Solve (3+9i)(-9+7i)=

$$( 3 + 9 i ) ( - 9 + 7 i ) =$$

Evaluate

$$-90-60i$$

Solution Steps

Multiply complex numbers 3+9i and -9+7i like you multiply binomials.

$$3\left(-9\right)+3\times \left(7i\right)+9i\left(-9\right)+9\times 7i^{2}$$

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

$$3\left(-9\right)+3\times \left(7i\right)+9i\left(-9\right)+9\times 7\left(-1\right)$$

Do the multiplications.

$$-27+21i-81i-63$$

Combine the real and imaginary parts.

$$-27-63+\left(21-81\right)i$$

Do the additions.

$$-90-60i$$

Real Part

$$-90$$

Solution Steps

Multiply complex numbers 3+9i and -9+7i like you multiply binomials.

$$Re(3\left(-9\right)+3\times \left(7i\right)+9i\left(-9\right)+9\times 7i^{2})$$

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

$$Re(3\left(-9\right)+3\times \left(7i\right)+9i\left(-9\right)+9\times 7\left(-1\right))$$

Do the multiplications in 3\left(-9\right)+3\times \left(7i\right)+9i\left(-9\right)+9\times 7\left(-1\right).

$$Re(-27+21i-81i-63)$$

Combine the real and imaginary parts in -27+21i-81i-63.

$$Re(-27-63+\left(21-81\right)i)$$

Do the additions in -27-63+\left(21-81\right)i.

$$Re(-90-60i)$$

The real part of -90-60i is -90.

$$-90$$