Prove that (a+ib)m\n+(a-ib)m\n=2(a2+b2)m\2n .cos((m\n)(tan-1b\a))

     

          Bs GREWAL

     EXERCISE SOLUTION   19.2

     A number of the form x+iy, where x and y are real numbers and i= (-1)^1/2 ,

  Is called a complex number.

    Every complex number x+iy can always be expressed in the form r(cosӨ+isinӨ).

           R(x+iy) .i.e. x=rcosӨ

             I(x+iy), .i.e. y=rsinӨ

                                                                                                                         And squaring and adding ,we get x^2+y^2=r^2 .i.e. r=(x^2+y^2)^1/2  (taking positive   roots only)            

    y/x=tanӨ   .i.e. Ө=tan-1(y/x).

                   

          

(a+ib)m\n+(a-ib)m\n=2(a2+b2)m\2n.cos(m\ntan-1b\a)



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