Solve (6+2i)(6-2i)-(3-4i)(3+4i)

$$( 6 + 2 i ) ( 6 - 2 i ) - ( 3 - 4 i ) ( 3 + 4 i )$$

Evaluate

$$15$$

Solution Steps

Multiply complex numbers 6+2i and 6-2i like you multiply binomials.

$$6\times 6+6\times \left(-2i\right)+2i\times 6+2\left(-2\right)i^{2}-\left(3-4i\right)\left(3+4i\right)$$

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

$$6\times 6+6\times \left(-2i\right)+2i\times 6+2\left(-2\right)\left(-1\right)-\left(3-4i\right)\left(3+4i\right)$$

Do the multiplications in 6\times 6+6\times \left(-2i\right)+2i\times 6+2\left(-2\right)\left(-1\right).

$$36-12i+12i+4-\left(3-4i\right)\left(3+4i\right)$$

Combine the real and imaginary parts in 36-12i+12i+4.

$$36+4+\left(-12+12\right)i-\left(3-4i\right)\left(3+4i\right)$$

Do the additions in 36+4+\left(-12+12\right)i.

$$40-\left(3-4i\right)\left(3+4i\right)$$

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

$$40-\left(3\times 3+3\times \left(4i\right)-4i\times 3-4\times 4i^{2}\right)$$

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

$$40-\left(3\times 3+3\times \left(4i\right)-4i\times 3-4\times 4\left(-1\right)\right)$$

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

$$40-\left(9+12i-12i+16\right)$$

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

$$40-\left(9+16+\left(12-12\right)i\right)$$

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

$$40-25$$

Subtract 25 from 40 to get 15.

$$15$$

Real Part

$$15$$

Solution Steps

Multiply complex numbers 6+2i and 6-2i like you multiply binomials.

$$Re(6\times 6+6\times \left(-2i\right)+2i\times 6+2\left(-2\right)i^{2}-\left(3-4i\right)\left(3+4i\right))$$

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

$$Re(6\times 6+6\times \left(-2i\right)+2i\times 6+2\left(-2\right)\left(-1\right)-\left(3-4i\right)\left(3+4i\right))$$

Do the multiplications in 6\times 6+6\times \left(-2i\right)+2i\times 6+2\left(-2\right)\left(-1\right).

$$Re(36-12i+12i+4-\left(3-4i\right)\left(3+4i\right))$$

Combine the real and imaginary parts in 36-12i+12i+4.

$$Re(36+4+\left(-12+12\right)i-\left(3-4i\right)\left(3+4i\right))$$

Do the additions in 36+4+\left(-12+12\right)i.

$$Re(40-\left(3-4i\right)\left(3+4i\right))$$

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

$$Re(40-\left(3\times 3+3\times \left(4i\right)-4i\times 3-4\times 4i^{2}\right))$$

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

$$Re(40-\left(3\times 3+3\times \left(4i\right)-4i\times 3-4\times 4\left(-1\right)\right))$$

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

$$Re(40-\left(9+12i-12i+16\right))$$

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

$$Re(40-\left(9+16+\left(12-12\right)i\right))$$

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

$$Re(40-25)$$

Subtract 25 from 40 to get 15.

$$Re(15)$$

The real part of 15 is 15.

$$15$$