The Complex Inverse Trigonometric and Hyperbolic Functions

8/12/2019 The Complex Inverse Trigonometric and Hyperbolic Functions

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Physics 116A Winter 2011

The complex inverse trigonometric and hyperbolic functions

In these notes, we examine the inverse trigonometric and hyperbolic functions, wherethe arguments of these functions can be complex numbers (see e.g. ref. 1). Theseare all multi-valued functions. We also carefully define the corresponding single-valued principal values of the inverse trigonometric and hyperbolic functions followingthe conventions employed by the computer algebra software system, Mathematica 8.These conventions are outlined in section 2.2.5 of ref. 2.

The principal value of a multi-valued complex functionf(z) of the complex vari-able z, which we denote by F(z), is continuous in all regions of the complex plane,except on a specific line (or lines) called branch cuts. The function F(z) has a dis-

continuity when zcrosses a branch cut. Branch cuts end at a branch point, whichis unambiguous for each function F(z). But the choice of branch cuts is a matterof convention.1 Thus, if mathematics software is employed to evaluate the functionF(z), you need to know the conventions of the software for the location of the branchcuts. The mathematical software needs to precisely define the principal value off(z)in order that it can produce a unique answer when the user types in F(z) for a par-ticular complex numberz. There are often different possible candidates for F(z) thatdiffer only in the values assigned to them when z lies on the branch cut(s). Thesenotes provide a careful discussion of these issues as they apply to the complex inversetrigonometric and hyperbolic functions.

1. The inverse trigonometric functions: arctan and arccot

We begin by examining the solution to the equation

z= tan w= sin w

cos w =

1

i

eiw eiweiw +eiw

=

1

i

e2iw 1e2iw + 1

.

We now solve for e2iw,

iz=e2iw 1e2iw + 1

=

e2iw =1 +iz

1 iz.

1In these notes, the principal value of the argument of the complex number z , denoted by Arg z,is defined to lie in the interval


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Taking the complex logarithm of both sides of the equation, we can solve for w,

w= 1

2iln

1 +iz

1 iz

.

The solution toz= tan w isw = arctan z. Hence,

arctan z= 1

2iln

1 +iz

1 iz

(1)

Since the complex logarithm is a multi-valued function, it follows that the arctangentfunction is also a multi-valued function. Using the definition of the multi-valuedcomplex logarithm,

arctan z= 1

2iLn

1 +iz

1

iz+12 Arg

1 +iz

1

iz + 2n , n= 0 ,1 ,2 , . . . , (2)where Arg is the principal value of the argument function.

Similarly,

z= cot w=cos w

sin w =

i(eiw +eiw

eiw eiw

=

i(e2iw + 1

e2iw 1

.

Again, we solve for e2iw,

iz= e2iw + 1

e2iw 1 = e2iw =

iz 1iz+ 1

.

Taking the complex logarithm of both sides of the equation, we conclude that

w= 1

2iln

iz 1iz+ 1

=

1

2iln

z+i

z i

,

after multiplying numerator and denominator by ito get a slightly more convenientform. The solution to z= cot w isw= arccotz. Hence,

arccotz= 1

2iln

z+i

z i

(3)

Thus, the arccotangent function is a multivalued function,

arccotz= 1

2iLn

z+iz i + 12

Arg

z+i

z i

+ 2n

, n= 0 ,1 ,2 , . . . , (4)

Using the definitions given by eqs. (1) and (3), the following relation is easilyderived:

arccot(z) = arctan

1

z

. (5)

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Note that eq. (5) can be used as the definition of the arccotangent function. Itis instructive to derive another relation between the arctangent and arccotangentfunctions. First, we first recall the property of the multi-valued complex logarithm,

ln(z1z2) = ln(z1) + ln(z2) , (6)

as a set equality. It is convenient to define a new variable,

v= i zi+z

, = 1v

=z+i

z i. (7)

It follows that:

arctan z+ arccot z= 1

2i

ln v+ ln

1

v

=

1

2iln

vv

=

1

2iln(1) .

