Solve i(1-1/1-i)

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

Evaluate

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

Solution Steps

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

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Multiply 1 and 1+i to get 1+i.

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

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

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

Combine the real and imaginary parts in numbers 1 and -\frac{1}{2}-\frac{1}{2}i.

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

Add 1 to -\frac{1}{2}.

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

Multiply i times \frac{1}{2}-\frac{1}{2}i.

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

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

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

Do the multiplications. Reorder the terms.

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

Real Part

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

Solution Steps

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

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Multiply 1 and 1+i to get 1+i.

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

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

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

Combine the real and imaginary parts in numbers 1 and -\frac{1}{2}-\frac{1}{2}i.

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

Add 1 to -\frac{1}{2}.

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

Multiply i times \frac{1}{2}-\frac{1}{2}i.

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

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

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

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

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

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

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