Solve =(1+i)^2+3(1-i)/2+i

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

Evaluate

$$1-i$$

Solution Steps

Calculate 1+i to the power of 2 and get 2i.

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

Multiply 3 and 1-i to get 3-3i.

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

Add 2i and 3-3i to get 3-i.

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

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-i.

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

Do the multiplications in \frac{\left(3-i\right)\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}.

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

Divide 5-5i by 5 to get 1-i.

$$1-i$$

Real Part

$$1$$

Solution Steps

Calculate 1+i to the power of 2 and get 2i.

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

Multiply 3 and 1-i to get 3-3i.

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

Add 2i and 3-3i to get 3-i.

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

Multiply both numerator and denominator of \frac{3-i}{2+i} by the complex conjugate of the denominator, 2-i.

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

Do the multiplications in \frac{\left(3-i\right)\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}.

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

Divide 5-5i by 5 to get 1-i.

$$Re(1-i)$$

The real part of 1-i is 1.

$$1$$