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In one of my calculations I arrived at the expression $$i\cdot(\arctan(\frac{y}{x})+\arctan(\frac{x}{y}))$$ I know that $$\arctan(\frac{y}{x})=Arg(z)$$ is there something similar for $\arctan(\frac{x}{y})$ ?

I have tried writing $\theta=\arctan(\frac{y}{x})$, taking $\tan$ of both sides, doing $1/$ of both sides and then take the $\arctan$of both sides to get to $\arctan(\frac{x}{y})$ but that didn't get me anything nice - it got me $\arctan(\frac{1}{\tan(\theta)})$

Can anyone please suggest a link between the argument of something related to $z$ and $\arctan(\frac{x}{y})$ ?

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    $\begingroup$ Here is an identity $\arctan(1/x)+\arctan(x)=\frac{\pi}{2}$. $\endgroup$ – Mhenni Benghorbal Apr 17 '13 at 20:16
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We know this result from analysis $$\arctan\left(\frac{y}{x}\right)+\arctan\left(\frac{x}{y}\right)=\mathrm{sign}\left(\frac{y}{x}\right)\frac{\pi}{2}=\pm\frac{\pi}{2}$$

To see this result with a geometric explication: Multiply a complex number $z$ by $e^{i\theta}$ rotate this number by an angle $\theta$.

If $z=x+iy$ then $iz=e^{i\pi/2}z=-y+ix$ so \begin{align}&\arctan\left(\frac{y}{x}\right)=Arg(z)\\ &\arctan\left(\frac{x}{y}\right)=-\arctan\left(\frac{x}{-y}\right)=-Arg(iz)=\left\{\begin{array}{l}-(\frac{\pi}{2}+Arg(z))\\ -(\frac{\pi}{2}+Arg(z)-\pi)\end{array} \right.\end{align} and the last result is justified by the fact $Arg(z)$ is the principal argument of $z$ so it belongs to $(-\pi,\pi]$, hence we find

$$\arctan\left(\frac{y}{x}\right)+\arctan\left(\frac{x}{y}\right)=\pm\frac{\pi}{2}$$

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  • $\begingroup$ One can compute the derivative of $f(x)=\arctan x+\arctan(1/x)$, which is zero; thus the function is constant in each connected component of its domain. Since $f(-1)=-\pi/2$ and $f(1)=\pi/2$ we have the result. $\endgroup$ – egreg Apr 17 '13 at 21:35
  • $\begingroup$ Yes this is a way to prove this equality. $\endgroup$ – user63181 Apr 17 '13 at 21:37
  • $\begingroup$ Some years ago, in the finals for high schools in my country (a nationwide exam), a question was "Prove that $\arctan x + \arctan\frac{1}{x}$ is constant". :( $\endgroup$ – egreg Apr 17 '13 at 21:39
  • $\begingroup$ The exam's questions are unforgettable :) $\endgroup$ – user63181 Apr 17 '13 at 21:46
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Let $z=x+iy$. Then $Arg(z)=\arctan(\frac{y}{x})$ , as you have written.

What is the relationship of $arctan(\frac{x}{y})$ with $Arg(z)$?

First, observe that it is the argument of the complex number $w=y+ix$, which the symmetric of z, with respect to the line $y=x$.

If you make a simple sketch, you will be able to see easily that:

$$Arg(w)=Arg(z)+2(\dfrac{\pi}{4}-Arg(z))$$ $$=\dfrac{\pi}{2}-Arg(z)$$

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If $x=y=0,$ from here, both the inverse functions are undefined.

If $x\ne y=0, \arctan \frac xy=\arctan \frac x0=$ sgn $(x)\cdot\frac\pi2$

If $xy\ne0,$

we know the general value of $\arctan a+\arctan b$ is $n\pi+\arctan\left(\frac{a+b}{1-ab}\right)$ where $n$ is any integer

So, the general value of $\arctan \frac xy+\arctan \frac yx$ will be

$n\pi+\arctan\left(\frac{\frac yx+\frac xy}{1-\frac yx \cdot\frac xy}\right)$ $=n\pi+\arctan \left(\frac{\frac{x^2+y^2}{xy}}0\right)$ $=n\pi+$sgn$(xy)\cdot\frac\pi2$

As the principal value lies in $\in(-\pi,\pi],$ the principal value of $\left(\arctan \frac xy+\arctan \frac yx\right)$ will be sgn$(xy)\cdot\frac\pi2$

Reference: A question about the arctangent addition formula.

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