COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
E-mail:  mkt@jbpub.com  
Internet:  http://www.jbpub.com/  
This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 5  ELEMENTARY FUNCTIONS

Needs["Graphics`ComplexMap`"]; 
Needs["Graphics`Colors`"]; 
a = b = d = t = u = v = w = x = y = z = Z = 
                            0; 
Clear[a, b, cos, cosZ, d, eq1, eq2, f, f2, n, r, set, sin, S, S12, 
        S13, t, t1, t2, table, θ, u, U, v, v1, v2, v3, v4, v5, v6, V, 
        w, w0, w1, w2, w3, w4, w5, w6, w7, w8, w9, x, x1, x2, y, y1, y2, z, 
        z0, z1, z2, z3, z4, zn, Z];

Section 5.1	The Complex Exponential Function

Section 5.2	Branches of the Complex Logarithm Function

Section 5.3	Complex Exponents

Section 5.4	Trigonometric and Hyperbolic Functions

Section 5.5  Inverse Trigonometric and Hyperbolic Functions

We expressed the trigonometric and Hyperbolic Functions in Section 5.4 in terms of the exponential function. In this section we look at their inverses. When we solve equations such as w = sin z, we will obtain formulas that involve the logarithm. Since trigonometric and hyperbolic functions are all periodic, they are many-to-one, hence their inverses are necessarily multivalued. The formulas for inverse hyperbolic functions are given by 

			\(arcsin \((z)\) = \(\(\(-\ i\)\ log[i\ z + \((1 - z\^2)\)\^\(1/2\)]\)\(
\)
        \)\)
\(arccos \((z)\) = \(\(\(-\ i\)\ log[
            z + i\ \((1 - z\^2)\)\^\(1/2\)]\)\(
\)
        \)\)
\(arctan \((z)\) = \(i\/2\) 
          log \((\(i\  + \ z\)\/\(i\  - \ z\))\)\)

And we will investigate the principal values for these functions

			\(ArcSin[
          z] = \(\(\(-\ i\)\ Log[i\ z + \@\(1 - z\^2\)]\)\(
\)
        \)\)
ArcCos[
          z] = - i Log[
            z + i (Sqrt[1 - (z^2)])]

        
ArcTan[z] = (i/2) 
          Log[((i  +  z)/(i  -  z))]

Formulas (4-6) on Page 153.  The derivatives of   arcsin(z), arccos(z), arctan(z)   

			(∂_z) ArcSin[z] = (1/(Sqrt[1  -  (z^2)])),  

			(∂_z) ArcCos[z] = ((-1)/(Sqrt[1  -  (z^2)])),  

			(∂_z) ArcTan[z] = (1/(1  +  (z^2))).

Solution (4-6) on Page 153.

Enter the formulas for arcsin(z), arccos(z), arctan(z) and find their derivatives.

Remove[z]; 
f1[
          z_] = - I Log[I z + (Sqrt[1 - (z^2)])]; 
g1[z_] = ArcSin[z]; 
Print["   ArcSin[z] = ", f1[z]]; 
Print[""]; 
\(\(Print[\*""\!\(∂\_z\) ArcSin[z] = "", \(f1'\)[
          z]];\)\ \)
\(\(Print[\*""\!\(∂\_z\) ArcSin[z] = "", 
        Simplify[\(f1'\)[z]]];\)\ \)
Print[""]; 
Print["   f[z] = ", g1[z]]; 
\(\(Print[\*""\!\(∂\_z\) f[z] = "", \(g1'\)[
          z]];\)\ \)

Remove[z]; 
f2[
          z_] = - I Log[
            z + I (Sqrt[1 - (z^2)])]; 
g2[z_] = ArcCos[z]; 
Print["   ArcCos[z] = ", f2[z]]; 
Print[""]; 
\(\(Print[\*""\!\(∂\_z\) ArcCos[z] = "", \(f2'\)[
          z]];\)\ \)
\(\(Print[\*""\!\(∂\_z\) ArcCos[z] = "", 
        Simplify[\(f2'\)[z]]];\)\ \)
Print[""]; 
Print["   f[z] = ", g2[z]]; 
\(\(Print[\*""\!\(∂\_z\) f[z] = "", \(g2'\)[
          z]];\)\ \)