Since ln(1) =i(+ 2n) for n = 0, 1, 2 . . ., we conclude that

arctan z+ arccot z= 12+n , for n= 0, 1, 2, . . . (8)

Finally, we mention two equivalent forms for the multi-valued complex arctangentand arccotangent functions. Recall that the complex logarithm satisfies

ln

z1z2

= ln z1 ln z2, (9)

where this equation is to be viewed as a set equality, where each set consists of allpossible results of the multi-valued function. Thus, the multi-valued arctangent andarccotangent functions given in eqs. (1) and (5), respectively, are equivalent to

arctan z= 1

2i

ln(1 +iz) ln(1 iz)

, (10)

arccot z= 1

2i

ln

1 +

i

z

ln

1 i

z

, (11)

2. The principal values Arctan and Arccot

It is convenient to define principal values of the inverse trigonometric functions,

which are single-valued functions, which will necessarily exhibit a discontinuity acrossbranch cuts in the complex plane. In Mathematica 8, the principal values of thecomplex arctangent and arccotangent functions, denoted by Arctan and Arccot re-spectively (with a capital A), are defined by employing the principal values of thecomplex logarithms in eqs. (10) and (11),

Arctan z= 1

2i

Ln(1 +iz) Ln(1 iz)

, z= i (12)

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and

Arccot z= Arctan

1

z

=

1

2i

Ln

1 +

i

z

Ln

1 i

z

, z= i , z= 0 (13)

Note that the pointsz= iare excluded from the above definitions, as the arctangentand arccotangent are divergent at these two points. The definition of the principalvalue of the arccotangent given in eq. (13) is deficient in one respect since it is notwell-defined at z = 0. We shall address this problem shortly. One useful feature ofthe definitions eqs. (12) and (13) is that they satisfy:

Arctan(z) = Arctan z , Arccot(z) = Arccot z .Because the principal value of the complex logarithm Ln does not satisfy eq. (9)in all regions of the complex plane, it follows that the definitions of the complexarctangent and arccotangent functions adopted by Mathematica 8 do not coincidewith alternative definitions that employ the principal value of the complex logarithmsin eqs. (1) and (4) [for further details, see Appendix A].

First, we shall identify the location of the branch cuts by identifying the linesof discontinuity of the principal values of the complex arctangent and arccotangentfunctions in the complex plane. The principal value of the complex arctangent func-tion is single-valued for all complex values ofz, excluding the two branch points atz= i. Moreover, the the principal-valued logarithms, Ln (1 iz) are discontinuousas z crosses the lines 1 iz < 0, respectively. We conclude that Arctanz must bediscontinuous when z= x+iy crosses lines on the imaginary axis such that

x= 0 and < y < 1 and 1< y < . (14)These two lines that lie along the imaginary axis are the branch cuts of Arctan z. Notethat Arctan zis single-valued on the branch cut itself, since it inherits this propertyfrom the principal value of the complex logarithm.

Likewise, the principal value of the complex arccotangent function is single-valuedfor all complex z, excluding the branch points z=i. Moreover, the the principal-valued logarithms, Ln

1 iz

are discontinuous as z crosses the lines 1 iz < 0,

respectively. We conclude that Arccotz must be discontinuous when z = x+ iycrosses the branch cut located on the imaginary axis such that

x= 0 and

1< y


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including points on the branch cut. Using eq. (12), one can easily show that ifz is anon-zero complex number infinitesimally close to 0, then

Arccot z =z0 , z=0

12 , for Rez >0 ,12 , for Rez= 0 and Imz


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the two logarithms that is equal to 2iN= 2i. The end result is,

Arctan() = 12i

[ln(1) 2i] = 12 ,

as required. As a final check, we can use the results of Tables 1 and 2 in the class hand-out,The Argument of a Complex Number, to conclude that Arg(a+bi) = Arctan(b/a)for a >0. Setting a= 1 and b = x then yields:

Arg (1 +ix) = Arctan x , Arg (1 ix) = Arctan(x) = Arctan x .Subtracting these two results yields eq. (19).

In contrast to the real arctangent function, there is no generally agreed definitionfor the principal range of the real-valued arccotangent function. However, a growingconsensus among computer scientists has led to the following choice for the principalrange of the real-valued arccotangent function,

12 0 ,12 , for Rez= 0, and Imz >1 or1< Im z 0 ,

12

, for Rez


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3. The inverse trigonometric functions: arcsin and arccos

The arcsine function is the solution to the equation:

z= sin w=

eiw

eiw

2i .

Lettingv eiw , we solve the equation

v 1v

= 2iz .

