Solve (1-i)(1(1-19i)

$$( 1 - i ) ( 1 ( 1 - 19 i )$$

Evaluate

$$-18-20i$$

Solution Steps

Multiply 1-i and 1 to get 1-i.

$$\left(1-i\right)\left(1-19i\right)$$

Multiply complex numbers 1-i and 1-19i like you multiply binomials.

$$1\times 1+1\times \left(-19i\right)-i-\left(-19i^{2}\right)$$

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

$$1\times 1+1\times \left(-19i\right)-i-\left(-19\left(-1\right)\right)$$

Do the multiplications.

$$1-19i-i-19$$

Combine the real and imaginary parts.

$$1-19+\left(-19-1\right)i$$

Do the additions.

$$-18-20i$$

Real Part

$$-18$$

Solution Steps

Multiply 1-i and 1 to get 1-i.

$$Re(\left(1-i\right)\left(1-19i\right))$$

Multiply complex numbers 1-i and 1-19i like you multiply binomials.

$$Re(1\times 1+1\times \left(-19i\right)-i-\left(-19i^{2}\right))$$

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

$$Re(1\times 1+1\times \left(-19i\right)-i-\left(-19\left(-1\right)\right))$$

Do the multiplications in 1\times 1+1\times \left(-19i\right)-i-\left(-19\left(-1\right)\right).

$$Re(1-19i-i-19)$$

Combine the real and imaginary parts in 1-19i-i-19.

$$Re(1-19+\left(-19-1\right)i)$$

Do the additions in 1-19+\left(-19-1\right)i.

$$Re(-18-20i)$$

The real part of -18-20i is -18.

$$-18$$