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According to my book, the principal value of $\tan^{-1}(\frac{-y}{-x})$ is $\tan^{-1}(\frac{y}{x})-\pi$. However, isn't $\tan^{-1}(\frac{-y}{-x})=\tan^{-1}(\frac{y}{x})$ and isn't $\tan^{-1}(\frac{y}{x})<\tan^{-1}(\frac{y}{x})-\pi$. So, isn't $\tan^{-1}(\frac{y}{x})$ the principal value of $\tan^{-1}(\frac{-y}{-x})$?

tryingtobeastoic
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4 Answers4

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The difference between $\tan^{-1}(\frac{y}{x}) $ and

$\tan^{-1}(\frac{-y}{-x})$ is that$$ \tan^{-1}(\frac{y}{x}) $$ is located in the first quadrant while

$$ \tan^{-1}(\frac{-y}{-x}) $$ is located in the third quadrant - in the sense that $(x,y),(-x,-y)$ does. Although they both give similar values when plugged in a calculator, the two differs by their location so you must compensate by $\pi$ accordingly to get the angle required starting from $0$

wd violet
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If $x,y>0$ then $$\tan^{-1} \frac{y}{x}= \tan^{-1}\frac{-y}{-x}= \pi/4$$ But primcipal value of $\arg (x+iy)=\pi/4$ but it is $-3\pi/4$ for $\arg(-x-iy).$

Z Ahmed
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You are correct, and your book is rubbish.

  1. It simultaneously and contradictorily claims:
    • the principal value of $\tan^{-1}\frac yx$ is $\tan^{-1}\frac yx$
    • the principal value of $\tan^{-1}\frac {-y}{-x}$ is $\tan^{-1}\frac yx-\pi.$
  2. Its second and fourth claims similarly contradict each other.
  3. In fact, the four claims are collectively a mess: they mix up clockwise and anticlockwise measurements, the concept of principal angle, and are notationally inconsistent.
ryang
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  • I get what they are trying to say, but their way of explaining it is pretty bad. They should've used the atan2 function as @HansLundmark mentions. – tryingtobeastoic Sep 21 '21 at 08:13
  • @tryingtobeastoic 1. That’s really not the only problem: they claims r still patently wrong even if u write arctan instead of $tan^{-1}$ and treat this as the principal. 2. There is no single interpretation under which the four claims r consistent. – ryang Sep 21 '21 at 08:39
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It is not just algebraic equivalence; the sign of numerator and denominator decides location of the tip of radius vector.

There are four possibilities to determine quadrant placement:

$$\dfrac{x}{y}=\dfrac{+}{+},\quad \dfrac{-}{+},\quad \dfrac{-}{-},\quad \dfrac{+}{-};\quad $$

For a point $(x,y)$ in first quadrant, its principal argument angle is in the first quadrant.

For a point $(-x,-y)$ in third quadrant its principal argument angle is in the third quadrant.

Co-terminal angle for any arctan function can be obtained by addition/subtraction of $k \pi$, meaning the point can always be taken to the vertically opposite quadrant.

Narasimham
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