Exercise 2.  Establish the following identities.

2 (a).   [Graphics:Images/ComplexFunTrigInverseModHome_gr_374.gif].  

Solution 2 (a).

Solution.   Start with  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_375.gif],

so that   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_376.gif],   or   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_377.gif].   

Solving for  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_378.gif]  gives   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_379.gif],

where the fractional power is the multivalued square root function.   

Taking the multivalued log of both sides gives  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_380.gif],

Hence,  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_381.gif].

Therefore,   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_382.gif].  

Aside.   In calculating the principal value of   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_383.gif],  Mathematica manipulates this formula as follows:   

                    [Graphics:../Images/ComplexFunTrigInverseModHome_gr_384.gif]   

Therefore,  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_385.gif].  

We are done.   

Aside.  The principal value of  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_386.gif]  can be calculated with Mathematica.

[Graphics:../Images/ComplexFunTrigInverseModHome_gr_387.gif]

[Graphics:../Images/ComplexFunTrigInverseModHome_gr_388.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell