Solve (8+5i)(3+2i)-(4+i)(4-i)

$$( 8 + 5 i ) ( 3 + 2 i ) - ( 4 + i ) ( 4 - i )$$

Evaluate

$$-3+31i$$

Solution Steps

Multiply complex numbers 8+5i and 3+2i like you multiply binomials.

$$8\times 3+8\times \left(2i\right)+5i\times 3+5\times 2i^{2}-\left(4+i\right)\left(4-i\right)$$

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

$$8\times 3+8\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right)-\left(4+i\right)\left(4-i\right)$$

Do the multiplications in 8\times 3+8\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right).

$$24+16i+15i-10-\left(4+i\right)\left(4-i\right)$$

Combine the real and imaginary parts in 24+16i+15i-10.

$$24-10+\left(16+15\right)i-\left(4+i\right)\left(4-i\right)$$

Do the additions in 24-10+\left(16+15\right)i.

$$14+31i-\left(4+i\right)\left(4-i\right)$$

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

$$14+31i-\left(4\times 4+4\left(-i\right)+4i-i^{2}\right)$$

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

$$14+31i-\left(4\times 4+4\left(-i\right)+4i-\left(-1\right)\right)$$

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

$$14+31i-\left(16-4i+4i+1\right)$$

Combine the real and imaginary parts in 16-4i+4i+1.

$$14+31i-\left(16+1+\left(-4+4\right)i\right)$$

Do the additions in 16+1+\left(-4+4\right)i.

$$14+31i-17$$

Subtract 17 from 14+31i by subtracting corresponding real and imaginary parts.

$$14-17+31i$$

Subtract 17 from 14 to get -3.

$$-3+31i$$

Real Part

$$-3$$

Solution Steps

Multiply complex numbers 8+5i and 3+2i like you multiply binomials.

$$Re(8\times 3+8\times \left(2i\right)+5i\times 3+5\times 2i^{2}-\left(4+i\right)\left(4-i\right))$$

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

$$Re(8\times 3+8\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right)-\left(4+i\right)\left(4-i\right))$$

Do the multiplications in 8\times 3+8\times \left(2i\right)+5i\times 3+5\times 2\left(-1\right).

$$Re(24+16i+15i-10-\left(4+i\right)\left(4-i\right))$$

Combine the real and imaginary parts in 24+16i+15i-10.

$$Re(24-10+\left(16+15\right)i-\left(4+i\right)\left(4-i\right))$$

Do the additions in 24-10+\left(16+15\right)i.

$$Re(14+31i-\left(4+i\right)\left(4-i\right))$$

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

$$Re(14+31i-\left(4\times 4+4\left(-i\right)+4i-i^{2}\right))$$

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

$$Re(14+31i-\left(4\times 4+4\left(-i\right)+4i-\left(-1\right)\right))$$

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

$$Re(14+31i-\left(16-4i+4i+1\right))$$

Combine the real and imaginary parts in 16-4i+4i+1.

$$Re(14+31i-\left(16+1+\left(-4+4\right)i\right))$$

Do the additions in 16+1+\left(-4+4\right)i.

$$Re(14+31i-17)$$

Subtract 17 from 14+31i by subtracting corresponding real and imaginary parts.

$$Re(14-17+31i)$$

Subtract 17 from 14 to get -3.

$$Re(-3+31i)$$

The real part of -3+31i is -3.

$$-3$$