Multiplying byv, one obtains a quadratic equation for v,

v2 2izv 1 = 0 . (23)The solution to eq. (23) is:

v= iz+ (1

z2)1/2 . (24)

Sincezis a complex variable, (1 z2)1/2 is the complex square-root function. This isa multi-valued function with two possible values that differ by an overall minus sign.Hence, we do not explicitly write out thesign in eq. (24). To avoid ambiguity, weshall write

v= iz+ (1 z2)1/2 =iz+e 12ln(1z2) =iz+e 12 [Ln|1z2|+i arg(1z2)]

=iz+ |1 z2|1/2e i2arg(1z2) .In particular, note that

ei

2arg(1z2) =ei

2Arg(1z2)ein = e i2Arg(1z2) , forn = 0, 1 ,which exhibits the two possible sign choices.

By definition, v eiw, from which it follows that

w=1

i ln v=

1

i ln

iz+ |1 z2|1/2e i2arg(1z2)

.

The solution toz= sin w is w = arcsin z. Hence,

arcsin z=1

i lniz+ |1 z

2|1/2e i2arg(1z2)The arccosine function is the solution to the equation:

z= cos w=eiw +eiw

2 .

Lettingv eiw , we solve the equation

v+1

v = 2z .

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v2 2zv + 1 = 0 . (25)The solution to eq. (25) is:

v= z+ (z2 1)1/2 .Following the same steps as in the analysis of arcsine, we write

w= arccos z=1

i ln v=

1

i ln

z+ (z2 1)1/2 , (26)where (z2 1)1/2 is the multi-valued square root function. More explicitly,

arccos z=1

i ln

z+ |z2 1|1/2e i2arg(z21)

. (27)

It is sometimes more convenient to rewrite eq. (27) in a slightly different form. Recall

thatarg(z1z2) = arg z+ arg z2, (28)

as a set equality. We now substitutez1 =zand z2 = 1 into eq. (28) and note thatarg(1) =+ 2n (for n = 0, 1, 2, . . .) and arg z= arg z+ 2n as a set equality.It follows that

arg(z) =+ arg z ,as a set equality. Thus,

ei2arg(z21) =ei/2 e

i2arg(1z2) =ie

i2arg(1z2) ,

and we can rewrite eq. (26) as follows:

arccos z=1

i ln

z+i

1 z2

, (29)

which is equivalent to the more explicit form,

arccos z=1

i ln

z+i|1 z2|1/2e i2arg(1z2)

The arcsine and arccosine functions are related in a very simple way. Usingeq. (24),

i

v=

i

iz+

1 z2 = i(iz+ 1 z2)

(iz+

1 z2)(iz+ 1 z2) =z+i

1 z2 ,

which we recognize as the argument of the logarithm in the definition of the arccosine[cf. eq. (29)]. Using eq. (6), it follows that

arcsin z+ arccos z=1

i

ln v+ ln

i

v

=

1

iln

iv

v

=

1

i ln i .

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Since ln i= i(12+ 2n) for n = 0, 1, 2 . . ., we conclude that

arcsin z+ arccos z= 12+ 2n , for n= 0, 1, 2, . . . (30)

4. The principal values Arcsin and Arccos

In Mathematica 8, the principal value of the arcsine function is obtained by em-ploying the principal value of the logarithm and the principle value of the square-rootfunction (which corresponds to employing the principal value of the argument). Thus,

Arcsin z=1

iLn

iz+ |1 z2|1/2e i2Arg(1z2)

. (31)

It is convenient to introduce some notation for the the principle value of the square-root function. Consider the multivalued square root function, denoted by z1/2. Hence-

forth, we shall employ the symbolzto denote the single-valued function,

z=

|z| e12Arg z , (32)where

|z| denotes the unique positive squared root of the real number|z|. In this

notation, eq. (31) is rewritten as:

Arcsin z=1

iLn

iz+

1 z2

(33)

One noteworthy property of the principal value of the arcsine function is

Arcsin(z) = Arcsin z . (34)To prove this result, it is convenient to define:

v= iz+

1 z2 , 1v

= 1

iz+

1 z2 = iz+

1 z2 . (35)

Then,

Arcsin z=1

iLn v , Arcsin(z) =1

iLn

1

v

.

