Solve (4-7i)(2-3i)

$$( 4 - 7 i ) ( 2 - 3 i )$$

Evaluate

$$-13-26i$$

Solution Steps

Multiply complex numbers 4-7i and 2-3i like you multiply binomials.

$$4\times 2+4\times \left(-3i\right)-7i\times 2-7\left(-3\right)i^{2}$$

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

$$4\times 2+4\times \left(-3i\right)-7i\times 2-7\left(-3\right)\left(-1\right)$$

Do the multiplications.

$$8-12i-14i-21$$

Combine the real and imaginary parts.

$$8-21+\left(-12-14\right)i$$

Do the additions.

$$-13-26i$$

Real Part

$$-13$$

Solution Steps

Multiply complex numbers 4-7i and 2-3i like you multiply binomials.

$$Re(4\times 2+4\times \left(-3i\right)-7i\times 2-7\left(-3\right)i^{2})$$

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

$$Re(4\times 2+4\times \left(-3i\right)-7i\times 2-7\left(-3\right)\left(-1\right))$$

Do the multiplications in 4\times 2+4\times \left(-3i\right)-7i\times 2-7\left(-3\right)\left(-1\right).

$$Re(8-12i-14i-21)$$

Combine the real and imaginary parts in 8-12i-14i-21.

$$Re(8-21+\left(-12-14\right)i)$$

Do the additions in 8-21+\left(-12-14\right)i.

$$Re(-13-26i)$$

The real part of -13-26i is -13.

$$-13$$