Solve operatorname{Re}(1/4+i)

$$\operatorname { Re } ( \frac { 1 } { 4 + i } )$$

Evaluate

$$\frac{4}{17}\approx 0.235294118$$

Solution Steps

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

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

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

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

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

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

Multiply 1 and 4-i to get 4-i.

$$Re(\frac{4-i}{17})$$

Divide 4-i by 17 to get \frac{4}{17}-\frac{1}{17}i.

$$Re(\frac{4}{17}-\frac{1}{17}i)$$

The real part of \frac{4}{17}-\frac{1}{17}i is \frac{4}{17}.

$$\frac{4}{17}$$

Real Part

$$\frac{4}{17} = 0.23529411764705882$$

Solution Steps

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

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

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

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

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

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

Multiply 1 and 4-i to get 4-i.

$$Re(Re(\frac{4-i}{17}))$$

Divide 4-i by 17 to get \frac{4}{17}-\frac{1}{17}i.

$$Re(Re(\frac{4}{17}-\frac{1}{17}i))$$

The real part of \frac{4}{17}-\frac{1}{17}i is \frac{4}{17}.

$$Re(\frac{4}{17})$$

The real part of \frac{4}{17} is \frac{4}{17}.

$$\frac{4}{17}$$