# Complex root: question about number of solution

I have this complex number:

• $z=(1-2i)^{2/4}$

I solved it, and according to wolfram, two solutions are ok, but I find 4 solution: isn't the 4th root = 4 solutions? It's like $z=(+-(1-2i))^{1/2}$ , or am I wrong? Why?

I think that by operator precedence rules $z=(1-2i)^{2/4}$ simplifies to $z=(1-2i)^{1/2}$ not $z=[(1-2i)^{2}]^{1/4}$.