$

Do the additions in 4-24+\left(-16-6\right)i.

$$Re(-20-22i)$$

The real part of -20-22i is -20.

$$-20$$