SOLUTION: What is the number of distinct solution of the equation {{{z^2+abs(z)}}} = {{{0}}} ?

Let us write the number z in Trigonometry form: z = .

Then  = , and your equation is

 = ,   or, which is the same,

 = 0.


The last equation deploys in two independent equations


1)  r = 0,   which means simply  z = 0.


2)  (r*(cos(2a) + i*sin(2a)) + 1 = 0,  which is the same as

    r*(cos(2a) + i*sin(2a) = -1.


    The last equation implies 

    r = 1,  cos(2a) = -1 and sin(2a) = 0,   which, in turn, implies

    r = 1  and  {2a = pi   OR  2a =  }.

    
    It means  that the solutions are  

           a) r = 1,  a = ,     OR/AND

           b) r = 1,  a = .


    In the rectangular form these solutions are  z = i   OR/AND  z = -i.


Answer.  The original equation has three solutions: z = 0;  z = i,  and  z = -i.