Solve (3-i)*(1+2i)

$$( 3 - i ) \times ( 1 + 2 i )$$

Evaluate

$$5+5i$$

Solution Steps

Multiply complex numbers 3-i and 1+2i like you multiply binomials.

$$3\times 1+3\times \left(2i\right)-i-2i^{2}$$

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

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

Do the multiplications.

$$3+6i-i+2$$

Combine the real and imaginary parts.

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

Do the additions.

$$5+5i$$

Real Part

$$5$$

Solution Steps

Multiply complex numbers 3-i and 1+2i like you multiply binomials.

$$Re(3\times 1+3\times \left(2i\right)-i-2i^{2})$$

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

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

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

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

Combine the real and imaginary parts in 3+6i-i+2.

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

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

$$Re(5+5i)$$

The real part of 5+5i is 5.

$$5$$