Solve (3+2i)(3-2i)?

$$( 3 + 2 i ) ( 3 - 2 i ) ?$$

Evaluate

$$13$$

Solution Steps

Multiply complex numbers 3+2i and 3-2i like you multiply binomials.

$$3\times 3+3\times \left(-2i\right)+2i\times 3+2\left(-2\right)i^{2}$$

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

$$3\times 3+3\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right)$$

Do the multiplications.

$$9-6i+6i+4$$

Combine the real and imaginary parts.

$$9+4+\left(-6+6\right)i$$

Do the additions.

$$13$$

Real Part

$$13$$

Solution Steps

Multiply complex numbers 3+2i and 3-2i like you multiply binomials.

$$Re(3\times 3+3\times \left(-2i\right)+2i\times 3+2\left(-2\right)i^{2})$$

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

$$Re(3\times 3+3\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right))$$

Do the multiplications in 3\times 3+3\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right).

$$Re(9-6i+6i+4)$$

Combine the real and imaginary parts in 9-6i+6i+4.

$$Re(9+4+\left(-6+6\right)i)$$

Do the additions in 9+4+\left(-6+6\right)i.

$$Re(13)$$

The real part of 13 is 13.

$$13$$