Remove[z]; 
f3[z_] = (I/2) 
          Log[((I  +  z)/(I  -  z))]; 
g3[z_] = ArcTan[z]; 
Print["   ArcTan[z] = ", f3[z]]; 
Print[""]; 
\(\(Print[\*""\!\(∂\_z\) ArcTan[z] = "", \(f3'\)[
          z]];\)\ \)
\(\(Print[\*""\!\(∂\_z\) ArcTan[z] = "", 
        Simplify[\(f3'\)[z]]];\)\ \)
Print[""]; 
Print["   f[z] = ", g3[z]]; 
\(\(Print[\*""\!\(∂\_z\) f[z] = "", \(g3'\)[
          z]];\)\ \)

The Inverse Sine  arcsin(z)

Formula (1) on Page 152.   Verify that the formula  \(ArcSin[
          z] = \(-\ i\)\ Log[i\ z + \@\(1 - z\^2\)]\)  is correct.  (At least for values of z in the upper half plane  Im[z] > 0.)

Solution (1) on Page 152.

First, use Mathematica to determine the formula for ArcSin[z].  Start with the identity  \(z = \(Sin[
            w] = \(e\^\(i\ w\)\  - \ e\^\(\(-\ i\)\ w\)\)\/\(2  i\)\)\).

Remove[w, z, Z]; 
Clear[f, f2, f3, g]; 
eqn = (((E^(I  w)) -  (E^(- I w)))/(2  I)) == 
          z; 
solset = Solve[eqn, w]; 
Print[Sin[w] == z]; 
Print[""]; 
Print[eqn]; 
Print[""]; 
Print[solset]; 
Print[""]; 
\(\(Print["ArcSin[z] = ", 
        solset\_\(\([[\)\(1, 1, 2\)\(]]\)\)];\)\ \)

And we can verify that  - i Log[i z + (Sqrt[1 - (z^2)])]  is the inverse.

\(
\(w = \(-\ I\)\ Log[I\ z + \@\(1 - z\^2\)];\)\ \)
Z = Sin[w]; 
Print["     ArcSin[z] = ", w]; 
Print[""]; 
\(\(Print["        Sin[z] = ", \(E\^\(I\ z\) - E\^\(\(-\ I\)\ z\)\)\/\(2  I\)];\)\ \)
Print[""]; 
Print["Sin[ArcSin[z]] = Sin[", 
        w, "]"]; 
Print["Sin[ArcSin[z]] = ", Z]; 
Print["Sin[ArcSin[z]] = ", 
        Simplify[Z]];

We can use Mathematica to verify the formula graphically, for values of z in the upper half plane  Im[z] > 0.

\(
\(f[
          z_]\  = \ \(-\ I\)\ Log[I\ z\  + \ \@\(1\  - \ z\^2\)];\)\ \)
g[z_]  =  ArcSin[z]; 
CartesianMap[ 
        f, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[-(π/2), (π/2), (π/4)], 
            Range[0, 2, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {DarkGreen, Green}]; 
Print["    The mapping  w = ", 
        f[z]]; 
CartesianMap[ 
        g, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[-(π/2), (π/2), (π/4)], 
            Range[0, 2, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["     The mapping  w = ", 
        g[z]];

Remark.  If you mess around with the square root it will be wrong.  The portion that is supposed to be in the first quadrant appears symmetrically in the fourth quadrant.

\(
\(f2[
          z_]\  = \ \(-\ I\)\ Log[I\ z\  + \ I\ \@\(z\^2\  - \ 1\)];\)\ \)
CartesianMap[ 
        f2, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[-π, π, (π/4)], 
            Range[-2, 2, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}];  
Print["Not so good mapping  w = ", f2[z]]; 
f3[
          z_]  =  - I Log[I z - (Sqrt[1 - (z^2)])]; 
CartesianMap[ 
        f3, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[-π, π, (π/4)], 
            Range[-2, 2, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}];  
Print["Not so good mapping  w = ", 
        f3[z]];

The Inverse Cosine  arccos(z)

Formula (2) on Page 152.   Verify that the formula  ArcCos[
          z] = - i Log[
            z + i (Sqrt[1 - (z^2)])]  is correct.  (At least for values of z in the upper half plane  Im[z] > 0.)

Solution (2) on Page 152.

First, use Mathematica to determine the formula for ArcCos[z].  Start with the identity  \(z = \(Cos[
            w]\  = \(e\^\(i\ w\) + \ e\^\(\(-\ i\)\ w\)\)\/2\)\).

