Solve (4+i/i-2+3i/1+2i)

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

Evaluate

$$-\frac{3}{5}-\frac{19}{5}i=-0.6-3.8i$$

Solution Steps

Multiply both numerator and denominator of \frac{4+i}{i} by imaginary unit i.

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

Divide -1+4i by -1 to get 1-4i.

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

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

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

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

$$1-4i-\frac{8-i}{5}$$

Divide 8-i by 5 to get \frac{8}{5}-\frac{1}{5}i.

$$1-4i+\left(-\frac{8}{5}+\frac{1}{5}i\right)$$

Add 1-4i and -\frac{8}{5}+\frac{1}{5}i to get -\frac{3}{5}-\frac{19}{5}i.

$$-\frac{3}{5}-\frac{19}{5}i$$

Real Part

$$-\frac{3}{5} = -0.6$$

Solution Steps

Multiply both numerator and denominator of \frac{4+i}{i} by imaginary unit i.

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

Divide -1+4i by -1 to get 1-4i.

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

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

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

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

$$Re(1-4i-\frac{8-i}{5})$$

Divide 8-i by 5 to get \frac{8}{5}-\frac{1}{5}i.

$$Re(1-4i+\left(-\frac{8}{5}+\frac{1}{5}i\right))$$

Add 1-4i and -\frac{8}{5}+\frac{1}{5}i to get -\frac{3}{5}-\frac{19}{5}i.

$$Re(-\frac{3}{5}-\frac{19}{5}i)$$

The real part of -\frac{3}{5}-\frac{19}{5}i is -\frac{3}{5}.

$$-\frac{3}{5}$$