Solve 6/9+i

$$\frac { 6 } { 9 + i }$$

Evaluate

$$\frac{27}{41}-\frac{3}{41}i\approx 0.658536585-0.073170732i$$

Solution Steps

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

$$\frac{6\left(9-i\right)}{\left(9+i\right)\left(9-i\right)}$$

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

$$\frac{6\left(9-i\right)}{9^{2}-i^{2}}$$

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

$$\frac{6\left(9-i\right)}{82}$$

Multiply 6 times 9-i.

$$\frac{6\times 9+6\left(-i\right)}{82}$$

Do the multiplications in 6\times 9+6\left(-i\right).

$$\frac{54-6i}{82}$$

Divide 54-6i by 82 to get \frac{27}{41}-\frac{3}{41}i.

$$\frac{27}{41}-\frac{3}{41}i$$

Real Part

$$\frac{27}{41} = 0.6585365853658537$$

Solution Steps

Multiply both numerator and denominator of \frac{6}{9+i} by the complex conjugate of the denominator, 9-i.

$$Re(\frac{6\left(9-i\right)}{\left(9+i\right)\left(9-i\right)})$$

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

$$Re(\frac{6\left(9-i\right)}{9^{2}-i^{2}})$$

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

$$Re(\frac{6\left(9-i\right)}{82})$$

Multiply 6 times 9-i.

$$Re(\frac{6\times 9+6\left(-i\right)}{82})$$

Do the multiplications in 6\times 9+6\left(-i\right).

$$Re(\frac{54-6i}{82})$$

Divide 54-6i by 82 to get \frac{27}{41}-\frac{3}{41}i.

$$Re(\frac{27}{41}-\frac{3}{41}i)$$

The real part of \frac{27}{41}-\frac{3}{41}i is \frac{27}{41}.

$$\frac{27}{41}$$