Solve (-4-6i)(2+9i)

$$( - 4 - 6 i ) ( 2 + 9 i )$$

Evaluate

$$46-48i$$

Solution Steps

Multiply complex numbers -4-6i and 2+9i like you multiply binomials.

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

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

$$-4\times 2-4\times \left(9i\right)-6i\times 2-6\times 9\left(-1\right)$$

Do the multiplications.

$$-8-36i-12i+54$$

Combine the real and imaginary parts.

$$-8+54+\left(-36-12\right)i$$

Do the additions.

$$46-48i$$

Real Part

$$46$$

Solution Steps

Multiply complex numbers -4-6i and 2+9i like you multiply binomials.

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

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

$$Re(-4\times 2-4\times \left(9i\right)-6i\times 2-6\times 9\left(-1\right))$$

Do the multiplications in -4\times 2-4\times \left(9i\right)-6i\times 2-6\times 9\left(-1\right).

$$Re(-8-36i-12i+54)$$

Combine the real and imaginary parts in -8-36i-12i+54.

$$Re(-8+54+\left(-36-12\right)i)$$

Do the additions in -8+54+\left(-36-12\right)i.

$$Re(46-48i)$$

The real part of 46-48i is 46.

$$46$$