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I'm reading a book on complex analysis. In one step while evaluating a path integral, the author makes the following substitution:

$$\arctan \left(\dfrac{1}{1-α} \right) + \arctan(1-α) = \dfrac{\pi}{2}$$

I'm not sure if it's relevant here, but $α$ is a complex number such that the absolute value of the real and imaginary parts are both less than $1$.

How did the author make this substitution? Is he using some sort of trigonometric identity I'm not aware of? If so, which one is it? And is there a list of such identities? I'd like to make sure I'm not confused next time I see something like this.

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If $\tan \theta = u$, then $\cot \theta = \tan (\tfrac{\pi}{2} - \theta) = \frac{1}{u}$. The "co-" in the cofunctions refers to complementary angles.

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How about a triangle trigonometry answer? The tangent of an angle in a right triangle is the length of the opposite side divided by the length of the adjacent side. If you choose the other acute angle of the right triangle instead, the opposite and adjacent sides reverse themselves, so the tangent of the second angle is the reciprocal of the tangent of the first. Since both acute angles sum to $\frac\pi2$ radians, it makes sense that

$$\tan^{-1}a+\tan^{-1}\frac1a=\frac\pi2$$

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Using $$\tan(a+b) = \frac{\tan a + \tan b}{1- \tan a \tan b}$$ Set $a$ equal to the first $\arctan$ and $b$ to the second $\arctan$. Clearly $ab =1$ so the two angles add to $\pm \pi/2$.

Sum could be $-\pi/2$ also, if the individual terms are negative (note $-\pi/2$ is also $3 \pi/2$.

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  • $\begingroup$ saw your comment about $\alpha$. In this case the final answer has to be $+\pi/2$. In general it could be either $\pm \pi/2$ $\endgroup$
    – user44197
    Jan 14 '14 at 5:47
  • $\begingroup$ Of course! At first I was confused becasue $tana tanb = 1$, which made the denominator $1 - 1 = 0$, which I thought was undefined. I forgot that $tan \frac{\pi}{2}$ itself is undefined. Thanks! $\endgroup$
    – modocache
    Jan 14 '14 at 6:09
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If you take the derivative of the function, you see it is constant. So plug in 0 for example to solve for the constant.

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