Solve 2/1+i

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

Evaluate

$$1-i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-i.

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

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

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

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

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

Multiply 2 times 1-i.

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

Do the multiplications in 2\times 1+2\left(-i\right).

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

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

$$1-i$$

Real Part

$$1$$

Solution Steps

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

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

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

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

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

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

Multiply 2 times 1-i.

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

Do the multiplications in 2\times 1+2\left(-i\right).

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

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

$$Re(1-i)$$

The real part of 1-i is 1.

$$1$$