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

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

Evaluate

$$100i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$100i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

$$Re(100i)$$

The real part of 100i is 0.

$$0$$