Remove[w, z, Z]; 
Clear[f, f2, g]; 
eqn = (((E^(I w)) + (E^(- I w)))/2) == 
          z; 
solset = Solve[eqn, w]; 
Print[Cos[w] == z]; 
Print[""]; 
Print[eqn]; 
Print[""]; 
Print[solset]; 
Print[""]; 
\(\(Print["ArcCos[z] = ", 
        solset\_\(\([[\)\(2, 1, 2\)\(]]\)\)];\)\ \)

The above formula looks different from ours.  However, the following simplification can be made. 

\(\(-\ i\)\ Log[
            z + \@\(\(-1\) + z\^2\)] = \(\(-\ i\)\ Log[
              z + \@\(\((\(-1\))\) \((1 - z\^2)\)\)] = \(-\ i\)\ Log[z + i \@\( 1 - z\^2\)]\)\)  

And we can verify that  - i Log[z + i (Sqrt[1 - (z^2)])]  is the inverse.

\(
\(w = \(-\ I\)\ Log[
            z + I\ \@\(1 - z\^2\)];\)\ \)
Z = Cos[w]; 
Print["     ArcCos[z] = ", w]; 
Print[""]; 
\(\(Print["        Cos[z] = ", \(E\^\(I\ z\) + E\^\(\(-\ I\)\ z\)\)\/2];\)\ \)
Print[""]; 
Print["Cos[ArcCos[z]] = Cos[", 
        w, "]"]; 
Print["Cos[ArcCos[z]] = ", Z]; 
Print["Cos[ArcCos[z]] = ", 
        Simplify[Z]];

We can use Mathematica to verify the formula graphically, for values of z in the upper half plane  Im[z] > 0.

\(
\(f[
          z_]\  = \ \(-\ I\)\ Log[
            z\  + \ I\ \@\(1\  - \ z\^2\)];\)\ \)
g[z_]  =  ArcCos[z]; 
CartesianMap[ 
        f, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[0, π, (π/4)], Range[-2, 0, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {DarkGreen, Green}]; 
Print["  The mapping  w = ", f[z]]; 
CartesianMap[ 
        g, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[0, π, (π/4)], Range[-2, 0, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["  The mapping  w = ", 
        g[z]];

Remark. If you mess around with the square root it will be wrong. A portion that is supposed to be in the fourth quadrant appears symmetrically in the second quadrant.

\(
\(f2[
          z_]\  = \ \(-\ I\)\ Log[
            z\  + \ \@\(z\^2\  - \ 1\)];\)\ \)
CartesianMap[ 
        f2, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[-π, π, (π/4)], 
            Range[-2, 2, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["Not so good mapping  w = ", f2[z]]; 
f3[
          z_]  =  -I Log[
            z - (Sqrt[-1 + (z^2)])]; 
CartesianMap[ 
        f3, {((-π)/2) + 0.001, (π/2) - 0.001}, {0.001, 4}, 
        Ticks -> {Range[-π, π, (π/4)], 
            Range[-2, 2, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["Not so good mapping  w = ", 
        f3[z]];

The Inverse Tangent  arctan(z)

Formula (3) on Page 152.  Verify that the formula  ArcTan[z] = (i/2) 
          Log[((i  +  z)/(i  -  z))]  is correct.  (At least for values of z in the upper half plane  Im[z] > 0.)

Solution (3) on Page 152.

First, use Mathematica to determine the formula for ArcTan[z].  Start with the identities  \(Cos[
          w] = \(\(e\^\(i\ w\) + \ e\^\(\(-\ i\)\ w\)\)\/2\ \ and\ \ Sin[
              w] = \(e\^\(i\ w\) - e\^\(\(-\ i\)\ w\)\)\/\(2  i\)\)\).

Remove[w, z, Z]; 
Clear[f, g, h]; 
eqn = (((E^(I w)) - (E^(- I w)))/(2  I))/(((E^(I w)) + (E^(- I w)))/2) == z; 
solset = Solve[eqn, w]; 
Print[Tan[w] == z]; 
Print[""]; 
Print[eqn]; 
Print[""]; 
Print[solset]; 
Print[""]; 
\(\(Print["ArcTan[z] = ", 
        solset\_\(\([[\)\(1, 1, 2\)\(]]\)\)];\)\ \)

The above formula looks different.  The following simplifications can be made and then the formulas will differ by the constant π.  Since Tan[z + π] = Tan[z]  both formulas are "right."  

