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

$$( 6 + 5 i ) \cdot ( 3 - 6 i )$$

Evaluate

$$48-21i$$

Solution Steps

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

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

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

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

Do the multiplications.

$$18-36i+15i+30$$

Combine the real and imaginary parts.

$$18+30+\left(-36+15\right)i$$

Do the additions.

$$48-21i$$

Real Part

$$48$$

Solution Steps

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

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

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

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

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

$$Re(18-36i+15i+30)$$

Combine the real and imaginary parts in 18-36i+15i+30.

$$Re(18+30+\left(-36+15\right)i)$$

Do the additions in 18+30+\left(-36+15\right)i.

$$Re(48-21i)$$

The real part of 48-21i is 48.

$$48$$