Solve 2-1/3+i

$$\frac { 2 - 1 } { 3 + i }$$

Evaluate

$$\frac{3}{10}-\frac{1}{10}i=0.3-0.1i$$

Solution Steps

Subtract 1 from 2 to get 1.

$$\frac{1}{3+i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-i.

$$\frac{1\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}$$

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

$$\frac{1\left(3-i\right)}{3^{2}-i^{2}}$$

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

$$\frac{1\left(3-i\right)}{10}$$

Multiply 1 and 3-i to get 3-i.

$$\frac{3-i}{10}$$

Divide 3-i by 10 to get \frac{3}{10}-\frac{1}{10}i.

$$\frac{3}{10}-\frac{1}{10}i$$

Real Part

$$\frac{3}{10} = 0.3$$

Solution Steps

Subtract 1 from 2 to get 1.

$$Re(\frac{1}{3+i})$$

Multiply both numerator and denominator of \frac{1}{3+i} by the complex conjugate of the denominator, 3-i.

$$Re(\frac{1\left(3-i\right)}{\left(3+i\right)\left(3-i\right)})$$

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

$$Re(\frac{1\left(3-i\right)}{3^{2}-i^{2}})$$

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

$$Re(\frac{1\left(3-i\right)}{10})$$

Multiply 1 and 3-i to get 3-i.

$$Re(\frac{3-i}{10})$$

Divide 3-i by 10 to get \frac{3}{10}-\frac{1}{10}i.

$$Re(\frac{3}{10}-\frac{1}{10}i)$$

The real part of \frac{3}{10}-\frac{1}{10}i is \frac{3}{10}.

$$\frac{3}{10}$$