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Question

If ziz+i(zi) is a purely imaginary number, then z.¯z is equal to


A
1
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B
2
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C
None of these 
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Solution

The correct option is B 1

Here ziz+i=x+i(y1x+i(y+1).xi(y+1)xi(y+1)
=(x2+y21)+i(2x)x2+(y+1)2
As ziz+i is purely imaginary, we get
x2+y21=0  x2+y2=1  z¯z=1.


Mathematics

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