\(\(-i\)\ Log[\(-\(\@\(i - z\)\/\@\(i + z\)\)\)] = \(i\ Log[\(-\(\@\(i + z\)\/\@\(i - z\)\)\)]

        \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(i\ Log[\((\(-1\))\) \@\(i + z\)\/\@\(i - z\)]

          \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(i\ \((Log[\@\(i + z\)\/\@\(i - z\)] + 
                  Log[\((\(-1\))\)])\)

            \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(i\ \((Log[\@\(i + z\)\/\@\(i - z\)] + i\ π)\)

              \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(i\ Log[\@\(i + z\)\/\@\(i - z\)] - \ π

                \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(i\ Log[\@\(\(i\  + \ z\)\/\(i\  - \ z\)\)] - \ π

                  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(\(i\/2\) Log[\(i\  + \ z\)\/\(i\  - \ z\)] - π\)\)\)\)\)\)\)\)\)\)\)\)\)\)   

And we can verify that  (i/2) 
        Log[((i  +  z)/(i  -  z))]  is the inverse.

\(
\(w = \(I\/2\) 
          Log[\(I\  + \ z\)\/\(I\  - \ z\)];\)\ \)
Z = Tan[w]; 
h[
          z_] = (((E^(I z)) - (E^(- I z)))/(2  I))/(((E^(I z)) + (E^(- I z)))/2); 
Print["     ArcTan[z] = ", w]; 
Print[""]; 
Print["        Tan[z] = ", h[z]]; 
Print[""]; 
Print["Tan[ArcTan[z]] = Tan[", 
        w, "]"]; 
Print["Tan[ArcTan[z]] = ", Z]; 
Print["Tan[ArcTan[z]] = ", h[w]]; 
Print["Tan[ArcTan[z]] = ", 
        FullSimplify[Z]];

We can use Mathematica to verify the formula graphically, for values of z in the upper half plane  Im[z] > 0.

\(
\(f[
          z_]\  = \ I\/2\ Log[\(I\  + \ z\)\/\(I\  - \ z\)];\)\ \)
g[z_]  = ArcTan[z]; 
CartesianMap[ f, {+0.00001, (π/2) - 0.00001}, {0.00001, 3.1}, 
        Ticks -> {Range[0, (π/2), (π/4)], Range[0, 2, (1/2)]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {DarkGreen, Green}]; 
Print["   The mapping  w = ", f[z]]; 
CartesianMap[ g, {+0.00001, (π/2) - 0.00001}, {0.00001, 3.1}, 
        Ticks -> {Range[0, (π/2), (π/4)], Range[0, 2, (1/2)]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["     The mapping  w = ", 
        g[z]];

Example 5.11,  Page 154.  Find the values of  arcsin ((Sqrt[2])).

Solution 5.11.

Use the formulas in the text to get the answers.

Remove[w, z]; 
(Subscript[w, 1]) = 
        ComplexExpand[- I (Log[((Sqrt[2]) + 
                        1) I] +  I 2  
                  n π)]; 
(Subscript[w, 2]) = 
        ComplexExpand[- I (Log[((Sqrt[2]) - 
                        1) I] +  I 2  
                  n π)]; 
\(\(Print[\*""arcsin(\!\(\@2\)) = "", 
        w\_1];\)\ \)
Print["       and"]; 
\(\(Print[\*""arcsin(\!\(\@2\)) = "", 
        w\_2];\)\ \)

We can check out the principal values.

Remove[d, w, z]; 
Clear[n]; 
(Subscript[w, 1]) = 
        ComplexExpand[- I (Log[((Sqrt[2]) + 
                      1) I])]; 
(Subscript[z, 1]) = Sin[(Subscript[w, 1])]; 
\(\(Print[\*""\!\(w\_1\) = ArcSin[\!\(\@2\)]""];\)\ \)
\(\(Print[\*""\!\(w\_1\) = "", 
        w\_1];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_1\) = Sin["", 
        w\_1, "]"];\)\ \)
\(\(Print[\*""\!\(z\_1\) = "", 
        z\_1];\)\ \)
(Subscript[z, 1]) = Together[Sin[(Subscript[w, 1])]]; 
(Subscript[n, 1]) = 
        Expand[Numerator[(Subscript[z, 1])] (1 - (Sqrt[2]))]; 
(Subscript[d, 1]) = 
        Expand[Denominator[
              (Subscript[z, 1])] (1 - (Sqrt[2]))]; 
\(\(Print[\*""\!\(z\_1\) = "", 
        z\_1];\)\ \)
\(\(Print[\*""\!\(z\_1\) = ("", n\_1, ")/(", 
        d\_1, ")"];\)\ \)
\(\(Print[\*""\!\(z\_1\) = "", 
        n\_1/d\_1];\)\ \)

