Solve (-3+5i)(-6+6i)

$$( - 3 + 5 i ) ( - 6 + 6 i )$$

Evaluate

$$-12-48i$$

Solution Steps

Multiply complex numbers -3+5i and -6+6i like you multiply binomials.

$$-3\left(-6\right)-3\times \left(6i\right)+5i\left(-6\right)+5\times 6i^{2}$$

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

$$-3\left(-6\right)-3\times \left(6i\right)+5i\left(-6\right)+5\times 6\left(-1\right)$$

Do the multiplications.

$$18-18i-30i-30$$

Combine the real and imaginary parts.

$$18-30+\left(-18-30\right)i$$

Do the additions.

$$-12-48i$$

Real Part

$$-12$$

Solution Steps

Multiply complex numbers -3+5i and -6+6i like you multiply binomials.

$$Re(-3\left(-6\right)-3\times \left(6i\right)+5i\left(-6\right)+5\times 6i^{2})$$

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

$$Re(-3\left(-6\right)-3\times \left(6i\right)+5i\left(-6\right)+5\times 6\left(-1\right))$$

Do the multiplications in -3\left(-6\right)-3\times \left(6i\right)+5i\left(-6\right)+5\times 6\left(-1\right).

$$Re(18-18i-30i-30)$$

Combine the real and imaginary parts in 18-18i-30i-30.

$$Re(18-30+\left(-18-30\right)i)$$

Do the additions in 18-30+\left(-18-30\right)i.

$$Re(-12-48i)$$

The real part of -12-48i is -12.

$$-12$$