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

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

Evaluate

$$-\frac{7}{25}-\frac{24}{25}i=-0.28-0.96i$$

Solution Steps

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

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

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

$$\frac{3-4i}{3+4i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-4i.

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

Do the multiplications in \frac{\left(3-4i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}.

$$\frac{-7-24i}{25}$$

Divide -7-24i by 25 to get -\frac{7}{25}-\frac{24}{25}i.

$$-\frac{7}{25}-\frac{24}{25}i$$

Real Part

$$-\frac{7}{25} = -0.28$$

Solution Steps

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

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

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

$$Re(\frac{3-4i}{3+4i})$$

Multiply both numerator and denominator of \frac{3-4i}{3+4i} by the complex conjugate of the denominator, 3-4i.

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

Do the multiplications in \frac{\left(3-4i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}.

$$Re(\frac{-7-24i}{25})$$

Divide -7-24i by 25 to get -\frac{7}{25}-\frac{24}{25}i.

$$Re(-\frac{7}{25}-\frac{24}{25}i)$$

The real part of -\frac{7}{25}-\frac{24}{25}i is -\frac{7}{25}.

$$-\frac{7}{25}$$