Solve (-8+6i)(6+8i)

$$( - 8 + 6 i ) ( 6 + 8 i )$$

Evaluate

$$-96-28i$$

Solution Steps

Multiply complex numbers -8+6i and 6+8i like you multiply binomials.

$$-8\times 6-8\times \left(8i\right)+6i\times 6+6\times 8i^{2}$$

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

$$-8\times 6-8\times \left(8i\right)+6i\times 6+6\times 8\left(-1\right)$$

Do the multiplications.

$$-48-64i+36i-48$$

Combine the real and imaginary parts.

$$-48-48+\left(-64+36\right)i$$

Do the additions.

$$-96-28i$$

Real Part

$$-96$$

Solution Steps

Multiply complex numbers -8+6i and 6+8i like you multiply binomials.

$$Re(-8\times 6-8\times \left(8i\right)+6i\times 6+6\times 8i^{2})$$

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

$$Re(-8\times 6-8\times \left(8i\right)+6i\times 6+6\times 8\left(-1\right))$$

Do the multiplications in -8\times 6-8\times \left(8i\right)+6i\times 6+6\times 8\left(-1\right).

$$Re(-48-64i+36i-48)$$

Combine the real and imaginary parts in -48-64i+36i-48.

$$Re(-48-48+\left(-64+36\right)i)$$

Do the additions in -48-48+\left(-64+36\right)i.

$$Re(-96-28i)$$

The real part of -96-28i is -96.

$$-96$$