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

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

Evaluate

$$100$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$100$$

Real Part

$$100$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(100)$$

The real part of 100 is 100.

$$100$$