Solve 1+i/1-i-(i/1-i)^2/1+i

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

Evaluate

$$\frac{3}{4}+\frac{3}{4}i=0.75+0.75i$$

Solution Steps

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

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

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

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

Divide 2i by 2 to get i.

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

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

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

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

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

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

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

Calculate -\frac{1}{2}+\frac{1}{2}i to the power of 2 and get -\frac{1}{2}i.

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

The opposite of -\frac{1}{2}i is \frac{1}{2}i.

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

Add i and \frac{1}{2}i to get \frac{3}{2}i.

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

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-i.

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

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

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

Divide \frac{3}{2}+\frac{3}{2}i by 2 to get \frac{3}{4}+\frac{3}{4}i.

$$\frac{3}{4}+\frac{3}{4}i$$

Real Part

$$\frac{3}{4} = 0.75$$

Solution Steps

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

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

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

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

Divide 2i by 2 to get i.

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

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

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

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

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

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

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

Calculate -\frac{1}{2}+\frac{1}{2}i to the power of 2 and get -\frac{1}{2}i.

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

The opposite of -\frac{1}{2}i is \frac{1}{2}i.

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

Add i and \frac{1}{2}i to get \frac{3}{2}i.

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

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

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

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

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

Divide \frac{3}{2}+\frac{3}{2}i by 2 to get \frac{3}{4}+\frac{3}{4}i.

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

The real part of \frac{3}{4}+\frac{3}{4}i is \frac{3}{4}.

$$\frac{3}{4}$$