Solve 1+3/1+i

$$1 + \frac { 3 } { 1 + i }$$

Evaluate

$$\frac{5}{2}-\frac{3}{2}i=2.5-1.5i$$

Solution Steps

Multiply both numerator and denominator of \frac{3}{1+i} by the complex conjugate of the denominator, 1-i.

$$1+\frac{3\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

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

$$1+\frac{3\left(1-i\right)}{1^{2}-i^{2}}$$

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

$$1+\frac{3\left(1-i\right)}{2}$$

Multiply 3 times 1-i.

$$1+\frac{3\times 1+3\left(-i\right)}{2}$$

Do the multiplications in 3\times 1+3\left(-i\right).

$$1+\frac{3-3i}{2}$$

Divide 3-3i by 2 to get \frac{3}{2}-\frac{3}{2}i.

$$1+\left(\frac{3}{2}-\frac{3}{2}i\right)$$

Combine the real and imaginary parts in numbers 1 and \frac{3}{2}-\frac{3}{2}i.

$$1+\frac{3}{2}-\frac{3}{2}i$$

Add 1 to \frac{3}{2}.

$$\frac{5}{2}-\frac{3}{2}i$$

Real Part

$$\frac{5}{2} = 2\frac{1}{2} = 2.5$$

Solution Steps

Multiply both numerator and denominator of \frac{3}{1+i} by the complex conjugate of the denominator, 1-i.

$$Re(1+\frac{3\left(1-i\right)}{\left(1+i\right)\left(1-i\right)})$$

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

$$Re(1+\frac{3\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(1+\frac{3\left(1-i\right)}{2})$$

Multiply 3 times 1-i.

$$Re(1+\frac{3\times 1+3\left(-i\right)}{2})$$

Do the multiplications in 3\times 1+3\left(-i\right).

$$Re(1+\frac{3-3i}{2})$$

Divide 3-3i by 2 to get \frac{3}{2}-\frac{3}{2}i.

$$Re(1+\left(\frac{3}{2}-\frac{3}{2}i\right))$$

Combine the real and imaginary parts in numbers 1 and \frac{3}{2}-\frac{3}{2}i.

$$Re(1+\frac{3}{2}-\frac{3}{2}i)$$

Add 1 to \frac{3}{2}.

$$Re(\frac{5}{2}-\frac{3}{2}i)$$

The real part of \frac{5}{2}-\frac{3}{2}i is \frac{5}{2}.

$$\frac{5}{2}$$