Solve (4-i)(6+2i)

$$( 4 - i ) ( 6 + 2 i )$$

Evaluate

$$26+2i$$

Solution Steps

Multiply complex numbers 4-i and 6+2i like you multiply binomials.

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

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

$$4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right)$$

Do the multiplications.

$$24+8i-6i+2$$

Combine the real and imaginary parts.

$$24+2+\left(8-6\right)i$$

Do the additions.

$$26+2i$$

Real Part

$$26$$

Solution Steps

Multiply complex numbers 4-i and 6+2i like you multiply binomials.

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

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

$$Re(4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right))$$

Do the multiplications in 4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right).

$$Re(24+8i-6i+2)$$

Combine the real and imaginary parts in 24+8i-6i+2.

$$Re(24+2+\left(8-6\right)i)$$

Do the additions in 24+2+\left(8-6\right)i.

$$Re(26+2i)$$

The real part of 26+2i is 26.

$$26$$