Solve (2+i)-(4-6i)(-3+3i)

$$( 2 + i ) - ( 4 - 6 i ) ( - 3 + 3 i )$$

Evaluate

$$-4-29i$$

Solution Steps

Multiply complex numbers 4-6i and -3+3i like you multiply binomials.

$$2+i-\left(4\left(-3\right)+4\times \left(3i\right)-6i\left(-3\right)-6\times 3i^{2}\right)$$

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

$$2+i-\left(4\left(-3\right)+4\times \left(3i\right)-6i\left(-3\right)-6\times 3\left(-1\right)\right)$$

Do the multiplications in 4\left(-3\right)+4\times \left(3i\right)-6i\left(-3\right)-6\times 3\left(-1\right).

$$2+i-\left(-12+12i+18i+18\right)$$

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

$$2+i-\left(-12+18+\left(12+18\right)i\right)$$

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

$$2+i-\left(6+30i\right)$$

Subtract 6+30i from 2+i by subtracting corresponding real and imaginary parts.

$$2-6+\left(1-30\right)i$$

Subtract 6 from 2. Subtract 30 from 1.

$$-4-29i$$

Real Part

$$-4$$

Solution Steps

Multiply complex numbers 4-6i and -3+3i like you multiply binomials.

$$Re(2+i-\left(4\left(-3\right)+4\times \left(3i\right)-6i\left(-3\right)-6\times 3i^{2}\right))$$

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

$$Re(2+i-\left(4\left(-3\right)+4\times \left(3i\right)-6i\left(-3\right)-6\times 3\left(-1\right)\right))$$

Do the multiplications in 4\left(-3\right)+4\times \left(3i\right)-6i\left(-3\right)-6\times 3\left(-1\right).

$$Re(2+i-\left(-12+12i+18i+18\right))$$

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

$$Re(2+i-\left(-12+18+\left(12+18\right)i\right))$$

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

$$Re(2+i-\left(6+30i\right))$$

Subtract 6+30i from 2+i by subtracting corresponding real and imaginary parts.

$$Re(2-6+\left(1-30\right)i)$$

Subtract 6 from 2. Subtract 30 from 1.

$$Re(-4-29i)$$

The real part of -4-29i is -4.

$$-4$$