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

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

Evaluate

$$\frac{5}{2}=2.5$$

Solution Steps

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

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

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

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

Do the multiplications.

$$\frac{9}{4}+\frac{3}{4}i-\frac{3}{4}i+\frac{1}{4}$$

Combine the real and imaginary parts.

$$\frac{9}{4}+\frac{1}{4}+\left(\frac{3}{4}-\frac{3}{4}\right)i$$

Do the additions.

$$\frac{5}{2}$$

Real Part

$$\frac{5}{2} = 2\frac{1}{2} = 2.5$$

Solution Steps

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

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

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

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

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

$$Re(\frac{9}{4}+\frac{3}{4}i-\frac{3}{4}i+\frac{1}{4})$$

Combine the real and imaginary parts in \frac{9}{4}+\frac{3}{4}i-\frac{3}{4}i+\frac{1}{4}.

$$Re(\frac{9}{4}+\frac{1}{4}+\left(\frac{3}{4}-\frac{3}{4}\right)i)$$

Do the additions in \frac{9}{4}+\frac{1}{4}+\left(\frac{3}{4}-\frac{3}{4}\right)i.

$$Re(\frac{5}{2})$$

The real part of \frac{5}{2} is \frac{5}{2}.

$$\frac{5}{2}$$