Question

Find all values of `(1+i)^(1/3)` & show that their continued
Product is (1+i).

Answer 1

Let           x=`(1+i)^(1/3)`

∴ `x^3=1+i=sqrt2(1/sqrt2+i/sqrt2)`

∴ `x^3=sqrt2[cos (pi/4)+isin (pi/4)]`

Add period 2k 𝝅 , 

`x^3=sqrt2[cos(1/3)(pi/4+2kpi)+isin(pi/4+2kpi)]`

By applying De Moivres theorem, 

`x=2sqrt2[cos(1/3)(pi/4+2kpi)+isin(1/3)(pi/4+2kpi)]`

where k =0,1,2. 

Roots are : 

Put k=0      ` x_0=2sqrt2_e i pi/12`

Put k=1     ` x_1=2sqrt2_ei(9pi)/12`

`Put  k=2    x_2=2sqrt2_ei(17pi)/12` 

The continued product of roots is given by , 

`x_0x_1x_2`= `2sqrt2eipi/12xx2sqrt2ei(9pi)/12xx2sqrt2ei(17pi)/12`

        =`16 sqrt2_e i(27pi)/12`

      = `sqrt2(1/sqrt2+i/sqrt2)` 

     = 1+i 

The continued product of roots is (1+i).