The second logarithm above can be simplified by making use of eq. (57) of the classhandout entitled, The complex logarithm, exponential and power functions,

Ln(1/z) =

Ln(z) + 2i , ifz is real and negative ,Ln(z) , otherwise . (36)

In Appendix C, we prove thatv can never be real and negative. Hence it follows fromeq. (36) that

Arcsin(z) =1i

Ln

1

v

= 1

iLn v= Arcsin z ,

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as asserted in eq. (34).We now examine the principal value of the arcsine for real-valued arguments such

that1 x 1. Setting z= x, where x is real and|x| 1,

Arcsin x=1

i Ln

ix+ 1 x2= 1i Ln ix+ 1 x2 +iArgix+ 1 x2= Arg

ix+

1 x2

, for|x| 1 , (37)

since ix+

1 x2 is a complex number with magnitude equal to 1 when x is realwith|x| 1. Moreover, ix+ 1 x2 lives either in the first or fourth quadrant ofthe complex plane, since Re(ix+

1 x2) 0. It follows that:

2 Arcsin x

2, for |x| 1 .

In Mathematica 8, the principal value of the arccosine is defined by:

Arccos z= 12 Arcsin z . (38)

We demonstrate below that this definition is equivalent to choosing the principalvalue of the complex logarithm and the principal value of the square root in eq. (29).That is,

Arccos z=1

iLn

z+i

1 z2

(39)

To verify that eq. (38) is a consequence of eq. (39), we employ the notation of eq. (35)to obtain:

Arcsin z+ Arccos z=1

i

Ln|v| + Ln

1

|v|

+iArg v+iArg

i

v

= Arg v+ Arg

i

v

. (40)

It is straightforward to check that:

Argv+ Arg i

v

= 12 , for Rev 0 .

However in Appendix C, we prove that Re v Re (iz+ 1 z2) 0 for all complexnumbers z. Hence, eq. (40) yields:

Arcsin z+ Arccos z= 12 ,

as claimed.

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We now examine the principal value of the arccosine for real-valued argumentssuch that1 x 1. Setting z= x, where x is real and|x| 1,

Arccos x=1

iLnx+i

1 x2=

1

i Ln x+i

1 x2 +iArgx+i

1 x2= Arg

x+i

1 x2

, for|x| 1 , (41)

since x+ i

1 x2 is a complex number with magnitude equal to 1 when x is realwith|x| 1. Moreover,x+i1 x2 lives either in the first or second quadrant ofthe complex plane, since Im(x+i

1 x2) 0. It follows that:

0 Arccos x , for |x| 1 .The principal value of the complex arcsine and arccosine functions are single-

valued for all complex z. The choice of branch cuts for Arcsinzand Arccosz mustcoincide in light of eq. (38). Moreover, due to the standard branch cut of the principal

value square root function,3 it follows that Arcsin z is discontinuous whenz= x+iycrosses lines on the real axis such that4

y= 0 and < x < 1 and 1< x < . (42)These two lines comprise the branch cuts of Arcsin zand Arccos z; each branch cutends at a branch point located at x =1 and x = 1, respectively (although thesquare root function is not divergent at these points).5

To obtain the relations between the single-valued and multi-valued functions, wefirst notice that the multi-valued nature of the logarithms imply that arcsin zcan takeon the values Arcsin z+ 2nand arccos zcan take on the values Arccosz+2n, where

n is any integer. However, we must also take into account the fact that (1 z2

)

1/2

can take on two values,1 z2. In particular,

arcsin z=1

iln(iz

1 z2) =1

i ln

1iz 1 z2

=

1

i

ln(1) ln(iz

1 z2)

= 1

iln(iz

1 z2) + (2n+ 1) ,

where n is any integer. Likewise,

arccos z=1

i ln(z i

1 z2) = 1

i ln

1

z i1 z2

= 1i

ln(z i

1 z2) + 2n ,3

One can check that the branch cut of the Ln function in eq. (33) is never encountered for anyfinite value ofz. For example, in the case of Arcsin z, the branch cut of Ln can only be reached ifiz+

1 z2 is real and negative. But this never happens since ifiz+ 1 z2 is real then z = iy

for some real value ofy , in which case iz +

1 z2 = y+

1 +y2 >0.4Note that for real w, we have| sin w| 1 and| cos w| 1. Hence, for both the functions

w = Arcsin z and w = Arccos z, it is desirable to choose the branch cuts to lie outside the intervalon the real axis where|Re z | 1.

5The functions Arcsin z and Arccos z also possess a branch point at the point of infinity (whichis defined more precisely in footnote 5). This can be verified by demonstrating that Arcsin(1 /z) andArccos(1/z) possess a branch point at z = 0. For further details, see e.g. Section 58 of ref 3.