(Subscript[w, 2]) = 
        ComplexExpand[- I (Log[((Sqrt[2]) - 
                      1) I])]; 
(Subscript[z, 2]) = Sin[(Subscript[w, 2])]; 
\(\(Print[\*""\!\(w\_2\) = ArcSin[\!\(\@2\)]""];\)\ \)
\(\(Print[\*""\!\(w\_2\) = "", 
        w\_2];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(z\_2\) = Sin["", 
        w\_2, "]"];\)\ \)
\(\(Print[\*""\!\(z\_2\) = "", 
        z\_2];\)\ \)
(Subscript[z, 2]) = Together[Sin[(Subscript[w, 2])]]; 
(Subscript[n, 2]) = 
        Expand[Numerator[(Subscript[z, 2])] (1 + (Sqrt[2]))]; 
(Subscript[d, 2]) = 
        Expand[Denominator[
              (Subscript[z, 2])] (1 + (Sqrt[2]))]; 
\(\(Print[\*""\!\(z\_2\) = "", 
        z\_2];\)\ \)
\(\(Print[\*""\!\(z\_2\) = ("", n\_2, ")/(", 
        d\_2, ")"];\)\ \)
\(\(Print[\*""\!\(z\_2\) = "", 
        n\_2/d\_2];\)\ \)

Example 5.12,  Page 154.  The principal value   Arcsin ((Sqrt[2]))  is  Arcsin ((Sqrt[2]))=(π/2)- i Log (1  +  (Sqrt[2])). 
Suppose that we make specific choices in equation (5-47) by selecting +i as the value of the square root (Sqrt[1 - ((((Sqrt[2])))^2)]) and using the principal value of the logarithm. With f (z) = Arcsin (z), the result is  

	f ((Sqrt[2])) = Arcsin ((Sqrt[2]))=\(\(\(\ \)\(\(\(-\ i\)\ Log[i\ \@2 + i]\)\(=\)\)\)\) (π/2)- i Log (1  +  (Sqrt[2])),  

and the corresponding value of the derivative is given by 

		f'[(Sqrt[2])] = (1/(Sqrt[1 - ((((Sqrt[2])))^2)])) =(1/i) = -i.

Solution 5.12.

Mathematica can find the solution.

f[z_] = ArcSin[z]; 
Print["f[z] = ", f[z] ];  
\(Print[\*""f[\!\(\@2\)] = "", f[\@2]\ ]; \ \)
\(Print[\*""f[\!\(\@2\)] = "", 
      ComplexExpand[f[\@2]]\ ]; \ \)
Print[""];  
Print["f'[z] = ", f'[z] ];  
Print[""]; 
\(\(Print[\*""(Sqrt[1 - ((((Sqrt[2])))^2)]) = "", \@\(1 - \((\@2)\)\^2\)];\)\ \)
Print[""]; 
\(\(Print[\*""f'[(Sqrt[2])] = (1/(Sqrt[1 - ((((Sqrt[2])))^2)]))""];\)\ \)
\(\(Print[\*""f'[\!\(\@2\)] = \!\(1\/I\)""];\)\ \)
\(\(Print[\*""f'[\!\(\@2\)] = "", \(f'\)[\@2]\ ];\)\ \)

The Inverse Hyperbolic Sine  arcsinh(z)

Formula (11) on Page 155.  Verify that the formula  arcsinh (z) = Log[z + (Sqrt[(z^2) + 1])]  is correct, we can verify this graphically.  (At least for values of z in the upper half plane first quadrant.)

Solution (11) on Page 155.

First, use Mathematica to determine the formula for ArcSinh[z].  Start with the identity  \(z = \(Sinh[
            w] = \(e\^w\  - \ e\^\(-\ w\)\)\/2\)\).

Remove[w, z, Z]; 
Clear[f, g]; 
eqn = (((E^w) - (E^(-w)))/2) == 
          z; 
solset = Solve[eqn, w]; 
Print[Sinh[w] == z]; 
Print[""]; 
Print[eqn]; 
Print[""]; 
Print[solset]; 
Print[""]; 
\(\(Print["ArcSinh[z] = ", 
        solset\_\(\([[\)\(2, 1, 2\)\(]]\)\)];\)\ \)

And we can verify that  Log[z + (Sqrt[(z^2) + 1])]  is the inverse.

