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

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

Evaluate

$$-2-i$$

Solution Steps

Calculate 2+i to the power of 2 and get 3+4i.

$$\frac{3+4i-\left(2+i\right)\left(2-i\right)}{\left(1-i\right)^{2}}$$

Multiply 2+i and 2-i to get 5.

$$\frac{3+4i-5}{\left(1-i\right)^{2}}$$

Subtract 5 from 3+4i to get -2+4i.

$$\frac{-2+4i}{\left(1-i\right)^{2}}$$

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

$$\frac{-2+4i}{-2i}$$

Multiply both numerator and denominator by imaginary unit i.

$$\frac{-4-2i}{2}$$

Divide -4-2i by 2 to get -2-i.

$$-2-i$$

Real Part

$$-2$$

Solution Steps

Calculate 2+i to the power of 2 and get 3+4i.

$$Re(\frac{3+4i-\left(2+i\right)\left(2-i\right)}{\left(1-i\right)^{2}})$$

Multiply 2+i and 2-i to get 5.

$$Re(\frac{3+4i-5}{\left(1-i\right)^{2}})$$

Subtract 5 from 3+4i to get -2+4i.

$$Re(\frac{-2+4i}{\left(1-i\right)^{2}})$$

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

$$Re(\frac{-2+4i}{-2i})$$

Multiply both numerator and denominator of \frac{-2+4i}{-2i} by imaginary unit i.

$$Re(\frac{-4-2i}{2})$$

Divide -4-2i by 2 to get -2-i.

$$Re(-2-i)$$

The real part of -2-i is -2.

$$-2$$