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

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

Evaluate

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

Solution Steps

Calculate i to the power of 2 and get -1.

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

The opposite of -1 is 1.

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

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

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

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

Do the additions.

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

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

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

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

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

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

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

Real Part

$$0$$

Solution Steps

Calculate i to the power of 2 and get -1.

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

The opposite of -1 is 1.

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

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The real part of -\frac{1}{2}i is 0.

$$0$$