Solve 1/1+i+i

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

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.

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

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

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

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

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

Multiply 1 and 1-i to get 1-i.

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

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

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

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

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

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

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

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.

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

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

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

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

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

Multiply 1 and 1-i to get 1-i.

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

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

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

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

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

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

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

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

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