Solve 1/1-i-1

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

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.

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

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

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

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

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

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

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

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

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

Subtract 1 from \frac{1}{2}+\frac{1}{2}i by subtracting corresponding real and imaginary parts.

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

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

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

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.

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

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

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

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

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

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

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

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

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

Subtract 1 from \frac{1}{2}+\frac{1}{2}i by subtracting corresponding real and imaginary parts.

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

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

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

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

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