In this section we investigate some conditions under which a bicomplex holomor- phic function reduces to a constant function.
Theorem 8.2.1. LetF : Ω→BCbe a bicomplex holomorphic function on a product- type domain Ω. Then F(Z) = 0 for all Z ∈ Ω if and only if F is a constant function.
Proof. If F is a constant function, its derivative is zero on Ω. Conversely, using the notation and result of Corollary 7.6.6, we are assured that
0 =F(Z) =G1(β1)e+G2(β2)e†,
for allZ=β1e+β2e† ∈Ω. This is equivalent toG1(β1) = 0 on Ω1andG2(β2) = 0 on Ω2. Because Ω1 and Ω2 are domains in the complex plane, this implies that G1, G2 are constant functions with respect to the variables β1, β2, respectively, i.e., G1(β1) =a1 ∈C(i) for all β1 ∈Ω1 and G2(β2) = a2∈C(i) for all β2 ∈Ω2, thereforeF(Z) =a1e+a2e†, a bicomplex constant for allZ ∈Ω.
Remark 8.2.2. The proof of the theorem above could have been obtained without using the idempotent representation of F. For example, writing F =f1+jf2 on Ω, the existence ofF onΩimplies the existence of the complex partial derivatives of f1 andf2 onΩwith respect to z1 andz2, related by the Cauchy-Riemann type conditions (7.55). Using the fact F(Z) = ∂F
∂Z = 0, we can derive easily that the partial derivatives of f1 and f2 with respect to both z1 and z2 are zero, which implies that f1 andf2 are constant on the domain Ω, soF is constant onΩ.
Now we are interested in examining under what conditions a bicomplex holo- morphic functionF has zero derivative on a domain Ω.
Theorem 8.2.3. LetF be a bicomplex holomorphic function on a product-type sim- ply connected domain ΩinBC. ThenF is a constant function if and only if any of the following equivalent conditions holds:
1. WritingF =f11+if12+jf21+kf22, one of the real functionsfk is constant.
In particular,F can be “missing” one (or more) of the real componentsfk , i.e.,fk = 0.
2. WritingF=f1+jf2, eitherf1orf2is a constantC(i)-function. Similarly for F =g1+kg2. In particular, F can be eitherC(i)-valued (i.e., f2=g2= 0), orjãC(i)-valued (i.e.f1= 0), orkãC(i)-valued (i.e., g1= 0).
3. WritingF =ρ1+iρ2, eitherρ1orρ2 is constant. Similarly forF =γ1+kγ2. In particular,F can be eitherC(j)-valued (i.e.,ρ2=γ2= 0), oriãC(j)-valued (i.e., ρ1= 0), or kãC(j)-valued (i.e., γ1= 0).
182 Chapter 8. Some Properties of Bicomplex Holomorphic Functions 4. WritingF =f1+if2, eitherf1 orf2 is constant. Similarly, forF =g1+jg2. In particular, F can be either D-valued (i.e., f2 = g2 = 0), or iãD-valued (i.e., f1= 0), orjãD-valued (i.e.,g1= 0).
5. F(Z) =Aãf(Z), whereA∈BCis an invertible bicomplex number andf is either a C(i),C(j),D,jC(i),kC(i),iC(j),kC(j), iD orjD-valued function.
Observe that if A∈S, then bothF(Z)andF(Z)are in S.
6. The function F is of the form F = f +jλf with a C(i)-valued f and λ∈ C(i)\{±i}. SimilarlyF can be of the formF=ρ+i(μρ)withμ∈C(j)\{±j}
and a C(j)-valued ρ; or F =f+i(af) with a ∈ D. Observe that if λ=±i, then the values ofF(Z)and of F(Z)are in S0.
7. Either one of the complex moduli |F(Z)|i or|F(Z)|j equals a non-zero con- stant (inC(i)orC(j)) for allZ ∈Ω. Observe that either one of the complex moduli is zero for allZ if and only ifF(Z)and soF(Z) are inS0. 8. The hyperbolic modulus|F(Z)|k equals an invertible hyperbolic constant for
all Z ∈ Ω. Observe that |F(Z)|k is a hyperbolic zero-divisor if and only if F(Z)and soF(Z)are in S0.