\(
\(w = Log[z + \@\(z\^2 + 1\)];\)\ \)
Z = Sinh[w]; 
Print["     ArcSinh[z] = ", w]; 
Print[""]; 
\(\(Print["        Sinh[z] = ", \(E\^z - E\^\(-\ z\)\)\/2];\)\ \)
Print[""]; 
Print["Sinh[ArcSinh[z]] = Sinh[", 
        w, "]"]; 
Print["Sinh[ArcSinh[z]] = ", Z]; 
Print["Sinh[ArcSinh[z]] = ", 
        Simplify[Z]];

We can use Mathematica to verify the formula graphically.   (At least for values of z in the upper half plane first quadrant.)

\(
\(f[z_]\  = \ 
        Log[z\  + \ \@\(z\^2\  + \ 1\)];\)\ \)
g[z_]  =   ArcSinh[z]; 
CartesianMap[f, {0.00001, 16.00001, 1}, {0.00001, 16.00001, 1}, 
        PlotRange -> {{0, 3.83}, {0, 1.6}}, 
        Ticks -> {Range[0, 4, 1], Range[0, (π/2), (π/4)]}, 
        AspectRatio -> (π/7.66), 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {DarkGreen, Green}]; 
Print["     The mapping  w = ", f[z]]; 
CartesianMap[g, {0.00001, 16.00001, 1}, {0.00001, 16.00001, 1}, 
        PlotRange -> {{0, 3.83}, {0, 1.6}}, 
        Ticks -> {Range[0, 4, 1], Range[0, (π/2), (π/4)]}, 
        AspectRatio -> (π/7.66), 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["     The mapping  w = ", 
        g[z]];

The Inverse Hyperbolic Cosine  arccosh(z)

Formula (12) on Page 155.  Verify that the formula  arccosh (z) =  Log[z  +  (Sqrt[(z^2)  -  1])]  is correct, we can verify this graphically.   But it is correct only in quadrants  I  and  IV.

Solution (12) on Page 155.

First, use Mathematica to determine the formula for ArcCosh[z].  Start with the identity  \(\(\(\ \)\(z = \(Cosh[
            w] = \(\(\(e\^w\  + \ e\^\(-\ w\)\)\/2\)\(.\)\)\)\)\)\).

Remove[w, z, Z]; 
Clear[f, f2, g]; 
eqn = (((E^w) + (E^(- w)))/2) == 
          z; 
solset = Solve[eqn, w]; 
Print[Cos[w] == z]; 
Print[""]; 
Print[eqn]; 
Print[""]; 
Print[solset]; 
Print[""]; 
\(\(Print["ArcCosh[z] = ", 
        solset\_\(\([[\)\(2, 1, 2\)\(]]\)\)];\)\ \)

And we can verify that  Log[z  +  (Sqrt[(z^2)  -  1])]  is the inverse.

\(
\(w = Log[z\  + \ \@\(z\^2\  - \ 1\)];\)\ \)
Z = Cosh[w]; 
Print["     ArcCosh[z] = ", w]; 
Print[""]; 
\(\(Print["        Cosh[z] = ", \(E\^\(\(\ \)\(z\)\) + E\^\(-\ z\)\)\/2];\)\ \)
Print[""]; 
Print["Cosh[ArcCosh[z]] = Cosh[", 
        w, "]"]; 
Print["Cosh[ArcCosh[z]] = ", Z]; 
Print["Cosh[ArcCosh[z]] = ", 
        Simplify[Z]];

We can use Mathematica to verify the formula graphically.   (But only for values of  z  in quadrants  I  and  IV.)

\(
\(f[z_]\  = \ 
        Log[z\  + \ \@\(z\^2\  - \ 1\)];\)\ \)
g[z_]  =  ArcCosh[z]; 
CartesianMap[ f, {0.0001, 16.0001, 1}, {0.0001, 16.0001, 1}, 
        PlotRange -> {{0, 3.83}, {0, 1.6}}, 
        Ticks -> {Range[0, 4, 1], Range[0, (π/2), (π/4)]}, 
        AspectRatio -> (π/7.66), 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["     The mapping  w = ", f[z]]; 
CartesianMap[ g, {0.0001, 16.0001, 1}, {0.0001, 16.0001, 1}, 
        PlotRange -> {{0, 3.83}, {0, 1.6}}, 
        Ticks -> {Range[0, 4, 1], Range[0, (π/2), (π/4)]}, 
        AspectRatio -> (π/7.66), 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["     The mapping  w = ", 
        g[z]];

Remark. However, for other places it might not agree!  For Example in Quadrant II and III.  Here we must use the other branch of square root!  We use  arccosh (z) = Log[z  - (Sqrt[(z^2) - 1])].

