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Question

Given z is a complex number satisfying z2z|z|2+64|z|5=0 and Re(z) 12 (where Re(z) denotes real part of z), then |z| is less than


A

1

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B

2

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C

3

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D

4

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Solution

The correct options are
C

3


D

4


z2z=|z2|64|z|5 = a purely real number
z2z=¯z2¯z (z2¯z2)(z¯z)=0
if z¯z=0 z=¯z
z is purely real number
So put z = x in given equation.
x2x|x2|+64x5=0 (if x>0)
( x < 0 not possible)


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