Solve 9/5-i

$$\frac { 9 } { 5 - i }$$

Evaluate

$$\frac{45}{26}+\frac{9}{26}i\approx 1.730769231+0.346153846i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5+i.

$$\frac{9\left(5+i\right)}{\left(5-i\right)\left(5+i\right)}$$

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

$$\frac{9\left(5+i\right)}{5^{2}-i^{2}}$$

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

$$\frac{9\left(5+i\right)}{26}$$

Multiply 9 times 5+i.

$$\frac{9\times 5+9i}{26}$$

Do the multiplications in 9\times 5+9i.

$$\frac{45+9i}{26}$$

Divide 45+9i by 26 to get \frac{45}{26}+\frac{9}{26}i.

$$\frac{45}{26}+\frac{9}{26}i$$

Real Part

$$\frac{45}{26} = 1\frac{19}{26} = 1.7307692307692308$$

Solution Steps

Multiply both numerator and denominator of \frac{9}{5-i} by the complex conjugate of the denominator, 5+i.

$$Re(\frac{9\left(5+i\right)}{\left(5-i\right)\left(5+i\right)})$$

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

$$Re(\frac{9\left(5+i\right)}{5^{2}-i^{2}})$$

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

$$Re(\frac{9\left(5+i\right)}{26})$$

Multiply 9 times 5+i.

$$Re(\frac{9\times 5+9i}{26})$$

Do the multiplications in 9\times 5+9i.

$$Re(\frac{45+9i}{26})$$

Divide 45+9i by 26 to get \frac{45}{26}+\frac{9}{26}i.

$$Re(\frac{45}{26}+\frac{9}{26}i)$$

The real part of \frac{45}{26}+\frac{9}{26}i is \frac{45}{26}.

$$\frac{45}{26}$$