Solve 9/7-i

$$\frac { 9 } { 7 - i }$$

Evaluate

$$\frac{63}{50}+\frac{9}{50}i=1.26+0.18i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 7+i.

$$\frac{9\left(7+i\right)}{\left(7-i\right)\left(7+i\right)}$$

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

$$\frac{9\left(7+i\right)}{7^{2}-i^{2}}$$

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

$$\frac{9\left(7+i\right)}{50}$$

Multiply 9 times 7+i.

$$\frac{9\times 7+9i}{50}$$

Do the multiplications in 9\times 7+9i.

$$\frac{63+9i}{50}$$

Divide 63+9i by 50 to get \frac{63}{50}+\frac{9}{50}i.

$$\frac{63}{50}+\frac{9}{50}i$$

Real Part

$$\frac{63}{50} = 1\frac{13}{50} = 1.26$$

Solution Steps

Multiply both numerator and denominator of \frac{9}{7-i} by the complex conjugate of the denominator, 7+i.

$$Re(\frac{9\left(7+i\right)}{\left(7-i\right)\left(7+i\right)})$$

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

$$Re(\frac{9\left(7+i\right)}{7^{2}-i^{2}})$$

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

$$Re(\frac{9\left(7+i\right)}{50})$$

Multiply 9 times 7+i.

$$Re(\frac{9\times 7+9i}{50})$$

Do the multiplications in 9\times 7+9i.

$$Re(\frac{63+9i}{50})$$

Divide 63+9i by 50 to get \frac{63}{50}+\frac{9}{50}i.

$$Re(\frac{63}{50}+\frac{9}{50}i)$$

The real part of \frac{63}{50}+\frac{9}{50}i is \frac{63}{50}.

$$\frac{63}{50}$$