Solve 1+7i/left(2-iright)^2

$$\frac{ 1+7i }{ { \left(2-i \right) }^{ 2 } }$$

Evaluate

$$-1+i$$

Solution Steps

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

$$\frac{1+7i}{3-4i}$$

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

$$\frac{\left(1+7i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}$$

Do the multiplications in \frac{\left(1+7i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}.

$$\frac{-25+25i}{25}$$

Divide -25+25i by 25 to get -1+i.

$$-1+i$$

Real Part

$$-1$$

Solution Steps

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

$$Re(\frac{1+7i}{3-4i})$$

Multiply both numerator and denominator of \frac{1+7i}{3-4i} by the complex conjugate of the denominator, 3+4i.

$$Re(\frac{\left(1+7i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)})$$

Do the multiplications in \frac{\left(1+7i\right)\left(3+4i\right)}{\left(3-4i\right)\left(3+4i\right)}.

$$Re(\frac{-25+25i}{25})$$

Divide -25+25i by 25 to get -1+i.

$$Re(-1+i)$$

The real part of -1+i is -1.

$$-1$$