Solve (-17/4-6i)*(2-5/2i)=

$$( - \frac { 17 } { 4 } - 6 i ) \cdot ( 2 - \frac { 5 } { 2 } i ) =$$

Evaluate

$$-\frac{47}{2}-\frac{11}{8}i=-23.5-1.375i$$

Solution Steps

Multiply complex numbers -\frac{17}{4}-6i and 2-\frac{5}{2}i like you multiply binomials.

$$-\frac{17}{4}\times 2-\frac{17}{4}\times \left(-\frac{5}{2}i\right)-6i\times 2-6\left(-\frac{5}{2}\right)i^{2}$$

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

$$-\frac{17}{4}\times 2-\frac{17}{4}\times \left(-\frac{5}{2}i\right)-6i\times 2-6\left(-\frac{5}{2}\right)\left(-1\right)$$

Do the multiplications.

$$-\frac{17}{2}+\frac{85}{8}i-12i-15$$

Combine the real and imaginary parts.

$$-\frac{17}{2}-15+\left(\frac{85}{8}-12\right)i$$

Do the additions.

$$-\frac{47}{2}-\frac{11}{8}i$$

Real Part

$$-\frac{47}{2} = -23\frac{1}{2} = -23.5$$

Solution Steps

Multiply complex numbers -\frac{17}{4}-6i and 2-\frac{5}{2}i like you multiply binomials.

$$Re(-\frac{17}{4}\times 2-\frac{17}{4}\times \left(-\frac{5}{2}i\right)-6i\times 2-6\left(-\frac{5}{2}\right)i^{2})$$

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

$$Re(-\frac{17}{4}\times 2-\frac{17}{4}\times \left(-\frac{5}{2}i\right)-6i\times 2-6\left(-\frac{5}{2}\right)\left(-1\right))$$

Do the multiplications in -\frac{17}{4}\times 2-\frac{17}{4}\times \left(-\frac{5}{2}i\right)-6i\times 2-6\left(-\frac{5}{2}\right)\left(-1\right).

$$Re(-\frac{17}{2}+\frac{85}{8}i-12i-15)$$

Combine the real and imaginary parts in -\frac{17}{2}+\frac{85}{8}i-12i-15.

$$Re(-\frac{17}{2}-15+\left(\frac{85}{8}-12\right)i)$$

Do the additions in -\frac{17}{2}-15+\left(\frac{85}{8}-12\right)i.

$$Re(-\frac{47}{2}-\frac{11}{8}i)$$

The real part of -\frac{47}{2}-\frac{11}{8}i is -\frac{47}{2}.

$$-\frac{47}{2}$$