Fz
1(Z0) = 1, Fz
2(Z0) =−j and that neitherFζ
1(Z0) nor Fζ
2(Z0) exist for anyZ0∈BC.
Remark 7.1.3. Formulas (1.6)–(1.8) suggest the introduction of six more partial derivatives: four complex and two hyperbolic. But since there exists a direct relation between the corresponding variables, namely,
w1=z1, w2=−iz2, ω1=ζ1, ω2=−jζ2, w1=z1, w2=−kz2, it is easy to see that we do not get anything essentially new. The situation with partial derivatives with respect to variables arising from the idempotent represen- tations of bicomplex numbers is on the other hand much more interesting, and will be discussed in detail later.
7.2 The bicomplex derivative and the bicomplex derivability
We continue now with the definition of the derivative of a bicomplex function F : Ω⊂BC→BCof one bicomplex variableZ as follows:
Definition 7.2.1. The derivativeF(Z0)of the function F at a pointZ0∈Ωis the limit, if it exists,
F(Z0) := lim
Z→Z0
F(Z)−F(Z0)
Z−Z0 = lim
S0 H→0
F(Z0+H)−F(Z0)
H , (7.8)
forZ in the domain ofF such thatH =Z−Z0is an invertible bicomplex number.
In this case, the functionF is called derivableatZ0.
Corollary 7.2.2. The function F is derivable at Z0 if and only if there exists a function αF,Z0 such that
S0lim H→0αF,Z0(H) = 0 and
F(Z0+H)−F(Z0) =F(Z0)ãH+αF,Z0(H)H for allH ∈S0. (7.9) Remark 7.2.3.It is necessary to make a comment here. Traditionally, see e.g.[103], p. 138 and p. 432, if h is either a real or a complex increment, the symbol o(h)
is used to indicate any expression of the form α(h)|h| with lim
h→0α(h) = 0. Since, both in the real and in the complex case, the expression |h|
h remains bounded when h→0, it is clear that one could replace α(h)|h| by α(h)hin the expression of o.
However, the situation is quite different in the bicomplex case. Here, the expression
|H|
H is not bounded whenH →0,and therefore we need to carefully distinguish the two expressions. In accordance with the usual notation, we will always use o(H) to denote a function of the form α(H)|H|, and therefore the expression in the previous corollaryαF,Z0(H)H is not, in general,o(H). This distinction is at the basis of the notions of weak and strong Stoltz conditions for bicomplex functions, which are used by G.B.Price in[56].
Remark 7.2.4. A bicomplex function F derivable at Z0 enjoys the following pro- perty:
lim
H /∈S0, H→0(F(Z0+H)−F(Z0)) = 0. (7.10) In other words, a function F, which is derivable at a pointZ0, enjoys some sort of “weakened” continuity at that point, in the sense thatF(Z)converges toF(Z0) as long asZ converges toZ0 in such a way that Z−Z0 is invertible. We will see later on that this restriction can be removed under reasonable assumptions.
It turns out that the arithmetic operations on derivable functions follow the usual rules of their real and complex antecedents.
Theorem 7.2.5. (Derivability and arithmetic operations) Let F and G be two bicomplex functions defined on Ω⊂BCand derivable at Z0∈Ω. Then:
1. The sum and difference ofF andGare derivable at Z0 and (F±G)(Z0) =F(Z0)±G(Z0).
2. For any bicomplex number C the function CãF is derivable at Z0 and (CãF)(Z0) =CãF(Z0).
3. The product ofF andGis derivable at Z0 and
(FãG)(Z0) =F(Z0)ãG(Z0) +F(Z0)ãG(Z0).
4. If Gis continuous at Z0 and G(Z0)∈S0, then the quotient F
G is derivable atZ0 and
F G
(Z0) = F(Z0)ãG(Z0)−F(Z0)ãG(Z0) (G(Z0))2 .
