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

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

Evaluate

$$90$$

Solution Steps

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

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

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

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

Do the multiplications.

$$9-27i+27i+81$$

Combine the real and imaginary parts.

$$9+81+\left(-27+27\right)i$$

Do the additions.

$$90$$

Real Part

$$90$$

Solution Steps

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

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

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

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

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

$$Re(9-27i+27i+81)$$

Combine the real and imaginary parts in 9-27i+27i+81.

$$Re(9+81+\left(-27+27\right)i)$$

Do the additions in 9+81+\left(-27+27\right)i.

$$Re(90)$$

The real part of 90 is 90.

$$90$$