(1)Can the proof by induction be in Quaternion?
(2a)Can be induction on complex number?
(2b)There is a general way to use induction on complex number?
"Proof by induction" can be used in any "inductive set". That is a set in which:
1) there is a "first" member.
2) given any member there is a well-defined way of selecting a "next" number.
While there certainly are "inductive" subsets of the complex numbers and quaternions, those sets themselves are not inductive (neither are the rational numbers nor the real numbers).
For (1) and (2), the obvious example is the set of positive integers which is an inductive subset of both the complex numbers and the quaternions (the "first" member is 1 and, given any positive integer, x, the "next" member is x+1). Almost as obvious is the set of negative integers, taking the "first" member to be -1 and, for each negative integer x, the "next" member to be x-1.
I don't know what you mean by "inductive expression". I did not use the word "expression". However, both the complex numbers and the quaternions include the real numbers as a subset and the real numbers cannot be "ordered" so cannot be an inductive set. If a set is not inductive neither is any superset.
You can read up on someone's favorite transfinite proofs here:
Interesting Proofs Using Transfinite Induction