Solve (3+i)(4-i)

$$( 3 + i ) ( 4 - i )$$

Evaluate

$$13+i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$12-3i+4i+1$$

Combine the real and imaginary parts.

$$12+1+\left(-3+4\right)i$$

Do the additions.

$$13+i$$

Real Part

$$13$$

Solution Steps

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

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

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

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

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

$$Re(12-3i+4i+1)$$

Combine the real and imaginary parts in 12-3i+4i+1.

$$Re(12+1+\left(-3+4\right)i)$$

Do the additions in 12+1+\left(-3+4\right)i.

$$Re(13+i)$$

The real part of 13+i is 13.

$$13$$