Question 1028076
Use the fact that *[tex \large 2i = 2e^{i \frac{\pi}{2}}]


Since *[tex \large z^2 = 2i] you can conclude that the magnitude of z is *[tex \large \sqrt{2}].


Then you can conclude that *[tex \large z = 1+i] and *[tex \large z = -1-i] are the solutions. Since the equation is quadratic, it must have at most two complex solutions, including multiple roots.