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.

$$-29$$