9. F(Z) is invertible for all Z ∈ Ω and either one of the complex arguments argi(F(Z))orargj(F(Z))are constant for allZ.
Proof. Assume the functionf11 is constant on Ω. Using relations (7.18) it follows that a = b = c = d = 0, i.e., all the other real functions fk have real partial derivatives zero with respect to all four real variables x1, y1, x2 and y2. Since Ω is a domain, it follows that all fk are constant functions, thus F is a constant bicomplex function which provides a proof of (1).
The proofs of the next items (2)–(4) are easily reduced to (1). For example iff2 is constant, thenf21 andf22 are constant, soF is constant, etc. But let us further investigate the casef2(z1, z2) =λ∈C(i) in the idempotent representation F =G1e+G2e†. We obtain for anyZ ∈Ω:
G1(Z) =f1(Z)−iλ, G2(Z) =f1(Z) +iλ .
Since F(Z) exists on all Ω, thenG1 depends only on β1 =Ze and G2 depends only onβ2=Ze†, i.e.,
0 = ∂G1
∂β2(Z), 0 = ∂G2
∂β1(Z).
Recalling the relation between the partial derivatives with respect to z1 and z2 and those with respect toβ1 andβ2 we have:
∂
∂β2 =Wφ◦ +1
2 ∂
∂z1 −i ∂
∂z2 ,
◦Wφ−1,
∂
∂β1 =Wφ◦ +1
2 ∂
∂z1 +i ∂
∂z2 ,
◦Wφ−1,
which implies that
1 2
∂f1
∂z1 −i∂f1
∂z2
= 0, 1
2 ∂f1
∂z1 +i∂f1
∂z2
= 0,
hence the partial derivatives of f1 with respect to z1 and z2 are zero, sof1 is a constant. Similar arguments hold for the other cases (2)–(4).
We use the same type of calculations for the proof of the next assertion: let F(Z) = Aãf(Z), where f is either a C(i)-, C(j)-, D-, etc. valued function, and A=a1e+a2e†, say, witha1, a2∈C(i),a1a2= 0. Then
F = (a1f)e+ (a2f)e†=:G1e+G2e†. Again, because F(Z) exists in Ω, then ∂G1
∂β2(Z) = ∂G2
∂β1(Z) = 0, which will force the partial derivatives off with respect to theC(i)-,C(j)- or hyperbolic variables to be zero, i.e.,f is a constant. ThereforeF is constant.
Note that (5) can also be proved directly, using the fact that if F is deriv- able on Ω, then multiplying it by any bicomplex constant (in our case A−1) will yield a bicomplex derivable function G(Z). Therefore, if F(Z) = Aãf(Z), then G(Z) =f(Z) is a complex, hyperbolic, etc. valued BC-holomorphic function, so it is constant.
The proofs of the case when, for example,F =f +j(λf) is easily obtained from the previous one: we write F = (1 +jλ)f, and note that A := 1 +jλ is invertible if and only if λ = ±i. We note also that a direct proof of this fact is obtained using the Cauchy-Riemann conditions forf1=f andf2=λf: at anyZ we obtain
∂f
∂z1 =∂f1
∂z1 = ∂f2
∂z2 =λ∂f
∂z2,
∂f
∂z2 =∂f1
∂z2 =−∂f2
∂z1 =−λ∂f
∂z1, from which we get:
(1 +λ2)∂f
∂z1 = (1 +λ2)∂f
∂z2 = 0, so ifλ=±i, thenf is constant, and so on.
Assume now that |F(Z)|i = a ∈ C(i) for any Z ∈ Ω. In the idempotent representation, we have:
a2=|F(Z)|2i =G1(β1)ãG2(β2),
for all Z ∈ Ω. If a = 0, then we get that G1(β1) = 0 or G2(β2) = 0 for all Z = β1e+β2e†. Since G1 and G2 are complex holomorphic functions on Ω1,
184 Chapter 8. Some Properties of Bicomplex Holomorphic Functions respectively Ω2, domains inC(i), their zeros are isolated, unless they are identically zero. Thus, it follows that either G1 ≡0 on Ω1 or G2≡0 on Ω2. In either case, the result is thatF(Z)∈S0 for allZ ∈Ω.
