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

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

Evaluate

$$25$$

Solution Steps

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

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

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

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

Do the multiplications.

$$9-12i+12i+16$$

Combine the real and imaginary parts.

$$9+16+\left(-12+12\right)i$$

Do the additions.

$$25$$

Real Part

$$25$$

Solution Steps

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

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

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

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

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

$$Re(9-12i+12i+16)$$

Combine the real and imaginary parts in 9-12i+12i+16.

$$Re(9+16+\left(-12+12\right)i)$$

Do the additions in 9+16+\left(-12+12\right)i.

$$Re(25)$$

The real part of 25 is 25.

$$25$$