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

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

Evaluate

$$12-5i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$12-5i$$

Real Part

$$12$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(12-5i)$$

The real part of 12-5i is 12.

$$12$$