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

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

Evaluate

$$9+2i$$

Solution Steps

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

$$4\times 2+4i-i\times 2-i^{2}$$

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

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

Do the multiplications.

$$8+4i-2i+1$$

Combine the real and imaginary parts.

$$8+1+\left(4-2\right)i$$

Do the additions.

$$9+2i$$

Real Part

$$9$$

Solution Steps

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

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

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

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

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

$$Re(8+4i-2i+1)$$

Combine the real and imaginary parts in 8+4i-2i+1.

$$Re(8+1+\left(4-2\right)i)$$

Do the additions in 8+1+\left(4-2\right)i.

$$Re(9+2i)$$

The real part of 9+2i is 9.

$$9$$