Solve 1+2i/3-i+2-i/5i

$$\frac { 1 + 2 i } { 3 - i } + \frac { 2 - i } { 5 i }$$

Evaluate

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

Solution Steps

Multiply both numerator and denominator of \frac{1+2i}{3-i} by the complex conjugate of the denominator, 3+i.

$$\frac{\left(1+2i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}+\frac{2-i}{5i}$$

Do the multiplications in \frac{\left(1+2i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}.

$$\frac{1+7i}{10}+\frac{2-i}{5i}$$

Divide 1+7i by 10 to get \frac{1}{10}+\frac{7}{10}i.

$$\frac{1}{10}+\frac{7}{10}i+\frac{2-i}{5i}$$

Multiply both numerator and denominator of \frac{2-i}{5i} by imaginary unit i.

$$\frac{1}{10}+\frac{7}{10}i+\frac{1+2i}{-5}$$

Divide 1+2i by -5 to get -\frac{1}{5}-\frac{2}{5}i.

$$\frac{1}{10}+\frac{7}{10}i+\left(-\frac{1}{5}-\frac{2}{5}i\right)$$

Add \frac{1}{10}+\frac{7}{10}i and -\frac{1}{5}-\frac{2}{5}i to get -\frac{1}{10}+\frac{3}{10}i.

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

Real Part

$$-\frac{1}{10} = -0.1$$

Solution Steps

Multiply both numerator and denominator of \frac{1+2i}{3-i} by the complex conjugate of the denominator, 3+i.

$$Re(\frac{\left(1+2i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}+\frac{2-i}{5i})$$

Do the multiplications in \frac{\left(1+2i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}.

$$Re(\frac{1+7i}{10}+\frac{2-i}{5i})$$

Divide 1+7i by 10 to get \frac{1}{10}+\frac{7}{10}i.

$$Re(\frac{1}{10}+\frac{7}{10}i+\frac{2-i}{5i})$$

Multiply both numerator and denominator of \frac{2-i}{5i} by imaginary unit i.

$$Re(\frac{1}{10}+\frac{7}{10}i+\frac{1+2i}{-5})$$

Divide 1+2i by -5 to get -\frac{1}{5}-\frac{2}{5}i.

$$Re(\frac{1}{10}+\frac{7}{10}i+\left(-\frac{1}{5}-\frac{2}{5}i\right))$$

Add \frac{1}{10}+\frac{7}{10}i and -\frac{1}{5}-\frac{2}{5}i to get -\frac{1}{10}+\frac{3}{10}i.

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

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

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