Solve (2-i)(2-i)

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

Evaluate

$$3-4i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$3-4i$$

Real Part

$$3$$

Solution Steps

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

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

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

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

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

$$Re(3-4i)$$

The real part of 3-4i is 3.

$$3$$