Solve 2+i/i-3/2-i

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

Evaluate

$$-\frac{1}{5}-\frac{13}{5}i=-0.2-2.6i$$

Solution Steps

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

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

Divide -1+2i by -1 to get 1-2i.

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

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

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

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

$$1-2i-\frac{6+3i}{5}$$

Divide 6+3i by 5 to get \frac{6}{5}+\frac{3}{5}i.

$$1-2i+\left(-\frac{6}{5}-\frac{3}{5}i\right)$$

Add 1-2i and -\frac{6}{5}-\frac{3}{5}i to get -\frac{1}{5}-\frac{13}{5}i.

$$-\frac{1}{5}-\frac{13}{5}i$$

Real Part

$$-\frac{1}{5} = -0.2$$

Solution Steps

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

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

Divide -1+2i by -1 to get 1-2i.

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

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

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

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

$$Re(1-2i-\frac{6+3i}{5})$$

Divide 6+3i by 5 to get \frac{6}{5}+\frac{3}{5}i.

$$Re(1-2i+\left(-\frac{6}{5}-\frac{3}{5}i\right))$$

Add 1-2i and -\frac{6}{5}-\frac{3}{5}i to get -\frac{1}{5}-\frac{13}{5}i.

$$Re(-\frac{1}{5}-\frac{13}{5}i)$$

The real part of -\frac{1}{5}-\frac{13}{5}i is -\frac{1}{5}.

$$-\frac{1}{5}$$