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

Product is (1+i).

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#### Solution

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).

Concept: D’Moivre’S Theorem

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