Solve (8-7i)(6-2i)

$$( 8 - 7 i ) ( 6 - 2 i )$$

Evaluate

$$34-58i$$

Solution Steps

Multiply complex numbers 8-7i and 6-2i like you multiply binomials.

$$8\times 6+8\times \left(-2i\right)-7i\times 6-7\left(-2\right)i^{2}$$

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

$$8\times 6+8\times \left(-2i\right)-7i\times 6-7\left(-2\right)\left(-1\right)$$

Do the multiplications.

$$48-16i-42i-14$$

Combine the real and imaginary parts.

$$48-14+\left(-16-42\right)i$$

Do the additions.

$$34-58i$$

Real Part

$$34$$

Solution Steps

Multiply complex numbers 8-7i and 6-2i like you multiply binomials.

$$Re(8\times 6+8\times \left(-2i\right)-7i\times 6-7\left(-2\right)i^{2})$$

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

$$Re(8\times 6+8\times \left(-2i\right)-7i\times 6-7\left(-2\right)\left(-1\right))$$

Do the multiplications in 8\times 6+8\times \left(-2i\right)-7i\times 6-7\left(-2\right)\left(-1\right).

$$Re(48-16i-42i-14)$$

Combine the real and imaginary parts in 48-16i-42i-14.

$$Re(48-14+\left(-16-42\right)i)$$

Do the additions in 48-14+\left(-16-42\right)i.

$$Re(34-58i)$$

The real part of 34-58i is 34.

$$34$$