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

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

Evaluate

$$12+5i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$12+5i$$

Real Part

$$12$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(12+5i)$$

The real part of 12+5i is 12.

$$12$$