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where nis any integer. Hence, it follows that

arcsin z= (1)nArcsin z+n , n= 0 ,1 ,2 , , (43)arccos z= Arccos z+ 2n , n= 0 ,1 ,2 , , (44)

where eitherArccoszcan be employed to obtain a possible value of arccos z. Inparticular, the choice ofn = 0 in eq. (44) implies that:

arccos z= arccos z , (45)which should be interpreted as a set equality. Note that one can use eqs. (43) and(44) along with eq. (38) to confirm the result obtained in eq. (30).

5. The inverse hyperbolic functions: arctanh and arccoth

Consider the solution to the equation

z= tanh w= sinh w

cosh w =

ew ewew +ew

=

e2w 1e2w + 1

.

We now solve for e2w,

z=e2w 1e2w + 1

= e2w =1 +z1 z.

Taking the complex logarithm of both sides of the equation, we can solve for w,

w=1

2ln1 +z

1 z

.

The solution toz= tanh w isw= arctanhz. Hence,

arctanh z=1

2ln

1 +z

1 z

(46)

Similarly, by considering the solution to the equation

z= coth w=cosh w

sinh w =

ew +ew

ew

ew=

e2w + 1

e2w

1 .

we end up with:

arccothz=1

2ln

z+ 1

z 1

(47)

The above results then yield:

arccoth(z) = arctanh

1

z

,

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as a set equality.Finally, we note the relation between the inverse trigonometric and the inverse

hyperbolic functions:

arctanh z= i arctan(iz) ,arccoth z= i arccot(iz) .

As in the discussion at the end of Section 1, one can rewrite eqs. (46) and (47) inan equivalent form:

arctanh z= 12[ln(1 +z) ln(1 z)] , (48)

arccoth z= 12

ln

1 +

1

z

ln

1 1

z

. (49)

6. The principal values Arctanh and Arccoth

Mathematica 8 defines the principal values of the inverse hyperbolic tangent andinverse hyperbolic cotangent, Arctanh and Arccoth, by employing the principal valueof the complex logarithms in eqs. (48) and (49). We can define the principal valueof the inverse hyperbolic tangent function by employing the principal value of thelogarithm,

Arctanh z= 12[Ln(1 +z) Ln(1 z)] (50)and

Arccoth z= Arctanh

1z

= 12

Ln

1 +1z Ln1 1z (51)

Note that the branch points at z =1 are excluded from the above definitions, asArctanhz and Arccothz are divergent at these two points. The definition of theprincipal value of the inverse hyperbolic cotangent given in eq. (51) is deficient in onerespect since it is not well-defined at z = 0. For this special case, Mathematica 8defines

Arccoth(0) = 12

i . (52)

Of course, this discussion parallels that of Section 2. Moreover, alternative def-initions of Arctanhz and Arccothzanalogous to those defined in Appendix A for

the corresponding inverse trigonometric functions can be found in ref. 4. There isno need to repeat the analysis of Section 2 since a comparison of eqs. (12) and (13)with eqs. (50) and (51) shows that the inverse trigonometric and inverse hyperbolictangent and cotangent functions are related by:

Arctanh z= iArctan(iz) , (53)Arccoth z= i Arccot(iz) . (54)

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Using these results, all other properties of the inverse hyperbolic tangent and cotan-gent functions can be easily derived from the properties of the corresponding arctan-gent and arccotangent functions.

For example the branch cuts of these functions are easily obtained from eqs. (14)

and (15). Arctanh zis discontinuous when z= x + iy crosses the branch cuts locatedon the real axis such that6

y= 0 and < x < 1 and 1< x < . (55)Arccothz is discontinuous when z = x+ iy crosses the branch cuts located on thereal axis such that

y= 0 and 1< x


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By definition, v ew, from which it follows that

w= ln v= ln

z+ |1 +z2|1/2e i2arg(1+z2)

.

The solution toz= sinh w isw= arcsinhz. Hence,

arcsinh z= ln

z+ |1 +z2|1/2e i2arg(1+z2)

(59)

The inverse hyperbolic cosine function is the solution to the equation:

z= cos w=ew +ew

2 .

