Solve 5/8-i

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

Evaluate

$$\frac{8}{13}+\frac{1}{13}i\approx 0.615384615+0.076923077i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 8+i.

$$\frac{5\left(8+i\right)}{\left(8-i\right)\left(8+i\right)}$$

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

$$\frac{5\left(8+i\right)}{8^{2}-i^{2}}$$

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

$$\frac{5\left(8+i\right)}{65}$$

Multiply 5 times 8+i.

$$\frac{5\times 8+5i}{65}$$

Do the multiplications in 5\times 8+5i.

$$\frac{40+5i}{65}$$

Divide 40+5i by 65 to get \frac{8}{13}+\frac{1}{13}i.

$$\frac{8}{13}+\frac{1}{13}i$$

Real Part

$$\frac{8}{13} = 0.6153846153846154$$

Solution Steps

Multiply both numerator and denominator of \frac{5}{8-i} by the complex conjugate of the denominator, 8+i.

$$Re(\frac{5\left(8+i\right)}{\left(8-i\right)\left(8+i\right)})$$

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

$$Re(\frac{5\left(8+i\right)}{8^{2}-i^{2}})$$

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

$$Re(\frac{5\left(8+i\right)}{65})$$

Multiply 5 times 8+i.

$$Re(\frac{5\times 8+5i}{65})$$

Do the multiplications in 5\times 8+5i.

$$Re(\frac{40+5i}{65})$$

Divide 40+5i by 65 to get \frac{8}{13}+\frac{1}{13}i.

$$Re(\frac{8}{13}+\frac{1}{13}i)$$

The real part of \frac{8}{13}+\frac{1}{13}i is \frac{8}{13}.

$$\frac{8}{13}$$