Solve {l}{(18+6i)}{*(5+4i)}

$$\left. \begin{array} { l } { ( 18 + 6 i ) } \\ { \times ( 5 + 4 i ) } \end{array} \right.$$

Evaluate

$$66+102i$$

Solution Steps

Multiply complex numbers 18+6i and 5+4i like you multiply binomials.

$$18\times 5+18\times \left(4i\right)+6i\times 5+6\times 4i^{2}$$

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

$$18\times 5+18\times \left(4i\right)+6i\times 5+6\times 4\left(-1\right)$$

Do the multiplications.

$$90+72i+30i-24$$

Combine the real and imaginary parts.

$$90-24+\left(72+30\right)i$$

Do the additions.

$$66+102i$$

Real Part

$$66$$

Solution Steps

Multiply complex numbers 18+6i and 5+4i like you multiply binomials.

$$Re(18\times 5+18\times \left(4i\right)+6i\times 5+6\times 4i^{2})$$

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

$$Re(18\times 5+18\times \left(4i\right)+6i\times 5+6\times 4\left(-1\right))$$

Do the multiplications in 18\times 5+18\times \left(4i\right)+6i\times 5+6\times 4\left(-1\right).

$$Re(90+72i+30i-24)$$

Combine the real and imaginary parts in 90+72i+30i-24.

$$Re(90-24+\left(72+30\right)i)$$

Do the additions in 90-24+\left(72+30\right)i.

$$Re(66+102i)$$

The real part of 66+102i is 66.

$$66$$