\(
\(f2[z_]\  = \ Log[z\  - \ \@\(z\^2\  - \ 1\)];\)\ \)
CartesianMap[ 
        f2, {-16.0001, -0.0001, 1}, {0.0001, 16.0001, 1}, 
        PlotRange -> {{0, 3.83}, {0, π}}, 
        Ticks -> {Range[0, 4, 1], Range[0, π, (π/4)]}, 
        AspectRatio -> (π/3.83), 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["     The mapping  w = ", 
        f2[z]]; 
Remove[f, z]; 
f[z_]  =  ArcCosh[z]; 
CartesianMap[ 
        f, {-16.0001, -0.0001, 1}, {0.0001, 16.0001, 1}, 
        PlotRange -> {{0, 3.83}, {0, π}}, 
        Ticks -> {Range[0, 4, 1], Range[0, π, (π/4)]}, 
        AspectRatio -> (π/3.83), 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["     The mapping  w = ", 
        f[z]];

The Inverse Hyperbolic Tangent  arctanh(z)

Formula (13) on Page 155.  Verify that the formula  arctanh (z) = (1/2) Log[((1  +  z)/(1 - z))]  is correct, we can verify this graphically.

Solution (13) on Page 155.

First, use Mathematica to determine the formula for ArcTan[z].  Start with the identities  \(Cosh[
          w] = \(\(e\^w\  + \ e\^\(-\ w\)\)\/2\ \ and\ \ Sinh[
              w] = \(e\^w\  - \ e\^\(-\ w\)\)\/2\)\).

Remove[w, z, Z]; 
Clear[f, f2, g, h]; 
eqn = (((E^w) - (E^(- w)))/2)/(((E^w) + (E^(- w)))/2) == 
          z; 
solset = Solve[eqn, w]; 
Print[Tan[w] == z]; 
Print[""]; 
Print[eqn]; 
Print[""]; 
Print[solset]; 
Print[""]; 
\(\(Print["ArcTanh[z] = ", 
        solset\_\(\([[\)\(1, 1, 2\)\(]]\)\)];\)\ \)

The above formula looks different.  The following simplifications can be made and then the formulas will differ by the constant i π.  Since  Tanh[z + i π] = Tanh[z]  both formulas are "right."  

\(Log[\(-\(\@\(\(-1\) - z\)\/\@\(\(-1\) + z\)\)\)] = \(Log[\(-\@\(\(\(-1\) - z\)\/\(\(-1\) + z\)\)\)]

        \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(Log[\((\(-1\))\) \@\(\(1\  + \ z\)\/\(1 - z\)\)]

          \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(Log[\@\(\(1\  + \ z\)\/\(1 - z\)\)] + Log[\((\(-1\))\)]

            \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(\(1\/2\) 
                  Log[\(1\  + \ z\)\/\(1 - z\)] + i\ π\)\)\)\)\)\)\)\)   

And we can verify that  (1/2) Log[((1  +  z)/(1 - z))]  is the inverse.

\(
\(w = \(1\/2\) Log[\(1\  + \ z\)\/\(1 - z\)];\)\ \)
Z = Tanh[w]; 
h[
          z_] = (((E^z) - (E^(- z)))/2)/(((E^z) + (E^(- z)))/2); 
Print["     ArcTanh[z] = ", w]; 
Print[""]; 
Print["        Tanh[z] = ", h[z]]; 
Print[""]; 
Print["Tanh[ArcTanh[z]] = Tanh[", 
        w, "]"]; 
Print["Tanh[ArcTanh[z]] = ", 
        h[w]]; 
Print["Tanh[ArcTanh[z]] = ", 
        FullSimplify[Z]];

We can use Mathematica to verify the formula graphically.   (At least for values of z in the upper half plane first quadrant x > 0  and  y > 0.)

\(
\(f[z_]\  = \ 
        1\/2\ Log[\(1\  + \ z\)\/\(1 - z\)];\)\ \)
f[z_]  =  ArcTanh[z]; 
CartesianMap[ f, {0.00001, 3.00001, 0.1}, {0.00001, 3.00001, 0.1}, 
        PlotRange -> {{0, 2}, {0, (π/2)}}, 
        Ticks -> {Range[0, 2, 1], Range[0, (π/2), (π/4)]}, 
        AspectRatio -> (π/4), AxesLabel -> {"u", "v"}, 
        PlotStyle -> {DarkGreen, Green}]; 
Print["     The mapping  w = ", f[z]]; 
CartesianMap[ f, {0.00001, 3.00001, 0.1}, {0.00001, 3.00001, 0.1}, 
        PlotRange -> {{0, 2}, {0, (π/2)}}, 
        Ticks -> {Range[0, 2, 1], Range[0, (π/2), (π/4)]}, 
        AspectRatio -> (π/4), AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Red}]; 
Print["     The mapping  w = ", 
        f[z]];

