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

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

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

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

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

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

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

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

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

Multiply both numerator and denominator of \frac{1}{2i} by imaginary unit i.

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

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

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

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

$$\frac{1}{2}$$

Real Part

$$\frac{1}{2} = 0.5$$

Solution Steps

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

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

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

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

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

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

Multiply both numerator and denominator of \frac{1}{2i} by imaginary unit i.

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

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

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

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

$$Re(\frac{1}{2})$$

The real part of \frac{1}{2} is \frac{1}{2}.

$$\frac{1}{2}$$