Solve 1/3-i--1-2i/4

$$\frac { 1 } { 3 - i } - \frac { - 1 - 2 i } { 4 }$$

Evaluate

$$\frac{11}{20}+\frac{3}{5}i=0.55+0.6i$$

Solution Steps

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

$$\frac{1\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}-\frac{-1-2i}{4}$$

Do the multiplications in \frac{1\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}.

$$\frac{3+i}{10}-\frac{-1-2i}{4}$$

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

$$\frac{3}{10}+\frac{1}{10}i-\frac{-1-2i}{4}$$

Divide -1-2i by 4 to get -\frac{1}{4}-\frac{1}{2}i.

$$\frac{3}{10}+\frac{1}{10}i+\left(\frac{1}{4}+\frac{1}{2}i\right)$$

Add \frac{3}{10}+\frac{1}{10}i and \frac{1}{4}+\frac{1}{2}i to get \frac{11}{20}+\frac{3}{5}i.

$$\frac{11}{20}+\frac{3}{5}i$$

Real Part

$$\frac{11}{20} = 0.55$$

Solution Steps

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

$$Re(\frac{1\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}-\frac{-1-2i}{4})$$

Do the multiplications in \frac{1\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}.

$$Re(\frac{3+i}{10}-\frac{-1-2i}{4})$$

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

$$Re(\frac{3}{10}+\frac{1}{10}i-\frac{-1-2i}{4})$$

Divide -1-2i by 4 to get -\frac{1}{4}-\frac{1}{2}i.

$$Re(\frac{3}{10}+\frac{1}{10}i+\left(\frac{1}{4}+\frac{1}{2}i\right))$$

Add \frac{3}{10}+\frac{1}{10}i and \frac{1}{4}+\frac{1}{2}i to get \frac{11}{20}+\frac{3}{5}i.

$$Re(\frac{11}{20}+\frac{3}{5}i)$$

The real part of \frac{11}{20}+\frac{3}{5}i is \frac{11}{20}.

$$\frac{11}{20}$$