Solve (2-6i)(-3+6i)

$$( 2 - 6 i ) ( - 3 + 6 i )$$

Evaluate

$$30+30i$$

Solution Steps

Multiply complex numbers 2-6i and -3+6i like you multiply binomials.

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

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

$$2\left(-3\right)+2\times \left(6i\right)-6i\left(-3\right)-6\times 6\left(-1\right)$$

Do the multiplications.

$$-6+12i+18i+36$$

Combine the real and imaginary parts.

$$-6+36+\left(12+18\right)i$$

Do the additions.

$$30+30i$$

Real Part

$$30$$

Solution Steps

Multiply complex numbers 2-6i and -3+6i like you multiply binomials.

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

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

$$Re(2\left(-3\right)+2\times \left(6i\right)-6i\left(-3\right)-6\times 6\left(-1\right))$$

Do the multiplications in 2\left(-3\right)+2\times \left(6i\right)-6i\left(-3\right)-6\times 6\left(-1\right).

$$Re(-6+12i+18i+36)$$

Combine the real and imaginary parts in -6+12i+18i+36.

$$Re(-6+36+\left(12+18\right)i)$$

Do the additions in -6+36+\left(12+18\right)i.

$$Re(30+30i)$$

The real part of 30+30i is 30.

$$30$$