Solve (3+4i)*(2+10i)

$$( 3 + 4 i ) \times ( 2 + 10 i )$$

Evaluate

$$-34+38i$$

Solution Steps

Multiply complex numbers 3+4i and 2+10i like you multiply binomials.

$$3\times 2+3\times \left(10i\right)+4i\times 2+4\times 10i^{2}$$

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

$$3\times 2+3\times \left(10i\right)+4i\times 2+4\times 10\left(-1\right)$$

Do the multiplications.

$$6+30i+8i-40$$

Combine the real and imaginary parts.

$$6-40+\left(30+8\right)i$$

Do the additions.

$$-34+38i$$

Real Part

$$-34$$

Solution Steps

Multiply complex numbers 3+4i and 2+10i like you multiply binomials.

$$Re(3\times 2+3\times \left(10i\right)+4i\times 2+4\times 10i^{2})$$

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

$$Re(3\times 2+3\times \left(10i\right)+4i\times 2+4\times 10\left(-1\right))$$

Do the multiplications in 3\times 2+3\times \left(10i\right)+4i\times 2+4\times 10\left(-1\right).

$$Re(6+30i+8i-40)$$

Combine the real and imaginary parts in 6+30i+8i-40.

$$Re(6-40+\left(30+8\right)i)$$

Do the additions in 6-40+\left(30+8\right)i.

$$Re(-34+38i)$$

The real part of -34+38i is -34.

$$-34$$