Let z=peiθ be a complex number. What is the argument of 1−z― in terms of θ? Is it π−θ because the argument of −z― is π−θ? Follow-up: Is there a way to simplify the argument: arctan⁡(ρsin⁡θ1−ρcos⁡θ)?

Question

Let z=peiθ be a complex number. What is the argument of 1−z―  in terms of θ? Is it π−θ because the argument of −z―  is  π−θ?  Follow-up: Is there a way to simplify the argument: arctan⁡(ρsin⁡θ1−ρcos⁡θ)?

Micheal Hensley · Accepted Answer

Since tan⁡(−θ2)=sin⁡{(−θ)}1+cos⁡{(−θ)}=−sin⁡θ1+cos⁡θ=−sin⁡θ1−(−cos⁡θ)if ρ=−1 , you would have ρsin⁡{θ}1−ρcos⁡θ which is the form you are looking for. I don&#039;t know if there is any other way to reduce for other values of  eho but in general, i believe it is as far as you can go.

Let z=pe^{i \theta} be a complex number. What is the

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