Solve (4-i)(6+2i)-(7-i)(4+3i)

$$( 4 - i ) ( 6 + 2 i ) - ( 7 - i ) ( 4 + 3 i )$$

Evaluate

$$-5-15i$$

Solution Steps

Multiply complex numbers 4-i and 6+2i like you multiply binomials.

$$4\times 6+4\times \left(2i\right)-i\times 6-2i^{2}-\left(7-i\right)\left(4+3i\right)$$

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

$$4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right)-\left(7-i\right)\left(4+3i\right)$$

Do the multiplications in 4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right).

$$24+8i-6i+2-\left(7-i\right)\left(4+3i\right)$$

Combine the real and imaginary parts in 24+8i-6i+2.

$$24+2+\left(8-6\right)i-\left(7-i\right)\left(4+3i\right)$$

Do the additions in 24+2+\left(8-6\right)i.

$$26+2i-\left(7-i\right)\left(4+3i\right)$$

Multiply complex numbers 7-i and 4+3i like you multiply binomials.

$$26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3i^{2}\right)$$

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

$$26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right)\right)$$

Do the multiplications in 7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right).

$$26+2i-\left(28+21i-4i+3\right)$$

Combine the real and imaginary parts in 28+21i-4i+3.

$$26+2i-\left(28+3+\left(21-4\right)i\right)$$

Do the additions in 28+3+\left(21-4\right)i.

$$26+2i-\left(31+17i\right)$$

Subtract 31+17i from 26+2i by subtracting corresponding real and imaginary parts.

$$26-31+\left(2-17\right)i$$

Subtract 31 from 26. Subtract 17 from 2.

$$-5-15i$$

Real Part

$$-5$$

Solution Steps

Multiply complex numbers 4-i and 6+2i like you multiply binomials.

$$Re(4\times 6+4\times \left(2i\right)-i\times 6-2i^{2}-\left(7-i\right)\left(4+3i\right))$$

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

$$Re(4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right)-\left(7-i\right)\left(4+3i\right))$$

Do the multiplications in 4\times 6+4\times \left(2i\right)-i\times 6-2\left(-1\right).

$$Re(24+8i-6i+2-\left(7-i\right)\left(4+3i\right))$$

Combine the real and imaginary parts in 24+8i-6i+2.

$$Re(24+2+\left(8-6\right)i-\left(7-i\right)\left(4+3i\right))$$

Do the additions in 24+2+\left(8-6\right)i.

$$Re(26+2i-\left(7-i\right)\left(4+3i\right))$$

Multiply complex numbers 7-i and 4+3i like you multiply binomials.

$$Re(26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3i^{2}\right))$$

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

$$Re(26+2i-\left(7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right)\right))$$

Do the multiplications in 7\times 4+7\times \left(3i\right)-i\times 4-3\left(-1\right).

$$Re(26+2i-\left(28+21i-4i+3\right))$$

Combine the real and imaginary parts in 28+21i-4i+3.

$$Re(26+2i-\left(28+3+\left(21-4\right)i\right))$$

Do the additions in 28+3+\left(21-4\right)i.

$$Re(26+2i-\left(31+17i\right))$$

Subtract 31+17i from 26+2i by subtracting corresponding real and imaginary parts.

$$Re(26-31+\left(2-17\right)i)$$

Subtract 31 from 26. Subtract 17 from 2.

$$Re(-5-15i)$$

The real part of -5-15i is -5.

$$-5$$