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

$$( 3 - i ) ( 1 + \frac { 3 } { 2 } i )$$

Evaluate

$$\frac{9}{2}+\frac{7}{2}i=4.5+3.5i$$

Solution Steps

Multiply complex numbers 3-i and 1+\frac{3}{2}i like you multiply binomials.

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

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

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

Do the multiplications.

$$3+\frac{9}{2}i-i+\frac{3}{2}$$

Combine the real and imaginary parts.

$$3+\frac{3}{2}+\left(\frac{9}{2}-1\right)i$$

Do the additions.

$$\frac{9}{2}+\frac{7}{2}i$$

Real Part

$$\frac{9}{2} = 4\frac{1}{2} = 4.5$$

Solution Steps

Multiply complex numbers 3-i and 1+\frac{3}{2}i like you multiply binomials.

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

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

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

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

$$Re(3+\frac{9}{2}i-i+\frac{3}{2})$$

Combine the real and imaginary parts in 3+\frac{9}{2}i-i+\frac{3}{2}.

$$Re(3+\frac{3}{2}+\left(\frac{9}{2}-1\right)i)$$

Do the additions in 3+\frac{3}{2}+\left(\frac{9}{2}-1\right)i.

$$Re(\frac{9}{2}+\frac{7}{2}i)$$

The real part of \frac{9}{2}+\frac{7}{2}i is \frac{9}{2}.

$$\frac{9}{2}$$