Solve (4-i)(6-7i)

$$( 4 - i ) ( 6 - 7 i )$$

Evaluate

$$17-34i$$

Solution Steps

Multiply complex numbers 4-i and 6-7i like you multiply binomials.

$$4\times 6+4\times \left(-7i\right)-i\times 6-\left(-7i^{2}\right)$$

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

$$4\times 6+4\times \left(-7i\right)-i\times 6-\left(-7\left(-1\right)\right)$$

Do the multiplications.

$$24-28i-6i-7$$

Combine the real and imaginary parts.

$$24-7+\left(-28-6\right)i$$

Do the additions.

$$17-34i$$

Real Part

$$17$$

Solution Steps

Multiply complex numbers 4-i and 6-7i like you multiply binomials.

$$Re(4\times 6+4\times \left(-7i\right)-i\times 6-\left(-7i^{2}\right))$$

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

$$Re(4\times 6+4\times \left(-7i\right)-i\times 6-\left(-7\left(-1\right)\right))$$

Do the multiplications in 4\times 6+4\times \left(-7i\right)-i\times 6-\left(-7\left(-1\right)\right).

$$Re(24-28i-6i-7)$$

Combine the real and imaginary parts in 24-28i-6i-7.

$$Re(24-7+\left(-28-6\right)i)$$

Do the additions in 24-7+\left(-28-6\right)i.

$$Re(17-34i)$$

The real part of 17-34i is 17.

$$17$$