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If z+1/z=2cos6^@ , then z^1000+1/[z^1000...

If `z+1/z=2cos6^@` , then `z^1000+1/[z^1000]` +1 is equal to

Text Solution

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Let `z = re^(itheta) = r(costheta+isintheta)`
Then, from De Moivre`'`s theorem,
`z^n = r^n(cosntheta+isinntheta)`
`:. z^1000 = r^1000(cos(1000theta)+isin(1000theta))`
As `z+1/z = 2cos6^@`
`:.z^1000+1/(z^1000) = 2cos(6**1000)^@` (Comparing real parts)
`= 2cos6000^@`
`= 2cos(33pi+pi/3) = 2cos(32pi+pi+pi/3)`
...
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