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Question
arg(z) + arg`barz (barz ≠ 0)` is ______.
Solution
arg (z) + arg `barz (barz ≠ 0)` is 0.
Explanation:
arg(z) + arg`(barz) (barz ≠ 0)`
If arg(z) = θ, then arg`(barz)` = –θ
So θ + (–θ) = 0
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