Solve 1+2i/3-4i+2-i/5i

$$\frac { 1 + 2 i } { 3 - 4 i } + \frac { 2 - i } { 5 i }$$

Evaluate

$$-\frac{2}{5}=-0.4$$

Solution Steps

Multiply both numerator and denominator of \frac{1+2i}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$\frac{\left(1+2i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}+\frac{2-i}{5i}$$

Do the multiplications in \frac{\left(1+2i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}.

$$\frac{-5+10i}{25}+\frac{2-i}{5i}$$

Divide -5+10i by 25 to get -\frac{1}{5}+\frac{2}{5}i.

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

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

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

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

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

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

$$-\frac{2}{5}$$

Real Part

$$-\frac{2}{5} = -0.4$$

Solution Steps

Multiply both numerator and denominator of \frac{1+2i}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$Re(\frac{\left(1+2i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}+\frac{2-i}{5i})$$

Do the multiplications in \frac{\left(1+2i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}.

$$Re(\frac{-5+10i}{25}+\frac{2-i}{5i})$$

Divide -5+10i by 25 to get -\frac{1}{5}+\frac{2}{5}i.

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

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

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

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

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

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

$$Re(-\frac{2}{5})$$

The real part of -\frac{2}{5} is -\frac{2}{5}.

$$-\frac{2}{5}$$