If z-cos2-15-isin2-15 where i - -1- then 1-z-z2-z3-z7-1-z8-z9-z14 - 2- - z- then - is

Z^15 = 1 1 + z^8 + z^9 + z^10 + .... + z^14 =1+1/z^7+1/z^6+....+1/z =z^7+z^6+z^5+....+z^1+1/z^7 So ( 1+z+z^2+....+z^7/1+z^8+z^9+....+z^14)= (z^7) =7 (z)=14π/15 ...

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Byju's Answer

Standard IX

History

Machiavelli

If z=cos2π/15...

Question

If $z = cos \frac{2 π}{15} + i sin \frac{2 π}{15}$ $(where i = \sqrt{- 1});$ then $arg (\frac{1 + z + z^{2} + z^{3} + . . . . . + z^{7}}{1 + z^{8} + z^{9} + . . . . . . + z^{14}})$ = $2 π + λ arg (¯ z)$ then $λ$ is

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Solution

$z^{15} = 1$
$1 + z^{8} + z^{9} + z^{10} + . . . . + z^{14}$
$= 1 + \frac{1}{z^{7}} + \frac{1}{z^{6}} + . . . . + \frac{1}{z}$
$= \frac{z^{7} + z^{6} + z^{5} + . . . . + z^{1} + 1}{z^{7}}$
So $arg (\frac{1 + z + z^{2} + . . . . + z^{7}}{1 + z^{8} + z^{9} + . . . . + z^{14}}) = arg (z^{7})$
$= 7 arg (z) = \frac{14 π}{15}$

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If z=cos2π15+isin2π15 (wherei=√−1); then arg(1+z+z2+z3+.....+z71+z8+z9+......+z14) = 2π+λarg(¯z) then λ is

z15=1 1+z8+z9+z10+....+z14 =1+1z7+1z6+....+1z =z7+z6+z5+....+z1+1z7 So arg(1+z+z2+....+z71+z8+z9+....+z14)=arg(z7) =7arg(z)=14π15

If $z = cos \frac{2 π}{15} + i sin \frac{2 π}{15}$ $(where i = \sqrt{- 1});$ then $arg (\frac{1 + z + z^{2} + z^{3} + . . . . . + z^{7}}{1 + z^{8} + z^{9} + . . . . . . + z^{14}})$ = $2 π + λ arg (¯ z)$ then $λ$ is