Solve 1*1/1+i

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

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.

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

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

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

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

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

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

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

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

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

Multiply 1 and \frac{1}{2}-\frac{1}{2}i to get \frac{1}{2}-\frac{1}{2}i.

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

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.

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

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

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

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

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

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

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

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

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

Multiply 1 and \frac{1}{2}-\frac{1}{2}i to get \frac{1}{2}-\frac{1}{2}i.

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

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

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