Solve (6+8i)(5-4i)

$$( 6 + 8 i ) ( 5 - 4 i )$$

Evaluate

$$62+16i$$

Solution Steps

Multiply complex numbers 6+8i and 5-4i like you multiply binomials.

$$6\times 5+6\times \left(-4i\right)+8i\times 5+8\left(-4\right)i^{2}$$

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

$$6\times 5+6\times \left(-4i\right)+8i\times 5+8\left(-4\right)\left(-1\right)$$

Do the multiplications.

$$30-24i+40i+32$$

Combine the real and imaginary parts.

$$30+32+\left(-24+40\right)i$$

Do the additions.

$$62+16i$$

Real Part

$$62$$

Solution Steps

Multiply complex numbers 6+8i and 5-4i like you multiply binomials.

$$Re(6\times 5+6\times \left(-4i\right)+8i\times 5+8\left(-4\right)i^{2})$$

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

$$Re(6\times 5+6\times \left(-4i\right)+8i\times 5+8\left(-4\right)\left(-1\right))$$

Do the multiplications in 6\times 5+6\times \left(-4i\right)+8i\times 5+8\left(-4\right)\left(-1\right).

$$Re(30-24i+40i+32)$$

Combine the real and imaginary parts in 30-24i+40i+32.

$$Re(30+32+\left(-24+40\right)i)$$

Do the additions in 30+32+\left(-24+40\right)i.

$$Re(62+16i)$$

The real part of 62+16i is 62.

$$62$$