First, what made you think, originally, that there were four square roots of i?
Second, while the modulus of z= i is 1, its argument is NOT [itex]\pi/4[/itex]. I don't know where you got that "1" in " ". Surely you don't think argument is the arctangent of the modulus? The argument of "a+ bi" is the arctangent of b/a, if a is not 0. Of course, if a= 0 then a+ bi= bi which has modulus if b> 0, if b< 0. (0 has no "argument").
So if z= i, . HALF of that is and half of is . Those are the arguments of the two square roots of i which, of course, have modulus , so the two square roots are and .