Exercise 11.  Suppose that  [Graphics:Images/ComplexAlgebraRevisitedModHome_gr_335.gif].  

11 (b).  Use 11 (a)  and De Moivre's formula to derive Lagrange's identity:  

    [Graphics:Images/ComplexAlgebraRevisitedModHome_gr_339.gif],  where  [Graphics:Images/ComplexAlgebraRevisitedModHome_gr_340.gif].  

Solution 11 (b).

See text and/or instructor's solution manual.

Substitute [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_341.gif] into the formula in part 11 (a) and write     

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_342.gif],  

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_343.gif],  
        
        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_344.gif].  

Now carefully work out the details for finding the real and imaginary parts of the left and right side of this equation.

For the left hand side use De Moivre's formula, and write  

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_345.gif]

On the right hand side and multiply numerator and denominator by [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_346.gif].  

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_347.gif]   

Then equate real parts of the left and right hand sides, and obtain

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_348.gif],

which is what we wanted to prove.

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_349.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_350.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_351.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_352.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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