Solve (2+i)(2-i)

$$( 2 + i ) ( 2 - i )$$

Evaluate

$$5$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$5$$

Real Part

$$5$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(5)$$

The real part of 5 is 5.

$$5$$