Solve 11i/1-i-(i/1-i)^2

$$\frac { 11 i } { 1 - i } - ( \frac { i } { 1 - i } ) ^ { 2 }$$

Evaluate

$$-\frac{11}{2}+6i=-5.5+6i$$

Solution Steps

Multiply both numerator and denominator of \frac{11i}{1-i} by the complex conjugate of the denominator, 1+i.

$$\frac{11i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}-\left(\frac{i}{1-i}\right)^{2}$$

Do the multiplications in \frac{11i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

$$\frac{-11+11i}{2}-\left(\frac{i}{1-i}\right)^{2}$$

Divide -11+11i by 2 to get -\frac{11}{2}+\frac{11}{2}i.

$$-\frac{11}{2}+\frac{11}{2}i-\left(\frac{i}{1-i}\right)^{2}$$

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

$$-\frac{11}{2}+\frac{11}{2}i-\left(\frac{i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{2}$$

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

$$-\frac{11}{2}+\frac{11}{2}i-\left(\frac{-1+i}{2}\right)^{2}$$

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

$$-\frac{11}{2}+\frac{11}{2}i-\left(-\frac{1}{2}+\frac{1}{2}i\right)^{2}$$

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

$$-\frac{11}{2}+\frac{11}{2}i-\left(-\frac{1}{2}i\right)$$

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

$$-\frac{11}{2}+\frac{11}{2}i+\frac{1}{2}i$$

Add -\frac{11}{2}+\frac{11}{2}i and \frac{1}{2}i to get -\frac{11}{2}+6i.

$$-\frac{11}{2}+6i$$

Real Part

$$-\frac{11}{2} = -5\frac{1}{2} = -5.5$$

Solution Steps

Multiply both numerator and denominator of \frac{11i}{1-i} by the complex conjugate of the denominator, 1+i.

$$Re(\frac{11i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}-\left(\frac{i}{1-i}\right)^{2})$$

Do the multiplications in \frac{11i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

$$Re(\frac{-11+11i}{2}-\left(\frac{i}{1-i}\right)^{2})$$

Divide -11+11i by 2 to get -\frac{11}{2}+\frac{11}{2}i.

$$Re(-\frac{11}{2}+\frac{11}{2}i-\left(\frac{i}{1-i}\right)^{2})$$

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

$$Re(-\frac{11}{2}+\frac{11}{2}i-\left(\frac{i\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{2})$$

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

$$Re(-\frac{11}{2}+\frac{11}{2}i-\left(\frac{-1+i}{2}\right)^{2})$$

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

$$Re(-\frac{11}{2}+\frac{11}{2}i-\left(-\frac{1}{2}+\frac{1}{2}i\right)^{2})$$

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

$$Re(-\frac{11}{2}+\frac{11}{2}i-\left(-\frac{1}{2}i\right))$$

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

$$Re(-\frac{11}{2}+\frac{11}{2}i+\frac{1}{2}i)$$

Add -\frac{11}{2}+\frac{11}{2}i and \frac{1}{2}i to get -\frac{11}{2}+6i.

$$Re(-\frac{11}{2}+6i)$$

The real part of -\frac{11}{2}+6i is -\frac{11}{2}.

$$-\frac{11}{2}$$