Solve 9/4-i

$$\frac { 9 } { 4 - i }$$

Evaluate

$$\frac{36}{17}+\frac{9}{17}i\approx 2.117647059+0.529411765i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4+i.

$$\frac{9\left(4+i\right)}{\left(4-i\right)\left(4+i\right)}$$

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

$$\frac{9\left(4+i\right)}{4^{2}-i^{2}}$$

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

$$\frac{9\left(4+i\right)}{17}$$

Multiply 9 times 4+i.

$$\frac{9\times 4+9i}{17}$$

Do the multiplications in 9\times 4+9i.

$$\frac{36+9i}{17}$$

Divide 36+9i by 17 to get \frac{36}{17}+\frac{9}{17}i.

$$\frac{36}{17}+\frac{9}{17}i$$

Real Part

$$\frac{36}{17} = 2\frac{2}{17} = 2.1176470588235294$$

Solution Steps

Multiply both numerator and denominator of \frac{9}{4-i} by the complex conjugate of the denominator, 4+i.

$$Re(\frac{9\left(4+i\right)}{\left(4-i\right)\left(4+i\right)})$$

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

$$Re(\frac{9\left(4+i\right)}{4^{2}-i^{2}})$$

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

$$Re(\frac{9\left(4+i\right)}{17})$$

Multiply 9 times 4+i.

$$Re(\frac{9\times 4+9i}{17})$$

Do the multiplications in 9\times 4+9i.

$$Re(\frac{36+9i}{17})$$

Divide 36+9i by 17 to get \frac{36}{17}+\frac{9}{17}i.

$$Re(\frac{36}{17}+\frac{9}{17}i)$$

The real part of \frac{36}{17}+\frac{9}{17}i is \frac{36}{17}.

$$\frac{36}{17}$$