Solve z=(1+i2)(2-i)+1/(2-i)^2-2+i

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

Solve for z

$$z=-\frac{2}{5}+\frac{9}{5}i=-0.4+1.8i$$

Steps for Solving Linear Equation

Multiply 1+2i and 2-i to get 4+3i.

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

Add 4+3i and 1 to get 5+3i.

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

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

$$z=\frac{5+3i}{3-4i-2+i}$$

Do the additions in 3-4i-2+i.

$$z=\frac{5+3i}{1-3i}$$

Multiply both numerator and denominator of \frac{5+3i}{1-3i} by the complex conjugate of the denominator, 1+3i.

$$z=\frac{\left(5+3i\right)\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}$$

Do the multiplications in \frac{\left(5+3i\right)\left(1+3i\right)}{\left(1-3i\right)\left(1+3i\right)}.

$$z=\frac{-4+18i}{10}$$

Divide -4+18i by 10 to get -\frac{2}{5}+\frac{9}{5}i.

$$z=-\frac{2}{5}+\frac{9}{5}i$$

Assign z

$$z≔-\frac{2}{5}+\frac{9}{5}i$$