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Question

lf z1,z2,...,zn lie on the circle |z|=2, then the value of |z1+z2++zn|41z1+1z2++1zn is

A
0
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B
n
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C
n
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D
1
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Solution

The correct option is A 0
|1/z1 + 1/z2 + 1/z3 + ... + 1/zn |

= |¯z1/|z1|2 + ¯z2/|z2|2 + ... + ¯zn/|zn|2 | (Multiplying and dividing by conjugate)
=14 |¯z1+¯z2+...+¯zn| (since |z| = 2 and z1 , z2 , ... lies on this circle)

Putting it into given equation gives result equals 0.
Hence, Option A is correct

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