7.2. The bicomplex derivative and the bicomplex derivability 139 Proof. From Corollary 7.2.2, there exist two bicomplex functionsαF,Z0 andαG,Z0 such that
S0lim H→0αF,Z0(H) = lim
S0 H→0αG,Z0(H) = 0 and
F(Z0+H)−F(Z0) =F(Z0)ãH+αF,Z0(H)H , (7.11) G(Z0+H)−G(Z0) =G(Z0)ãH+αG,Z0(H)H , (7.12) for all H ∈ S0. The first statement about the sum or difference F ±G is ob- tained easily by adding (or subtracting) formulas (7.11) and (7.12), and noting that the functions αF±G,Z0 :=αF,Z0±αG,Z0 respect the properties from Corol- lary 7.2.2 (which is an “if and only if” statement). Similarly, if we multiply (7.11) by the numberC, we obtain statement (2). Note thatC needs not necessarily be an invertible bicomplex number. For example, if C = ce, where c ∈ C(i), then from (7.11) (multiplied byC) we obtain that
ceãF(Z0) = (c F(Z0))e∈BCe
is the derivative ofceãF atZ0, which is a zero-divisor. The non-invertibility ofC does not affect any arguments, asH is always invertible.
If we multiply the termsF(Z0+H) andG(Z0+H) obtained from the left- hand side of formulas (7.11) and (7.12) we obtain:
F(Z0+H)ãG(Z0+H) =F(Z0)G(Z0) + (F(Z0)ãG(Z0) +F(Z0)ãG(Z0))ãH + (F(Z0)αG,Z0(H) +G(Z0)αF,Z0(H))ãH
+ (F(Z0)G(Z0) +F(Z0)αG,Z0(H) +G(Z0)αF,Z0(H) +αF,Z0(H)αG,Z0(H))ãH2.
Now, writing
αF G,Z0(H) :=F(Z0)αG,Z0(H) +G(Z0)αF,Z0(H) + (F(Z0)G(Z0) +F(Z0)αG,Z0(H)+
+G(Z0)αF,Z0(H) +αF,Z0(H)αG,Z0(H))ãH, we obtain:
S0lim H→0αF G,Z0(H) = 0 and
F(Z0+H)ãG(Z0+H)−F(Z0)G(Z0)
= (F(Z0)ãG(Z0) +F(Z0)ãG(Z0))ãH+αF G,Z0(H)H , which proves that the product F ãG is derivable at Z0 and the usual product formula holds.
We assume now that G(Z0) ∈ S0 and that G is continuous at Z0, which implies thatG(Z)∈S0 in a neighborhood ofZ0. All the computations below are performed in this neighborhood. Then we write:
F(Z0+H)
G(Z0+H)−F(Z0)
G(Z0) = F(Z0+H)G(Z0)−G(Z0+H)F(Z0) G(Z0+H)G(Z0)
= (F(Z0+H)−F(Z0))G(Z0)−(G(Z0+H)−G(Z0))F(Z0) G(Z0+H)G(Z0)
= (F(Z0)ãG(Z0)−G(Z0)ãF(Z0))ãH+ (αF,Z0(H)G(Z0) +αG,Z0(H)F(Z0))ãH
G(Z0+H)G(Z0) .
(7.13) Now note that if we write
1
G(Z0+H)− 1 G(Z0) =
−G(Z0) +αG,Z0(H) G(Z0+H)G(Z0)
ãH =:α1(H), then the functionα1(H) has the property
S0lim H→0α1(H) = 0. Thus using the formula:
1
G(Z0+H) = 1
G(Z0)+α1(H), rewriting (7.13) leads to the expression:
F(Z0+H)
G(Z0+H)−F(Z0)
G(Z0) =F(Z0)ãG(Z0)−G(Z0)ãF(Z0)
(G(Z0))2 ãH+αF
G,Z0(H)ãH , whereαF
G,Z0(H) is the bicomplex function containing all the remaining terms and which enjoys the property
S0lim H→0αF
G,Z0(H) = 0.
Using again Corollary 7.2.2, the proof of the derivability of the quotient and of
the corresponding formula is finished.