Lettingv ew , we solve the equation

v+1

v = 2z .


v2 2zv + 1 = 0 . (60)The solution to eq. (60) is:

v= z+ (z2 1)1/2 .Following the same steps as in the analysis of inverse hyperbolic sine function, wewrite

w= arccosh z= ln v= ln

z+ (z2 1)1/2 , (61)where (z2

1)1/2 is the multi-valued square root function. More explicitly,

arccosh z= ln

z+ |z2 1|1/2e i2arg(z21)

The multi-valued square root function satisfies:

(z2 1)1/2 = (z+ 1)1/2(z 1)1/2 .Hence, an equivalent form for the multi-valued inverse hyperbolic cosine function is:

arccosh z= ln

z+ (z+ 1)1/2(z 1)1/2 ,or equivalently,

arccosh z= ln

z+ |z2 1|1/2e i2arg(z+1)e i2arg(z1)

. (62)

Finally, we note the relations between the inverse trigonometric and the inversehyperbolic functions:

arcsinh z= i arcsin(iz) , (63)arccosh z= iarccos z , (64)

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where the equalities in eqs. (63) and (64) are interpreted as set equalities for themulti-valued functions. The in eq. (64) indicates that both signs are employedin determining the members of the set of all possible arccoshz values. In derivingeq. (64), we have employed eqs. (26) and (61). In particular, the origin of the two

possible signs in eq. (64) is a consequence of eq. (45) [and its hyperbolic analog,eq. (73)].

8. The principal values Arcsinh and Arccosh

The principal value of the inverse hyperbolic sine function, Arcsinh z, is definedby Mathematica 8 by replacing the complex logarithm and argument functions ofeq. (59) by their principal values. That is,

Arcsinh z= Ln

z+

1 +z2

(65)

For the principal value of the inverse hyperbolic cosine function Arccosh z, Mathemat-ica 8 chooses eq. (62) with the complex logarithm and argument functions replacedby their principal values. That is,

Arccosh z= Ln

z+

z+ 1

z 1 (66)In eqs. (65) and (66), the principal values of the square root functions are employedfollowing the notation of eq. (32).

The relation between the principal values of the inverse trigonometric and theinverse hyperbolic sine functions is given by

Arcsinh z= iArcsin(iz) , (67)as one might expect in light of eq. (63). A comparison of eqs. (39) and (66) revealsthat

Arccosh z=

iArccos z , for either Imz >0 or for Im z= 0 and Rez 1 ,iArccos z , for either Imz 0.

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crosses lines on the imaginary axis such that

x= 0 and < y < 1 and 1< y < . (69)These two lines comprise the branch cuts of Arcsinh z, and each branch cut endsat a branch point located at z=

i and z= i, respectively, due to the square root

function in eq. (65), although the square root function is not divergent at these points.The function Arcsinhzalso possesses a branch point at the point of infinity, whichcan be verified by examining the behavior of Arcsinh(1/z) at the point z= 0.8

The branch cut for Arccosh zderives from the standard branch cuts of the squareroot function and the branch cut of the complex logarithm. In particular, for real zsatisfying|z|


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which should be interpreted as a set equality.This completes our survey of the multi-valued complex inverse trigonometric and

hyperbolic functions and their single-valued principal values.

APPENDIX A: Alternative definitions for Arctan and Arccot

The well-known reference book for mathematical functions by Abramowitz andStegun (see ref. 1) defines the principal values of the complex arctangent and arc-cotangent functions by employing the principal values of the logarithms in eqs. (1)and (4). This yields,11

Arctan z= 1

2iLn

1 +iz

1 iz

, (74)

Arccot z= Arctan1z= 1

2i

Lnz+iz i. (75)

Note that with these definitions, the branch cuts are still given by eqs. (14) and (15),respectively. Comparing the above definitions with those of eqs. (12) and (13), onecan check that the two definitions differ only on the branch cuts and at certain pointsof infinity. In particular, there is no longer any ambiguity in how to define Arccot(0).Pluggingz= 0 into eq. (75) yields

Arccot(0) = 1

2iLn(1) = 12 .

It is convenient to define a new variable,

v= i zi+z

, = 1v

=z+iz i. (76)

Then, we can write:

Arctan z+ Arccot z= 1

2i

Ln v+ Ln

1

v

= 1

2i

Ln|v| + Ln

1

|v|

+iArg v+iArg

1

v

=1

2

Arg v+ Arg1v . (77)

11A definition of the principal value of the arccotangent function that is equivalent to eq. (75) forall complex numbers z is:

Arccot z= 1

2i[Ln(iz 1) Ln(iz+ 1)] .

A proof of the equivalence of this form and that of eq. (75) can be found in Appendix C of the firstreference in ref. 5.

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It is straightforward to check that for any non-zero complex number v,

Arg v+ Arg

1

v

=

, for Imv 0 , , for Imv


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Note that we have excluded the pointsx= 0,y = 1, which correspond to the branchpoints where the arctangent function diverges.

