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I am studying undergraduate complex analysis, and in my Textbook the author claimed that


when he was doing an example regarding to principle argument. The origin of his claim is definitely not layed in his book so I was wondering if there is a rigorous proof that can explain this equality.

I have tried to prove this statement using Taylor series but I just can't reduce the term to the equation above.

Any hint would be much appreciated

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The simplest way to see this is to draw a right triangle with an angle having opposite side 2 and adjacent side 1. The other acute angle will be the complement of this angle, and it will have opposite side 1 and adjacent side 2. – user84413 Sep 4 '13 at 0:29
up vote 7 down vote accepted

Note that $$ \tan(\pi/2-x)=\frac{\sin(\pi/2-x)}{\cos(\pi/2-x)}. $$ But $\sin(\pi/2-x)=\cos(x)$ and $\cos(\pi/2-x)=\sin(x)$. So, we have $$ \tan(\pi/2-x)=\frac{\cos(x)}{\sin(x)}=\frac{1}{\tan(x)}. $$ So, if you take $$ \tan\left(\frac{\pi}{2}-\arctan(2)\right)=\frac{1}{\tan(\arctan(2))}=\frac{1}{2}. $$ Since $-\frac{\pi}{2}<\frac{\pi}{2}-\arctan(2)<\frac{\pi}{2}$, this implies that $$ \frac{\pi}{2}-\arctan(2)=\arctan\left(\frac{1}{2}\right), $$ as claimed.

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Let $f:(0,\infty)\to\mathbb R$ be defined by $f(x)=\arctan x+\arctan \frac1x$. Then $$f'(x)=\dfrac{1}{1+x^2}+\dfrac{1}{1+\left(\frac{1}{x}\right)^2}\cdot\dfrac{-1}{x^2}=0$$ for all $x>0$. Hence $f$ is constant on $(0,\infty)$ and $f(x)=f(1)=\frac\pi2$ for all $x>0$, including $x=2$.

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If you want a proof without (overt) appeal to geometry, you can observe that $\arctan 2$ is the principal argument of $1+2i$, and $\arctan \frac 12$ is the principal argument of $1+\frac12 i$. Therefore their sum is an argument of $(1+2i)(1+\frac12 i) = \frac52 i$, so it must be $\pi/2+2\pi n$ for some $n$. But by considering the range of $\arctan$ we can see that $n=0$ is the only possibility, so $$ \arctan 2 + \arctan\frac12 = \frac\pi2 $$

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In fact arctanx=pi/2-arctan1/x, there are few ways of proving this, one way is the right-angled triangle way suggested by user84413 who just commented on your post, or you could tan both sides and carefully show they are both equivalent.

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