Solve -2+i/2i

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

Evaluate

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

Solution Steps

Multiply both numerator and denominator by imaginary unit i.

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply -2+i times i.

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

By definition, i^{2} is -1.

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

Reorder the terms.

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

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

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

Real Part

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

Solution Steps

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

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

By definition, i^{2} is -1. Calculate the denominator.

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

Multiply -2+i times i.

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

By definition, i^{2} is -1.

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

Reorder the terms.

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

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

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

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

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