Solve 6/4-i

$$\frac { 6 } { 4 - i }$$

Evaluate

$$\frac{24}{17}+\frac{6}{17}i\approx 1.411764706+0.352941176i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4+i.

$$\frac{6\left(4+i\right)}{\left(4-i\right)\left(4+i\right)}$$

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

$$\frac{6\left(4+i\right)}{4^{2}-i^{2}}$$

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

$$\frac{6\left(4+i\right)}{17}$$

Multiply 6 times 4+i.

$$\frac{6\times 4+6i}{17}$$

Do the multiplications in 6\times 4+6i.

$$\frac{24+6i}{17}$$

Divide 24+6i by 17 to get \frac{24}{17}+\frac{6}{17}i.

$$\frac{24}{17}+\frac{6}{17}i$$

Real Part

$$\frac{24}{17} = 1\frac{7}{17} = 1.411764705882353$$

Solution Steps

Multiply both numerator and denominator of \frac{6}{4-i} by the complex conjugate of the denominator, 4+i.

$$Re(\frac{6\left(4+i\right)}{\left(4-i\right)\left(4+i\right)})$$

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

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

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

$$Re(\frac{6\left(4+i\right)}{17})$$

Multiply 6 times 4+i.

$$Re(\frac{6\times 4+6i}{17})$$

Do the multiplications in 6\times 4+6i.

$$Re(\frac{24+6i}{17})$$

Divide 24+6i by 17 to get \frac{24}{17}+\frac{6}{17}i.

$$Re(\frac{24}{17}+\frac{6}{17}i)$$

The real part of \frac{24}{17}+\frac{6}{17}i is \frac{24}{17}.

$$\frac{24}{17}$$