replacing complex numbers

Question

I know: I'm going to make a poor showing, but really I can't understand this:

a is an expressione whose FullForm is

Power[Plus[Subscript[u,x],Times[Complex[0,-1],Subscript[u,y]],Times[Complex[0,1],Subscript[v,x]],Subscript[v,y]],2]

Why does the following code return True, instead of the expected substitution ?

b = Replace[a, {Complex[a_, b_] -> a + H b}]
a == b

I have also tried

b = Replace[a, {Complex[a_, b_] -> a + H b},levSpec]
a == b

(even Infinity included) but without succeeding.

Thanks in advance !!

Answer 1

The original formulation was close, and the level spec of Infinity "almost" worked. As was noted in comments, it does work if Rule is replaced by RuleDelayed. The reason it otherwise causes trouble is from a "variable capture" in scoping. The pattern variables, a_ and b_, have the same names as expressions under consideration. With Rule the rhs is evaluated immediately and the replacement becomes something unwanted. So yet another way to go about this is as below.

Clear[a, b, aa, bb]
a = Power[
   Plus[Subscript[u, x], Times[Complex[0, -1], Subscript[u, y]], 
    Times[Complex[0, 1], Subscript[v, x]], Subscript[v, y]], 2];
b = Replace[a, {Complex[aa_, bb_] -> aa + H bb}, Infinity]

(* Out[534]= (Subscript[u, x] - H Subscript[u, y] + H Subscript[v, x] + 
  Subscript[v, y])^2 *)

Moral: Be careful with pattern variable naming.

Answer 2

a = Power[
  Plus[Subscript[u, x], Times[Complex[0, -1], Subscript[u, y]], 
   Times[Complex[0, 1], Subscript[v, x]], Subscript[v, y]], 2]

b = ComplexExpand@a /. I -> H

Answer 3

Assuming this is what you are going for,

You can get there two ways,

Replace[a, {I x_ :> H x, -I x_ :> - H x}, Infinity]

or

a /. {I x_ :> H x, -I x_ :> - H x}