Solve (12i)(-11+11i)(8+2i)

$$( 12 i ) ( - 11 + 11 i ) ( 8 + 2 i )$$

Evaluate

$$-792-1320i$$

Solution Steps

Multiply 12i times -11+11i.

$$\left(12i\left(-11\right)+12\times 11i^{2}\right)\left(8+2i\right)$$

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

$$\left(12i\left(-11\right)+12\times 11\left(-1\right)\right)\left(8+2i\right)$$

Do the multiplications. Reorder the terms.

$$\left(-132-132i\right)\left(8+2i\right)$$

Multiply complex numbers -132-132i and 8+2i like you multiply binomials.

$$-132\times 8-132\times \left(2i\right)-132i\times 8-132\times 2i^{2}$$

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

$$-132\times 8-132\times \left(2i\right)-132i\times 8-132\times 2\left(-1\right)$$

Do the multiplications.

$$-1056-264i-1056i+264$$

Combine the real and imaginary parts.

$$-1056+264+\left(-264-1056\right)i$$

Do the additions.

$$-792-1320i$$

Real Part

$$-792$$

Solution Steps

Multiply 12i times -11+11i.

$$Re(\left(12i\left(-11\right)+12\times 11i^{2}\right)\left(8+2i\right))$$

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

$$Re(\left(12i\left(-11\right)+12\times 11\left(-1\right)\right)\left(8+2i\right))$$

Do the multiplications in 12i\left(-11\right)+12\times 11\left(-1\right). Reorder the terms.

$$Re(\left(-132-132i\right)\left(8+2i\right))$$

Multiply complex numbers -132-132i and 8+2i like you multiply binomials.

$$Re(-132\times 8-132\times \left(2i\right)-132i\times 8-132\times 2i^{2})$$

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

$$Re(-132\times 8-132\times \left(2i\right)-132i\times 8-132\times 2\left(-1\right))$$

Do the multiplications in -132\times 8-132\times \left(2i\right)-132i\times 8-132\times 2\left(-1\right).

$$Re(-1056-264i-1056i+264)$$

Combine the real and imaginary parts in -1056-264i-1056i+264.

$$Re(-1056+264+\left(-264-1056\right)i)$$

Do the additions in -1056+264+\left(-264-1056\right)i.

$$Re(-792-1320i)$$

The real part of -792-1320i is -792.

$$-792$$