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pratikanand (574)

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1.) if z1 and z2 be the nth roots of unity which subtend right angle at the origin. then n must be of the form
A)4k+1
B)4k+2
C)4k+3
D)4k
 
10000010100000010110101100110010001001111010101110
11011000010100000110010101110110111010111001011001
10000011011111010101100111000101111100011001011111
01111100100000011111000011001010000001110111010010
10101011001100101011010111010101001110011100000100
10011001011010111010010100011101101010110001010101
00111100010111111110001100100110110100000011001100
11011010111010001111011101101100001111111001111011
10110111110111001101100000100111010111111101111011
01001001110011011010110011100011101110000101100010
01001110001011000001111001111100010101111110001000
10000010101110010010001000011100001011101100010110
10010100100001011001000001110001111111010101101001
11000110101111000010101111000101111000001100111111
01011100000001001011110000000010110011100111010110
11101010000000001001110111110011101111010011000110
10110111010110000111110000110011001010001011101100
01110100101000000111000001100111100011101111011110
10010110101110001100010001011000000010010100001010
11100110110001110101110110011011100010111010110101
01001111000000110001111011001111110010101000110010
00000010100111100101111011110011011000101010111011
01000010100110101001110010111011110100111001110100
11101110101000000011001000100110010111010010011010
11110010001101011010101000000000100111101101101111
01011011110101111111101100010001010101100110010101
01110111100111111100111111000001100010111001101100
11010011001101111100110010111111001001010100001110
00110000000111111110101010101100111011100000110001
01001110110010111101010000001100110000100000001001
11000010011010011001111011101000011111000111100100
11110111011000001001101010100110111110111111100110
have a nice day

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And miles to go before i sleep

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spideyunlimited (3079)

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4k + 2.


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nadeemoidu (1184)

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2 consecutive nth roots of unity will subtend an angle of 2pi/ n .
Here pi/2 is a multiple of 2pi/n  . So n should be of the form 4k
Answer is (D).
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sabarish_13 (0)

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the answer is 4K+3

N.SABARISH
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pratikanand (574)

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the answer is 4k. i did it myself later but anyway thanks for helping.
 
Solution:  z1=e^i2pir/n
               z2=e^i2pis/n
 
so    (z2-0)/(z1-0) =e^2pi(r-s)/n
 
so argument should be equal to pi/2
 
    2pi(r-s)/n = pi/2
=> n= 4(r-s) i.e. 4k
hence , (D)

Woods are lovely dark and deep
But i have promises to keep
And miles to go before i sleep
And miles to go before i sleep

" HAVE A NICE DAY "






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