Question( 5 points) Find the solution set of the equation $$(1+i) z^{2}-2 z=(-1+i)$$

Studdy Solution
Simplify the expression for $z$.
Separate into two parts: $$z_1 = \frac{2 + 2i}{2+2i}$$ $$z_2 = \frac{2 - 2i}{2+2i}$$
Simplify $z_1$: $$z_1 = 1$$
Simplify $z_2$ using the conjugate: Multiply numerator and denominator by the conjugate of the denominator: $$z_2 = \frac{(2-2i)(2-2i)}{(2+2i)(2-2i)}$$ $$z_2 = \frac{4 - 8i + 4i^2}{4 + 4}$$ $$z_2 = \frac{4 - 8i - 4}{8}$$ $$z_2 = \frac{-8i}{8}$$ $$z_2 = -i$$
The solution set for $z$ is: $$\boxed{\{1, -i\}}$$