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Question

If z=a+ib then its conjugate is aib. If 1,ω,ω2 are cube roots of unity then (i) 1+ω+ω2=0 (ii) ω3=1 The conjugate of 63i7+i is

A
3927i50
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B
39+27i50
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C
39+27i50
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D
3927i50
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Solution

The correct option is D 39+27i50
Z=63i7+i

Z=(63i)(7i)(7+i)(7i)

Z=426i21i3(7)2(i)2

Z=3927i49+1

Z=3927i50

¯¯¯¯Z=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯3927i50

¯¯¯¯Z=39+27i50

Conjugate of 63i7+i=39+27i50

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