Solve 1+1/1-i+2

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

Evaluate

$$3+i$$

Solution Steps

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

Multiply 2 times 1+i.

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

Do the multiplications in 2\times 1+2i.

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

Divide 2+2i by 2 to get 1+i.

$$1+i+2$$

Combine the real and imaginary parts in numbers 1+i and 2.

$$1+2+i$$

Add 1 to 2.

$$3+i$$

Real Part

$$3$$

Solution Steps

Add 1 and 1 to get 2.

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

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

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

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

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

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

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

Multiply 2 times 1+i.

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

Do the multiplications in 2\times 1+2i.

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

Divide 2+2i by 2 to get 1+i.

$$Re(1+i+2)$$

Combine the real and imaginary parts in numbers 1+i and 2.

$$Re(1+2+i)$$

Add 1 to 2.

$$Re(3+i)$$

The real part of 3+i is 3.

$$3$$