Solve 17-21/2-1i

$$\frac { 17 - 21 } { 2 - 1 i }$$

Evaluate

$$-\frac{8}{5}-\frac{4}{5}i=-1.6-0.8i$$

Solution Steps

Subtract 21 from 17 to get -4.

$$\frac{-4}{2-i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 2+i.

$$\frac{-4\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{-4\left(2+i\right)}{2^{2}-i^{2}}$$

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

$$\frac{-4\left(2+i\right)}{5}$$

Multiply -4 times 2+i.

$$\frac{-4\times 2-4i}{5}$$

Do the multiplications in -4\times 2-4i.

$$\frac{-8-4i}{5}$$

Divide -8-4i by 5 to get -\frac{8}{5}-\frac{4}{5}i.

$$-\frac{8}{5}-\frac{4}{5}i$$

Real Part

$$-\frac{8}{5} = -1\frac{3}{5} = -1.6$$

Solution Steps

Subtract 21 from 17 to get -4.

$$Re(\frac{-4}{2-i})$$

Multiply both numerator and denominator of \frac{-4}{2-i} by the complex conjugate of the denominator, 2+i.

$$Re(\frac{-4\left(2+i\right)}{\left(2-i\right)\left(2+i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{-4\left(2+i\right)}{2^{2}-i^{2}})$$

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

$$Re(\frac{-4\left(2+i\right)}{5})$$

Multiply -4 times 2+i.

$$Re(\frac{-4\times 2-4i}{5})$$

Do the multiplications in -4\times 2-4i.

$$Re(\frac{-8-4i}{5})$$

Divide -8-4i by 5 to get -\frac{8}{5}-\frac{4}{5}i.

$$Re(-\frac{8}{5}-\frac{4}{5}i)$$

The real part of -\frac{8}{5}-\frac{4}{5}i is -\frac{8}{5}.

$$-\frac{8}{5}$$