Solve left(3+i5right)left(2+i6right)

$$\left( 3+i5 \right) \left( 2+i6 \right)$$

Evaluate

$$-24+28i$$

Solution Steps

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

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

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

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

Do the multiplications.

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

Combine the real and imaginary parts.

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

Do the additions.

$$-24+28i$$

Real Part

$$-24$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$Re(-24+28i)$$

The real part of -24+28i is -24.

$$-24$$