Formulas (14-16) on Page 155.  The derivatives of   arcsinh(z), arccosh(z), arctanh(z)  

			(∂_z) ArcSinh[z] == (1/(Sqrt[1  +  (z^2)])),  

			(∂_z) ArcCosh[z] == (1/(Sqrt[(z^2) - 1])),  

			(∂_z) ArcTanh[z] == (1/(1  -  (z^2))).

Solution (14-16) on Page 155.

Remove[z]; 
f1[z_] = Log[z + (Sqrt[(z^2) + 1])]; 
g1[z_] = ArcSinh[z]; 
Print["   ArcSinh[z] = ", f1[z]]; 
Print[""]; 
\(\(Print[\*""\!\(∂\_z\) ArcSinh[z] = "", \(f1'\)[
          z]];\)\ \)
\(\(Print[\*""\!\(∂\_z\) ArcSinh[z] = "", 
        Simplify[\(f1'\)[z]]];\)\ \)
Print[""]; 
Print["   f[z] = ", g1[z]]; 
\(\(Print[\*""\!\(∂\_z\) f[z] = "", \(g1'\)[
          z]];\)\ \)

Remove[z]; 
f2[z_] = Log[z + (Sqrt[(z^2) - 1])]; 
g2[z_] = ArcCosh[z]; 
Print["   ArcCosh[z] = ", f2[z]]; 
Print[""]; 
\(\(Print[\*""\!\(∂\_z\) ArcCosh[z] = "", \(f2'\)[
          z]];\)\ \)
\(\(Print[\*""\!\(∂\_z\) ArcCosh[z] = "", 
        Simplify[\(f2'\)[z]]];\)\ \)
Print[""]; 
Print["   f[z] = ", g2[z]]; 
\(\(Print[\*""\!\(∂\_z\) f[z] = "", \(g2'\)[
          z]];\)\ \)

Remove[z]; 
f3[z_] = (1/2) 
          Log[((1 + z)/(1 - z))]; 
g3[z_] = ArcTanh[z]; 
Print["   ArcTanh[z] = ", f3[z]]; 
Print[""]; 
\(\(Print[\*""\!\(∂\_z\) ArcTanh[z] = "", \(f3'\)[
          z]];\)\ \)
\(\(Print[\*""\!\(∂\_z\) ArcTanh[z] = "", 
        Simplify[\(f3'\)[z]]];\)\ \)
Print[""]; 
Print["   f[z] = ", g3[z]]; 
\(\(Print[\*""\!\(∂\_z\) f[z] = "", \(g3'\)[
          z]];\)\ \)

Example 5.13,  Page 155.  Find the values of  arctanh (1  +  2  i).

Solution 5.13.

Use the formula in the text to get the solutions.

Remove[w]; 
w = 
        Simplify[
          ComplexExpand[
            (1/2) (Log[-1  +  I] + 
                  2  π n)]]; 
Print["arctanh[1+2\[ImaginaryI]] = ", 
        w];

Remark.  The principal value can be found.

(Subscript[w, 1]) = 
        Simplify[
          ComplexExpand[
            ArcTanh[1 + 2  I]]]; 
(Subscript[w, 2]) = 
        Simplify[
          ComplexExpand[(1/2) 
              Log[-1  +  I]]]; 
Print["ArcTanh[1+2\[ImaginaryI]] = ", 
        (Subscript[w, 1])]; 
Print["ArcTanh[1+2\[ImaginaryI]] = ", 
        Expand[(Subscript[w, 1])]]; 
Print[""]; 
\(\(Print[\*""\!\(1\/2\)Log[-1 + I] = "", 
        w\_2];\)\ \)
\(\(Print[\*""\!\(1\/2\)Log[-1 + I] = "", 
        Expand[w\_2]];\)\ \)

Library Research Experience for Undergraduates

Project I. Write a report on Inverse Trigonometric and Hyperbolic Functions.

1. Pennisi, Louis L. and Sjoblom, L., (1959), ''Remarks on the Inverse Trigonometric and Hyperbolic Functions of a Complex Variable,'' Math. Mag., Vol. 33 pp. 39-42.

Section 5.5  Exercises for Inverse Trigonometric and Hyperbolic Functions
See textbook page 198.

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