Solve 7/3-i

$$\frac { 7 } { 3 - i }$$

Evaluate

$$\frac{21}{10}+\frac{7}{10}i=2.1+0.7i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+i.

$$\frac{7\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}$$

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

$$\frac{7\left(3+i\right)}{3^{2}-i^{2}}$$

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

$$\frac{7\left(3+i\right)}{10}$$

Multiply 7 times 3+i.

$$\frac{7\times 3+7i}{10}$$

Do the multiplications in 7\times 3+7i.

$$\frac{21+7i}{10}$$

Divide 21+7i by 10 to get \frac{21}{10}+\frac{7}{10}i.

$$\frac{21}{10}+\frac{7}{10}i$$

Real Part

$$\frac{21}{10} = 2\frac{1}{10} = 2.1$$

Solution Steps

Multiply both numerator and denominator of \frac{7}{3-i} by the complex conjugate of the denominator, 3+i.

$$Re(\frac{7\left(3+i\right)}{\left(3-i\right)\left(3+i\right)})$$

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

$$Re(\frac{7\left(3+i\right)}{3^{2}-i^{2}})$$

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

$$Re(\frac{7\left(3+i\right)}{10})$$

Multiply 7 times 3+i.

$$Re(\frac{7\times 3+7i}{10})$$

Do the multiplications in 7\times 3+7i.

$$Re(\frac{21+7i}{10})$$

Divide 21+7i by 10 to get \frac{21}{10}+\frac{7}{10}i.

$$Re(\frac{21}{10}+\frac{7}{10}i)$$

The real part of \frac{21}{10}+\frac{7}{10}i is \frac{21}{10}.

$$\frac{21}{10}$$