Solve 5+2/-2-i

$$\frac { 5 + 2 } { - 2 - i }$$

Evaluate

$$-\frac{14}{5}+\frac{7}{5}i=-2.8+1.4i$$

Solution Steps

Add 5 and 2 to get 7.

$$\frac{7}{-2-i}$$

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

$$\frac{7\left(-2+i\right)}{\left(-2-i\right)\left(-2+i\right)}$$

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

$$\frac{7\left(-2+i\right)}{\left(-2\right)^{2}-i^{2}}$$

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

$$\frac{7\left(-2+i\right)}{5}$$

Multiply 7 times -2+i.

$$\frac{7\left(-2\right)+7i}{5}$$

Do the multiplications in 7\left(-2\right)+7i.

$$\frac{-14+7i}{5}$$

Divide -14+7i by 5 to get -\frac{14}{5}+\frac{7}{5}i.

$$-\frac{14}{5}+\frac{7}{5}i$$

Real Part

$$-\frac{14}{5} = -2\frac{4}{5} = -2.8$$

Solution Steps

Add 5 and 2 to get 7.

$$Re(\frac{7}{-2-i})$$

Multiply both numerator and denominator of \frac{7}{-2-i} by the complex conjugate of the denominator, -2+i.

$$Re(\frac{7\left(-2+i\right)}{\left(-2-i\right)\left(-2+i\right)})$$

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

$$Re(\frac{7\left(-2+i\right)}{\left(-2\right)^{2}-i^{2}})$$

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

$$Re(\frac{7\left(-2+i\right)}{5})$$

Multiply 7 times -2+i.

$$Re(\frac{7\left(-2\right)+7i}{5})$$

Do the multiplications in 7\left(-2\right)+7i.

$$Re(\frac{-14+7i}{5})$$

Divide -14+7i by 5 to get -\frac{14}{5}+\frac{7}{5}i.

$$Re(-\frac{14}{5}+\frac{7}{5}i)$$

The real part of -\frac{14}{5}+\frac{7}{5}i is -\frac{14}{5}.

$$-\frac{14}{5}$$