Solve (5-7i)(4-8i)

$$( 5 - 7 i ) ( 4 - 8 i )$$

Evaluate

$$-36-68i$$

Solution Steps

Multiply complex numbers 5-7i and 4-8i like you multiply binomials.

$$5\times 4+5\times \left(-8i\right)-7i\times 4-7\left(-8\right)i^{2}$$

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

$$5\times 4+5\times \left(-8i\right)-7i\times 4-7\left(-8\right)\left(-1\right)$$

Do the multiplications.

$$20-40i-28i-56$$

Combine the real and imaginary parts.

$$20-56+\left(-40-28\right)i$$

Do the additions.

$$-36-68i$$

Real Part

$$-36$$

Solution Steps

Multiply complex numbers 5-7i and 4-8i like you multiply binomials.

$$Re(5\times 4+5\times \left(-8i\right)-7i\times 4-7\left(-8\right)i^{2})$$

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

$$Re(5\times 4+5\times \left(-8i\right)-7i\times 4-7\left(-8\right)\left(-1\right))$$

Do the multiplications in 5\times 4+5\times \left(-8i\right)-7i\times 4-7\left(-8\right)\left(-1\right).

$$Re(20-40i-28i-56)$$

Combine the real and imaginary parts in 20-40i-28i-56.

$$Re(20-56+\left(-40-28\right)i)$$

Do the additions in 20-56+\left(-40-28\right)i.

$$Re(-36-68i)$$

The real part of -36-68i is -36.

$$-36$$