Solve 2/1+i+2/1

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

Evaluate

$$3-i$$

Solution Steps

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

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

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

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

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

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

Multiply 2 times 1-i.

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

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

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

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

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

Anything divided by one gives itself.

$$1-i+2$$

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

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

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

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

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

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

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

Multiply 2 times 1-i.

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

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

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

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

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

Anything divided by one gives itself.

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

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.

$$3$$