Therefore, it follows that in the finite complex plane excluding the branch pointsat z= i,

N=

1 , if Rez= 0 and Im z < 1 ,0 , otherwise.

This means that in the finite complex plane, the two possible definitions for theprincipal value of the arctangent function given by eqs. (12) and (74) differ only onthe branch cut along the negative imaginary axis below z =i. That is, for finitevalues ofz= i,

1

2iLn

1 +iz

1 iz

=

+ 1

2i[Ln(1 +iz) Ln(1 iz)] , if Rez= 0 and Im z < 1 ,

1

2i[Ln(1 +iz) Ln(1 iz)], otherwise .(83)

In order to compare the two definitions of Arctan z in the limit of|z| , we canemploy the relation Arccot z= Arctan(1/z) [which holds for both sets of definitions],and examine the behavior of Arccot z in the limit ofz 0.

The difference of the two definitions of Arccot zgiven by eqs. (13) and (75) followsimmediately from eq. (83). For z= iand z= 0,13

1

2iLn

z+i

z

i=

+ 1

2i

Ln

1 +

i

z

Ln

1 i

z

, if Rez= 0 and 0


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This is in contrast to the behavior of Arccot as defined in Mathematica 8 [cf. eq. (13)],where the value of Arccot is discontinuous at z= 0 on the branch cut, in which caseone must separately define the value of Arccot(0).

So which set of conventions is best? Of course, there is no one right or wrong

answer to this question. The authors of refs. 46 argue for choosing eq. (12) to definethe principal value of the arctangent and eq. (75) to define the principal value ofthe arccotangent. This has the benefit of ensuring that eq. (85) is satisfied so thatArccot(0) is unambiguously defined. But, it will lead to corrections to the relationArccot z= Arctan(1/z) for a certain range of complex numbers that lie on the branchcuts. In particular, with the definitions of Arctanzand Arccotzgiven by eqs. (12)and (85), we immediately find from eq. (84) that

Arccot z=

+ Arctan

1

z

, if Rez= 0 and 0 1 ,1

2 , for Rez


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differently via the relation,

Arccot z= 12

Arctan z . (88)This relation should be compared with the corresponding relations, eqs. (22) and

(79), which are satisfied with the definitions of the principal value of the arccotangentintroduced in eqs. (13) and (75), respectively. Eq. (88) is adopted by the Maple 14computer algebra system, which is one of the main competitors to Mathematica.

The main motivation for eq. (88) is that the principal interval for real values x is

0 Arccot x ,instead of the interval quoted in eq. (20). One advantage of this latter definitionis that the real-valued arccotangent function, Arccotx, is continuous at x = 0, incontrast to eq. (87) which exhibits a discontinuity at x = 0. Note that if one adoptseq. (88) as the the definition of the principal value of the arccotangent, then thebranch cuts of Arccotz are the same as those of Arctanz, namely eq. (14). Thedisadvantages of the definition given in eq. (88) are discussed in refs. 4 and 5.

Which convention does your calculator and/or your favorite mathematics softwareuse? Try evaluating Arccot(1). In the convention of eq. (13) or eq. (75), we haveArccot(1) = 14, whereas in the convention of eq. (88), we have Arccot(1) = 34.

APPENDIX B: Derivation of eq. (22)

To derive eq. (22), we will make use of the computations provided in Appendix A.Start from eq. (79), which is based on the definitions of the principal values of the

arctangent and arccotangent given in eqs. (74) and (75), respectively. We then useeqs. (83) and (84) which allow us to translate between the definitions of eqs. (74)and (75) and the Mathematica 8 definitions of the principal values of the arctangentand arccotangent given in eqs. (12) and (13), respectively. Eqs. (83) and (84) implythat the result for Arctan z+ Arccot zdoes not change if Re z= 0. For the case ofRez= 0, Arctan z+Arccotzchanges from 12to 12 if 0< Im z 0

It is convenient to define:

v= iz+

1 z2 , 1v

= 1

iz+

1 z2 = iz+

1 z2 .

