Solve (5-2i)(5+2i)

$$( 5 - 2 i ) ( 5 + 2 i )$$

Evaluate

$$29$$

Solution Steps

Multiply complex numbers 5-2i and 5+2i like you multiply binomials.

$$5\times 5+5\times \left(2i\right)-2i\times 5-2\times 2i^{2}$$

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

$$5\times 5+5\times \left(2i\right)-2i\times 5-2\times 2\left(-1\right)$$

Do the multiplications.

$$25+10i-10i+4$$

Combine the real and imaginary parts.

$$25+4+\left(10-10\right)i$$

Do the additions.

$$29$$

Real Part

$$29$$

Solution Steps

Multiply complex numbers 5-2i and 5+2i like you multiply binomials.

$$Re(5\times 5+5\times \left(2i\right)-2i\times 5-2\times 2i^{2})$$

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

$$Re(5\times 5+5\times \left(2i\right)-2i\times 5-2\times 2\left(-1\right))$$

Do the multiplications in 5\times 5+5\times \left(2i\right)-2i\times 5-2\times 2\left(-1\right).

$$Re(25+10i-10i+4)$$

Combine the real and imaginary parts in 25+10i-10i+4.

$$Re(25+4+\left(10-10\right)i)$$

Do the additions in 25+4+\left(10-10\right)i.

$$Re(29)$$

The real part of 29 is 29.

$$29$$