Question

Show that all roots of `(x+1)^6+(x-1)^6=0` are given by -icot`((2k+1)n)/12`where k=0,1,2,3,4,5.

Answer 1

`(x+1)^6+(x-1)^6=0` 

∴ `(x+1)^6=(-x-1)^6` 

∴ `(x+1)^6/(x-1)^6=-1`

∴ `((x+1)/(x-1))^6=e^(ipi)`         `{∵ e^(ipi)=cospi+isinpi=-1+i(0)=-1}`      (Principal value) 

∴` ((x+1)/(x-1)^6=e^(i(pi+2kpi))`         , k=0,1,2,3,4,5 

∴ `(x+1)/(x-1)=e(ipi(1+2k)/6)`

Let `2θ= (pi(1+2k))/6`

∴ From (1) & (2),` (x+1)/(x-1)=e^(i2θ)`

∴ By Componendo – Dividendo,`( (x+1)+(x-1))/((x+1)-(x-1))=(e^12θ+1)/(e^12θ-1)` 

∴ `2x/2= (e^(iθ)[e^(iθ)+e^(-iθ)])/(e^(iθ)[e^(iθ)-e^(-iθ)])` 

∴ `x=(2cosθ)/(2isinθ)`      {∵`sinθ=(e^(iθ)-e^(-iθ))/(2i) and cos θ= (e^(iθ)+e^(-iθ))/(2) }`

∴` x=1/i cotθ ` 

∴ `x=-icot  ((2k+1)n)/12`(From 2) where k = 0, 1, 2, 3, 4, 5

 

 

 

 

 

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