In this Appendix, we shall prove that:

Rev 0 , and Re

1

v

0 . (89)

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Using the fact that Re (iz) = Im z for any complex number z,

Rev = Im z+ |1 z2|1/2 cos 12

Arg(1 z2) , (90)Re

1v

= Imz+ |1 z2

|1/2

cos12Arg(1 z

2

)

. (91)

One can now prove that

Re v 0 , and Re

1

v

0 , (92)

for any finite complex number z by considering separately the cases of Im z < 0,Im z = 0 and Im z > 0. The case of Imz = 0 is the simplest, since in this caseRe v = 0 for|z| 1 and Re v > 0 for|z| > 1 (since the principal value of thesquare root of a positive number is always positive). In the case of Im z

= 0, we

first note that that 0. Likewise, ifIm z >0, then it immediately follows from eq. (91) that Re (1/v)>0. However, thesign of the real part of any complex number zis the sameas the sign of the real partof 1/z, since

1

x+iy =

x iyx2 +y2

.

Hence, it follows that both Re v 0 and Re (1/v) 0, and eq. (89) is proven.

APPENDIX D: Derivation of eq. (68)

We begin with the definitions given in eqs. (39) and (66),14

iArccos z= Ln

z+i

1 z2

, (94)

Arccosh z= Ln

z+

z+ 1

z 1

, (95)

where the principal values of the square root functions are employed following thenotation of eq. (32). Our first task is to relate

z+ 1

z 1 to z2 1. Of course,

14We caution the reader that some authors employ different choices for the definitions of theprincipal values of arccos z and arccoshz and their branch cuts. The most common alternativedefinitions are:

Arccosh z = i Arccosz = Ln(z+

z2 1) , (93)which differ from the definitions, eqs. (94) and (95), employed by Mathematica 8 and these notes.In particular, with the alternative definitions given by eq. (93), Arccos z now possesses the sameset of branch cuts as Arccoshz given by eq. (70), in contrast to eq. (42). Moreover, Arccosz nolonger satisfies eq. (38) if either (Re z )(Imz )< 0 or if|Rez | >1 and Im z = 0 [cf. eq. (99)]. Otherdisadvantages of the alternative definitions of Arccos z and Arccosh z are discussed in ref. 4.

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these two quantities are equal for all real numbersz 1. But, as these quantities areprincipal values of the square roots of complex numbers, one must be more careful inthe general case. We shall make use of eqs. (13) and (77) of the class handout enti-tled, The complex logarithm, exponential and power functions, in which the following

formula is obtained:z1z2= e

1

2Ln(z1z2) =e

1

2(Lnz1+Lnz2+2iN+) =

z1

z2e

iN+ ,

where

N+=

1 , if Arg z1+ Arg z2 > ,0 , if 0 ,Im z= 0 or Re z= 0 ,Im z >0 ,1 , if Rez


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It follow that Arccosh z= Ln(zz2 1), where the sign is identified with above.Noting that z z2 1 = [z+ z2 1]1, where z+ z2 1 is real and negativeif and only if Im z= 0 and Rez 1,15 one finds after applying eq. (36) that:

Ln(z z2 1) = 2i Ln(z+ z2 1) , for Imz= 0 and Re z 1 ,Ln(z+ z2 1) , otherwise .To complete this part of the analysis, we must consider separately the points z= 1.At these two points, eq. (95) yields Arccosh(1) = 0 and Arccosh(1) = Ln(1) =i.Collecting all of the above results then yields:

Arccoshz=

Ln(z+

z2 1) , if Imz >0 , Re z 0 o r I m z= 0 , Re z 1or Im z 0 ,

Ln(z+ z2 1) , if Imz >0 , Re z 0 ,

assuming that z=1. If we put z=x +iy, then 1 z2

= 1 x2

+y2

2ixy, and15Letw = z+

z2 1, and assume that Im w = 0 and Re w = 0. That is,w is real and nonzero,

in which case Im w2 = 0. But

0 = Im w2 = Im

2z2 1 + 2z

z2 1

= Im (2zw 1) = 2wIm z ,

which confirms that Im z = 0, i.e. z must be real. If we require in addition that that Re w < 0,then we also must have Re z 1.

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we deduce that

Arg(1 z2) is

positive , either ifxy 1 ,zero , either ifx = 0 or if y = 0 and|x| 0 .

We exclude the pointsz= 1 (corresponding toy = 0 andx = 1) where Arg(1z2)is undefined. Treating these two points separately, eq. (94) yields Arccos(1) = 0 andiArccos(1) = Ln(1) =i. Collecting all of the above results then yields:

iArccosz=

Ln(z+

z2 1) , if Imz >0 ,Rez 0 o r I mz 0 ,Rez

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The Complex Inverse Trigonometric and Hyperbolic Functions