If a = 0, then the following argument applies: the partial derivatives with respect to bothβ1 andβ2 of the productG1ãG2 are zero:
0 = ∂(a2)
∂β1 =∂(G1(β1)ãG2(β2))
∂β1 =G1(β1)ãG2(β2), 0 = ∂(a2)
∂β2 =∂(G1(β1)ãG2(β2))
∂β2 =G1(β1)ãG2(β2),
for allβ1∈Ω1and β2∈Ω2, so in the case that neither G1norG2are identically zero, then bothG1andG2are constant functions on their domains. Therefore,F is constant. Similarly in theC(j) case.
A special computation is necessary when the hyperbolic modulus ofF equals a hyperbolic constant a=a1+ka2, where nowa1, a2∈R. LetF =f1+if2, and write each hyperbolic functionf1=s1e+t1e† andf2=s2e+t2e† in idempotent components (in the intrinsic hyperbolic writing), wheres1, s2, t1, t2are real-valued functions of the bicomplex variableZ. Then
F = (s1e+t1e†) +i(s2e+t2e†) = (s1+is2)e+ (t1+it2)e†=:G1e+G2e†. Then the square of the hyperbolic modulus ofF is given by
a21e+a22e† =a2=|F(Z)|2k= (s21(Z) +s22(Z))e+ (t21(Z) +t22(Z))e†, therefore
a21=s21(Z) +s22(Z) =G1(Z)ãG1(Z) and
a22=t21(Z) +t22(Z) =G2(Z)ãG2(Z)
for allZ. Becauses1, s2, t1andt2are real-valued functions, thena1= 0 ora2= 0 is equivalent to s1 = s2 ≡0 or t1 =t2 ≡0, respectively, which is equivalent to F(Z) being a zero-divisor for allZ.
The functionG1(Z) is a complex holomorphic function which depends onβ1 only:G1=G1(β1), thus the functionG1 is anti-holomorphic inβ1, so that in the identitya21=G1(β1)ãG1(β1) we cannot take the derivatives of both sides but we can apply the Cauchy–Riemann operator ∂
∂β1 which gives:
0 = ∂(G1ãG1)
∂β1 (β1) =∂G1
∂β1(β1)ãG1(β1) =G1(β1)ãG1(β1), where we used the fact that ∂G1
∂β1 = ∂G1
∂β1 = 0. Since this happens for allZ, so for allβ1 ∈Ω1, it follows that either G1≡0 so G1 ≡0, or that G1 is constant. An
analogous argument holds forG2. Note thatG1≡0 if and only ifs1=s2= 0, so F(Z)∈S0for allZ, and similarly forG2. Otherwise, it follows thatF is constant.
Consider now F(Z) to be an invertible bicomplex number for all Z ∈ Ω.
Assume that, in itsC(i)-complex trigonometric form
F(Z) =|F(Z)|iejΘ=|F(Z)|i(cos Θ +jsin Θ),
the principal argument Θ = argi(F(Z)) is constant for allZ. Then the expression ejΘ is an invertible bicomplex number, say,A. ThereforeF(Z) =Aãf(Z), where f(Z) =|F(Z)|iis aC(i)-valued function. BecauseF is bicomplex holomorphic on Ω, it follows thatF is constant. Similarly for theC(j) case.
Corollary 8.2.4. Let F be a bicomplex holomorphic function on a product-type domainΩ. Each of the following conditions implies thatF(Z)∈Sfor all Z∈Ω:
1. F(Z)∈S for allZ∈Ω.
2. If we write F =G1e+G2e†, then eitherG1 or G2 is a constant function.
This happens if and only ifF(Z)∈S for allZ.
3. If F(Z) =Aãf(Z), where A∈S andf is either aC(i)-, C(j)-, D-,jC(i)-, kC(i)-,iC(j)-,kC(j)-,iD- orjD-valued function.
4. |F(Z)|i= 0for allZ ∈Ω.
5. |F(Z)|j= 0for all Z∈Ω.
6. |F(Z)|k∈S for allZ∈Ω.