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|z(1)|=|z(2)|" and "arg z(1)+argz(2)=0 t...

`|z_(1)|=|z_(2)|" and "arg z_(1)+argz_(2)=0` then

A

`z_(1)=z_(2)`

B

`z_(1)= -z_(2)`

C

`z_(1)=bar(z_2)`

D

`z_(1)= -bar(z_2)`

Text Solution

Verified by Experts

The correct Answer is:
C
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