Solve (2-i)^2/(3-i)^2

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

Evaluate

$$\frac{12}{25}-\frac{7}{50}i=0.48-0.14i$$

Solution Steps

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

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

Calculate 3-i to the power of 2 and get 8-6i.

$$\frac{3-4i}{8-6i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 8+6i.

$$\frac{\left(3-4i\right)\left(8+6i\right)}{\left(8-6i\right)\left(8+6i\right)}$$

Do the multiplications in \frac{\left(3-4i\right)\left(8+6i\right)}{\left(8-6i\right)\left(8+6i\right)}.

$$\frac{48-14i}{100}$$

Divide 48-14i by 100 to get \frac{12}{25}-\frac{7}{50}i.

$$\frac{12}{25}-\frac{7}{50}i$$

Real Part

$$\frac{12}{25} = 0.48$$

Solution Steps

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

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

Calculate 3-i to the power of 2 and get 8-6i.

$$Re(\frac{3-4i}{8-6i})$$

Multiply both numerator and denominator of \frac{3-4i}{8-6i} by the complex conjugate of the denominator, 8+6i.

$$Re(\frac{\left(3-4i\right)\left(8+6i\right)}{\left(8-6i\right)\left(8+6i\right)})$$

Do the multiplications in \frac{\left(3-4i\right)\left(8+6i\right)}{\left(8-6i\right)\left(8+6i\right)}.

$$Re(\frac{48-14i}{100})$$

Divide 48-14i by 100 to get \frac{12}{25}-\frac{7}{50}i.

$$Re(\frac{12}{25}-\frac{7}{50}i)$$

The real part of \frac{12}{25}-\frac{7}{50}i is \frac{12}{25}.

$$\frac{12}{25}$$