Theorem 7.2.6. LetF andGbe two bicomplex functions, F is defined on an open setΩ andGis defined onF(Ω)⊂BC. Assume that there is a pointZ0∈Ωsuch thatW0=F(Z0)is an interior point of F(Ω) and such that
F(Z)−F(Z0)∈S0, implies Z−Z0∈S0, Z∈Ω.
Assume now that F is derivable at Z0 andG is derivable at W0. Then the com- positionG◦F is derivable at Z0 and
(G◦F)(Z0) =G(F(Z0))ãF(Z0).
7.2. The bicomplex derivative and the bicomplex derivability 141 Moreover, if F(Z0) ∈ S0 and if F is a bijective function around Z0, then its inverseF−1 is derivable at W0=F(Z0)and
(F−1)(W0)ãF(Z0) = 1.
Proof. Via Corollary 7.2.2, there exist two bicomplex functionsαF,Z0 andαG,W0 such that
S0lim H→0αF,Z0(H) = lim
S0 K→0αG,W0(K) = 0 and
F(Z0+H)−F(Z0) =F(Z0)ãH+αF,Z0(H)H , (7.14) G(W0+K)−G(W0) =G(W0)ãK+αG,W0(K)K , (7.15) for allH, K∈S0.
Denote by VW0 a neighborhood of W0 contained in F(Ω) which exists by hypothesis. Let thenVW∗0 be the set obtained fromVW0 by eliminating the points W for which W −W0 are zero-divisors. Let UZ∗0 be the set of all Z ∈ UZ0 :=
F−1(VW0) such thatZ−Z0 is not a zero-divisor.
Now we note that formula (7.15) is true for those invertible K such that WK :=W0+K∈ VW0; suchKexist by takingWK ∈ VW∗0. Then by the properties ofF, for any suchKthere exists aZK ∈ UZ∗0 such thatWK =W0+K=F(ZK).
Now we write:
ZK =Z0+H:=Z0+ (ZK−Z0),
whereHis invertible by the choice ofZK. Therefore we replaceW0+K=F(ZK) = F(Z0+H),W0=F(Z0) andK=F(Z0+H)−F(Z0) in formula (7.15), and we obtain:
G(F(Z0+H))−G(F(Z0)) =G(F(Z0))ã(F(Z0+H)−F(Z0)) +α"G,F(Z0)(H)ã(F(Z0+H)−F(Z0)), where α"G,F(Z0)(H) := αG,F(Z0)(F(Z0 +H)−F(Z0)) is a function having the property from Corollary 7.2.2, as whenK→0 so doesH →0, in both cases along invertible bicomplex numbers. Now we use formula (7.14) and we get:
G(F(Z0+H))−G(F(Z0)) =G(F(Z0))ã(F(Z0)ãH+αF,Z0(H)H) +α"G,F(Z0)(H)ã(F(Z0)ãH+αF,Z0(H)H)
=:G(F(Z0))ãF(Z0)ãH+αG◦F(H)H ,
where the functionαG◦F(H) has obviously the desired property from Corollary 7.2.2.
The last statement of the theorem can be proved quite analogously to the cases of real and complex functions. If one assumes beforehand the existence of the derivative of the inverse function atWo, then the corresponding formula becomes a particular case of the chain rule above:
1 = (F−1◦F)(Z0) = (F−1)(F(Z0))ãF(Z0),
where we used that the derivative of the identity function is one which is obvious.
Theorem 7.2.7 (The derivability of bicomplex elementary functions). All bicom- plex elementary functions introduced in Chapter 6 are derivable at any point Z=Z0 where they are defined. In more detail:
1. Any constant function is derivable with the derivative zero.
2. Bicomplex polynomials are derivable for anyZ ∈BC. In particular, we have:
(Zn) =nZn−1.
