Solve 8/9-i

$$\frac { 8 } { 9 - i }$$

Evaluate

$$\frac{36}{41}+\frac{4}{41}i\approx 0.87804878+0.097560976i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 9+i.

$$\frac{8\left(9+i\right)}{\left(9-i\right)\left(9+i\right)}$$

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

$$\frac{8\left(9+i\right)}{9^{2}-i^{2}}$$

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

$$\frac{8\left(9+i\right)}{82}$$

Multiply 8 times 9+i.

$$\frac{8\times 9+8i}{82}$$

Do the multiplications in 8\times 9+8i.

$$\frac{72+8i}{82}$$

Divide 72+8i by 82 to get \frac{36}{41}+\frac{4}{41}i.

$$\frac{36}{41}+\frac{4}{41}i$$

Real Part

$$\frac{36}{41} = 0.8780487804878049$$

Solution Steps

Multiply both numerator and denominator of \frac{8}{9-i} by the complex conjugate of the denominator, 9+i.

$$Re(\frac{8\left(9+i\right)}{\left(9-i\right)\left(9+i\right)})$$

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

$$Re(\frac{8\left(9+i\right)}{9^{2}-i^{2}})$$

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

$$Re(\frac{8\left(9+i\right)}{82})$$

Multiply 8 times 9+i.

$$Re(\frac{8\times 9+8i}{82})$$

Do the multiplications in 8\times 9+8i.

$$Re(\frac{72+8i}{82})$$

Divide 72+8i by 82 to get \frac{36}{41}+\frac{4}{41}i.

$$Re(\frac{36}{41}+\frac{4}{41}i)$$

The real part of \frac{36}{41}+\frac{4}{41}i is \frac{36}{41}.

$$\frac{36}{41}$$