TargetIIT

let Z[ss]r[/ss],r=1,2,3.....n are n distinct roots of eq [im]http://codecogs.izyba.com/gif.latex?%5E%7Bn%7DC_%7B1%7Dx+%5E%7Bn%7DC_%7B2%7Dx%5E%7B2%7D+%5E%7Bn%7DC_%7B3%7Dx%5E%7B3%7D+.......%5E%7Bn%7DC_%7Bn%7Dx%5E%7Bn%7D%3D0[/im] in argand plane.If der exist exactly one Z[ss]r[/ss], rε{1,2....n} such that [im]http://codecogs.izyba.com/gif.latex?arg%5Cleft%28%5Cfrac%7BZ_%7Br%7D-%28-1+%5Csqrt%7B2%7Di%7D%7B%28-1%29-%28-1+%5Csqrt%7B2%7Di%29%7D%20%5Cright%29%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D[/im] then n can be ??

pls check the denominator of the arg expression

i checked the denominator......no prob with it.

[image]39118368.jpg[/image] is the same as (x+1)[p]n[/p]=1

The solution to this lies on the red circle in the fig.

[image]39119092.jpg[/image]


z[ss]r[/ss] is a point such that, arg(-1+[sqrt]2[/sqrt]i - z[ss]r[/ss])=∩/4

This when solved, along with the eqn of circle, |z+1|=1  or by using simple geometry gives z[ss]r[/ss]= e[p]i3∩/4[/p]-1

This when substituted in the eqn gives, (e[p]i3∩/4[/p])[p]n/r[/p]=1

=> 3∩n/4 = 2∩ ,   or n/r=8/3,

This gives n=[b]8[/b] and e[p]i3∩/4[/p]-1 is the third root anti-clockwise...........

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yes correct !