When i calculated it by $\tan^{-1}\dfrac{y}{x}$, I got $\dfrac{\pi}{4}$, then i added $\pi$ to make it in the right quadrant, so my final answer is $\dfrac{5\pi}{4}$
However, the correct answer is $-\dfrac{3\pi}{4}$...why is that?
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3The "principal argument" is often taken to be normalized to lie in $(-\pi,\pi]$. Then instead of adding $\pi$, you should have subtracted $\pi$. – Daniel Fischer Jan 12 '14 at 15:32
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Now thats a good answer...thanks! :) – Mohdak Jan 12 '14 at 15:37
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Using the definition of atan2, $$\arg(-5-5i)=\arctan\left(\frac{-5}{-5}\right)-\pi=\arctan1-\pi=\frac\pi4-\pi$$
lab bhattacharjee
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Wish I know the mistake here that caused the down-vote. The link contains all the information why the subtraction is done and also the range, so is not included in the answer – lab bhattacharjee Jan 12 '14 at 15:35
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because saying that the principal argument must be between -pi and pi is a much much better and simpler answer, thanks anyway. – Mohdak Jan 12 '14 at 15:39
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@MohammedAlKhatib, please find "This produces results in the range" in the link – lab bhattacharjee Jan 12 '14 at 15:40
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Your answer is absolutely correct except the notation that you need to use $\mathrm{Arg}$. – NasuSama Jan 12 '14 at 15:44
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@NasuSama, thanks for your feedback. But, its really unfortunate that OP has concern with the answer – lab bhattacharjee Jan 12 '14 at 15:47
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@NasuSama http://math.stackexchange.com/questions/635542/operatornamearg-and-operatornamearg-whats-the-difference – Mohammad Jan 12 '14 at 15:50