Solve (5-3i)(6+2i)

$$( 5 - 3 i ) ( 6 + 2 i )$$

Evaluate

$$36-8i$$

Solution Steps

Multiply complex numbers 5-3i and 6+2i like you multiply binomials.

$$5\times 6+5\times \left(2i\right)-3i\times 6-3\times 2i^{2}$$

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

$$5\times 6+5\times \left(2i\right)-3i\times 6-3\times 2\left(-1\right)$$

Do the multiplications.

$$30+10i-18i+6$$

Combine the real and imaginary parts.

$$30+6+\left(10-18\right)i$$

Do the additions.

$$36-8i$$

Real Part

$$36$$

Solution Steps

Multiply complex numbers 5-3i and 6+2i like you multiply binomials.

$$Re(5\times 6+5\times \left(2i\right)-3i\times 6-3\times 2i^{2})$$

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

$$Re(5\times 6+5\times \left(2i\right)-3i\times 6-3\times 2\left(-1\right))$$

Do the multiplications in 5\times 6+5\times \left(2i\right)-3i\times 6-3\times 2\left(-1\right).

$$Re(30+10i-18i+6)$$

Combine the real and imaginary parts in 30+10i-18i+6.

$$Re(30+6+\left(10-18\right)i)$$

Do the additions in 30+6+\left(10-18\right)i.

$$Re(36-8i)$$

The real part of 36-8i is 36.

$$36$$