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

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

Evaluate

$$32+7i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$30+12i-5i+2$$

Combine the real and imaginary parts.

$$30+2+\left(12-5\right)i$$

Do the additions.

$$32+7i$$

Real Part

$$32$$

Solution Steps

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

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

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

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

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

$$Re(30+12i-5i+2)$$

Combine the real and imaginary parts in 30+12i-5i+2.

$$Re(30+2+\left(12-5\right)i)$$

Do the additions in 30+2+\left(12-5\right)i.

$$Re(32+7i)$$

The real part of 32+7i is 32.

$$32$$