Solve 9/1-4i

$$\frac { 9 } { 1 - 4 i }$$

Evaluate

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

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+4i.

$$\frac{9\left(1+4i\right)}{\left(1-4i\right)\left(1+4i\right)}$$

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

$$\frac{9\left(1+4i\right)}{1^{2}-4^{2}i^{2}}$$

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

$$\frac{9\left(1+4i\right)}{17}$$

Multiply 9 times 1+4i.

$$\frac{9\times 1+9\times \left(4i\right)}{17}$$

Do the multiplications in 9\times 1+9\times \left(4i\right).

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

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

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

Real Part

$$\frac{9}{17} = 0.5294117647058824$$

Solution Steps

Multiply both numerator and denominator of \frac{9}{1-4i} by the complex conjugate of the denominator, 1+4i.

$$Re(\frac{9\left(1+4i\right)}{\left(1-4i\right)\left(1+4i\right)})$$

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

$$Re(\frac{9\left(1+4i\right)}{1^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{9\left(1+4i\right)}{17})$$

Multiply 9 times 1+4i.

$$Re(\frac{9\times 1+9\times \left(4i\right)}{17})$$

Do the multiplications in 9\times 1+9\times \left(4i\right).

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

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

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

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

$$\frac{9}{17}$$