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

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

Solve for z

$$z=\frac{48}{25}i=1.92i$$

Steps for Solving Linear Equation

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

$$z=\left(\frac{\left(2+i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}\right)^{2}-\left(\frac{2-i}{2+i}\right)^{2}$$

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

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

Divide 3+4i by 5 to get \frac{3}{5}+\frac{4}{5}i.

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

Calculate \frac{3}{5}+\frac{4}{5}i to the power of 2 and get -\frac{7}{25}+\frac{24}{25}i.

$$z=-\frac{7}{25}+\frac{24}{25}i-\left(\frac{2-i}{2+i}\right)^{2}$$

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

$$z=-\frac{7}{25}+\frac{24}{25}i-\left(\frac{\left(2-i\right)\left(2-i\right)}{\left(2+i\right)\left(2-i\right)}\right)^{2}$$

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

$$z=-\frac{7}{25}+\frac{24}{25}i-\left(\frac{3-4i}{5}\right)^{2}$$

Divide 3-4i by 5 to get \frac{3}{5}-\frac{4}{5}i.

$$z=-\frac{7}{25}+\frac{24}{25}i-\left(\frac{3}{5}-\frac{4}{5}i\right)^{2}$$

Calculate \frac{3}{5}-\frac{4}{5}i to the power of 2 and get -\frac{7}{25}-\frac{24}{25}i.

$$z=-\frac{7}{25}+\frac{24}{25}i-\left(-\frac{7}{25}-\frac{24}{25}i\right)$$

The opposite of -\frac{7}{25}-\frac{24}{25}i is \frac{7}{25}+\frac{24}{25}i.

$$z=-\frac{7}{25}+\frac{24}{25}i+\left(\frac{7}{25}+\frac{24}{25}i\right)$$

Add -\frac{7}{25}+\frac{24}{25}i and \frac{7}{25}+\frac{24}{25}i to get \frac{48}{25}i.

$$z=\frac{48}{25}i$$

Assign z

$$z≔\frac{48}{25}i$$