Exercise 2.  Establish the following identities.

2 (i).   [Graphics:Images/ComplexFunTrigInverseModHome_gr_452.gif].  

Solution 2 (i).

Solution.   Start with  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_453.gif]  and obtain   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_454.gif].   

This gives  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_455.gif].   Multiplying by  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_456.gif]  results in   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_457.gif].  

Use the quadratic formula to solve for  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_458.gif]  and obtain  

                    [Graphics:../Images/ComplexFunTrigInverseModHome_gr_459.gif],  

where the fractional power is the multivalued square root function.   

Taking the multivalued log of both sides gives the desired result:

                    [Graphics:../Images/ComplexFunTrigInverseModHome_gr_460.gif].

Therefore,   [Graphics:../Images/ComplexFunTrigInverseModHome_gr_461.gif].   

We are done.   

Aside.  The principal value of  [Graphics:../Images/ComplexFunTrigInverseModHome_gr_462.gif]  can be calculated with Mathematica.

[Graphics:../Images/ComplexFunTrigInverseModHome_gr_463.gif]

[Graphics:../Images/ComplexFunTrigInverseModHome_gr_464.gif]


[Graphics:../Images/ComplexFunTrigInverseModHome_gr_465.gif]

[Graphics:../Images/ComplexFunTrigInverseModHome_gr_466.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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