3. The derivative of the bicomplex exponential function eZ iseZ. 4. (sinZ)= cosZ and(cosZ) =−sinZ.
5. (tanZ) = 1
(cosZ)2 for any Z such that cosZ is an invertible bicomplex number, i.e., using the idempotent representationZ=β1e+β2e†, the complex numbersβ1 andβ2 are not equal to π
2 plus integer multiples ofπ.
Similarly, (cotZ) = −1
(sinZ)2 for any Z such that sinZ is an invertible bicomplex number, i.e., whenever β1 andβ2 are not integer multiples of π.
6. (coshZ)= sinhZ and (sinhZ)= coshZ.
7. The (m, n)-branch logarithmic function F(Z) = lnm,n(Z), which is defined for all invertible bicomplex numbersZ, is derivable for all suchZ, and
(lnm,n)(Z) = 1 Z. Proof. Item (1) is obviously true.
Note that the formula
An−Bn= (A−B)(An−1+An−2B+ã ã ã+Bn−1)
holds for anyA, B ∈BC, due to the algebraic properties of addition and multi- plication of bicomplex numbers. Applying this for A =Z +H and B = Z, we obtain:
S0lim H→0
(Z+H)n−Zn H
= lim
S0 H→0
H
(Z+H)n−1+ (Z+H)n−2Z+ã ã ã+Zn−1 H
= lim
S0 H→0
(Z+H)n−1+ (Z+H)n−2Z+ã ã ã+Zn−1
=nZn−1.
7.2. The bicomplex derivative and the bicomplex derivability 143 It follows that any bicomplex polynomial p(Z) =
$n k=0
AkZk is derivable at any Z∈BC, and its derivative is a polynomial of degree one less, given by
p(Z) =
$n k=1
kAkZk−1.
Let us consider the bicomplex exponential function eZ. For an arbitrary Z∈BCand an invertible bicomplex numberH, we have:
eZ+H−eZ
H =eZãeH−1 H . We show now that
S0lim H→0
eH−1 H = 1.
For this, let us write H = he+ke† in the idempotent representation, where h, k∈C(i) are such thathk= 0. TheneH =ehe+eke†, thus
eH−1
H =ehe+eke†
he+ke† = eh−1
h e+ek−1 k e†. Therefore
S0lim H→0
eH−1
H = lim
h→0
eh−1 h
e+ lim
k→0
ek−1 k
e†=e+e†= 1. It follows that
S0lim H→0
eZ+H−eZ
H =eZã lim
S0 H→0
eH−1 H =eZ, so (eZ) =eZ, for allZ∈BC.
Recall that the bicomplex trigonometric functions are defined by:
cosZ= ejZ+e−jZ
2 , sinZ= ejZ−e−jZ 2j .
Since the exponential function is derivable for allZ∈BC, it follows that cosZ is derivable for allZ and
(cosZ)= jejZ−je−jZ
2 =−ejZ−e−jZ
2j = sinZ . Similarly the function sinZ is derivable for allZ and (sinZ)= cosZ.
The bicomplex tangent function is defined as the quotient sinZ
cosZ, whenever cosZ is invertible. Using the quotient rule, we obtain:
(tanZ)= cos2Z+ sin2Z
cos2Z = 1
cos2Z. A similar argument applies for the function cotZ.
The hyperbolic functions of a bicomplex variable are defined as coshZ= eZ+e−Z
2 , sinZ= eZ−e−Z
2 .
Again, using that eZ is derivable for anyZ, it follows that coshZ and sinhZ are derivable for anyZ and
(coshZ) =eZ−e−Z
2 = sinhZ, (sinhZ)= eZ+e−Z
2 = coshZ . Recall that the (m, n)-branch bicomplex logarithmic function lnm,n, defined for all invertible bicomplex numbers, is the inverse function of eZ (at least on a subset of its domain of definition). Then we have:
1 =
elnm,n(Z)
=elnm,n(Z)ã(lnm,n)(Z) =Zã(lnm,n)(Z). SinceZ runs over all invertible bicomplex numbers, we obtain
(lnm,n)(Z) = 1 Z,
which concludes our proof.