Solve (6+9i)(3-4i)

$$( 6 + 9 i ) ( 3 - 4 i )$$

Evaluate

$$54+3i$$

Solution Steps

Multiply complex numbers 6+9i and 3-4i like you multiply binomials.

$$6\times 3+6\times \left(-4i\right)+9i\times 3+9\left(-4\right)i^{2}$$

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

$$6\times 3+6\times \left(-4i\right)+9i\times 3+9\left(-4\right)\left(-1\right)$$

Do the multiplications.

$$18-24i+27i+36$$

Combine the real and imaginary parts.

$$18+36+\left(-24+27\right)i$$

Do the additions.

$$54+3i$$

Real Part

$$54$$

Solution Steps

Multiply complex numbers 6+9i and 3-4i like you multiply binomials.

$$Re(6\times 3+6\times \left(-4i\right)+9i\times 3+9\left(-4\right)i^{2})$$

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

$$Re(6\times 3+6\times \left(-4i\right)+9i\times 3+9\left(-4\right)\left(-1\right))$$

Do the multiplications in 6\times 3+6\times \left(-4i\right)+9i\times 3+9\left(-4\right)\left(-1\right).

$$Re(18-24i+27i+36)$$

Combine the real and imaginary parts in 18-24i+27i+36.

$$Re(18+36+\left(-24+27\right)i)$$

Do the additions in 18+36+\left(-24+27\right)i.

$$Re(54+3i)$$

The real part of 54+3i is 54.

$$54$$