Solve (4-5i)(1-4i)

$$( 4 - 5 i ) ( 1 - 4 i )$$

Evaluate

$$-16-21i$$

Solution Steps

Multiply complex numbers 4-5i and 1-4i like you multiply binomials.

$$4\times 1+4\times \left(-4i\right)-5i-5\left(-4\right)i^{2}$$

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

$$4\times 1+4\times \left(-4i\right)-5i-5\left(-4\right)\left(-1\right)$$

Do the multiplications.

$$4-16i-5i-20$$

Combine the real and imaginary parts.

$$4-20+\left(-16-5\right)i$$

Do the additions.

$$-16-21i$$

Real Part

$$-16$$

Solution Steps

Multiply complex numbers 4-5i and 1-4i like you multiply binomials.

$$Re(4\times 1+4\times \left(-4i\right)-5i-5\left(-4\right)i^{2})$$

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

$$Re(4\times 1+4\times \left(-4i\right)-5i-5\left(-4\right)\left(-1\right))$$

Do the multiplications in 4\times 1+4\times \left(-4i\right)-5i-5\left(-4\right)\left(-1\right).

$$Re(4-16i-5i-20)$$

Combine the real and imaginary parts in 4-16i-5i-20.

$$Re(4-20+\left(-16-5\right)i)$$

Do the additions in 4-20+\left(-16-5\right)i.

$$Re(-16-21i)$$

The real part of -16-21i is -16.

$$-16$$