Solve (8-i)(3+2i)

$$( 8 - i ) ( 3 + 2 i )$$

Evaluate

$$26+13i$$

Solution Steps

Multiply complex numbers 8-i and 3+2i like you multiply binomials.

$$8\times 3+8\times \left(2i\right)-i\times 3-2i^{2}$$

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

$$8\times 3+8\times \left(2i\right)-i\times 3-2\left(-1\right)$$

Do the multiplications.

$$24+16i-3i+2$$

Combine the real and imaginary parts.

$$24+2+\left(16-3\right)i$$

Do the additions.

$$26+13i$$

Real Part

$$26$$

Solution Steps

Multiply complex numbers 8-i and 3+2i like you multiply binomials.

$$Re(8\times 3+8\times \left(2i\right)-i\times 3-2i^{2})$$

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

$$Re(8\times 3+8\times \left(2i\right)-i\times 3-2\left(-1\right))$$

Do the multiplications in 8\times 3+8\times \left(2i\right)-i\times 3-2\left(-1\right).

$$Re(24+16i-3i+2)$$

Combine the real and imaginary parts in 24+16i-3i+2.

$$Re(24+2+\left(16-3\right)i)$$

Do the additions in 24+2+\left(16-3\right)i.

$$Re(26+13i)$$

The real part of 26+13i is 26.

$$26$$