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Find the argument of $1+i\tan(x)$.

My approach (assuming I solve for Principal argument):

$\arg(1+i\tan(x))= \arctan\left(\frac{\tan(x)}{1}\right)=x$.

However, the answer in the textbook is $x-\pi$ (for Principal argument) otherwise, $x+k\pi$, where $k$ is an integer.

What did I do wrong?

EDIT: I don't think it adjusts the angle. Consider the following: It will adjust the angle for example for $tan(\frac{2\pi}{3})$ - the complex number makes an angle $-\frac{\pi}{3}$ with the real axis, and $\frac{2\pi}{3}-\pi=\frac{-\pi}{3}$ indeed adjusts it. But for $tan(\frac{\pi}{3})$ it "tries" to adjust the angle ( $\frac{\pi}{3}-\pi=\frac{-2\pi}{3}$ ) but it shouldn't, because the resulting complex number would make an angle $\frac{\pi}{3}$ with the real axis, not $\frac{-2\pi}{3}$

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  • $\begingroup$ In your textbook $Arg(z)$ belongs in $(-\pi,\pi]$, or in a different set? $\endgroup$ – Nikolaos Skout Mar 24 '16 at 13:51
  • $\begingroup$ If tan isnt in first quadrant or third quadrant itll be negative so $+k\pi$ adjusts it $\endgroup$ – Archis Welankar Mar 24 '16 at 13:53
  • $\begingroup$ @NikolaosSkout - in $(-\pi, \pi]$ $\endgroup$ – Richard Smith Mar 24 '16 at 14:01
  • $\begingroup$ @ArchisWelankar - EDIT: I think I got what you mean. It will adjust the angle for example for $tan(\frac{2\pi}{3})$ - the complex number makes an angle $-\frac{\pi}{3}$ with the real axis, and $\frac{2\pi}{3}-\pi=\frac{-\pi}{3}$ indeed adjusts it. But for $tan(\frac{\pi}{3})$ it "tries" to adjust the angle $\frac{\pi}{3}-\pi=\frac{-2\pi}{3}$ but it shouldn't, because the resulting complex number would make an angle $\frac{\pi}{3}$ with the real axis, not $\frac{-2\pi}{3}$ $\endgroup$ – Richard Smith Mar 24 '16 at 14:44
  • $\begingroup$ So k can be $0$?? $\endgroup$ – Archis Welankar Mar 24 '16 at 14:50
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Compute the modulus: $$ |1+i\tan x|^2=1+\tan^2x=\frac{1}{\cos^2x} $$ Depending on whether $\cos x>0$ or $\cos x<0$, the modulus is $1/\cos x$ or $-1/\cos x$.

Case $\cos x>0$

$$ 1+i\tan x=\frac{1}{\cos x}(\cos x+i\sin x) $$ and the argument is $x$ (reduced to whatever interval you choose as the principal one).

Case $\cos x<0$

$$ 1+i\tan x=\frac{-1}{\cos x}(-\cos x-i\sin x)= \frac{-1}{\cos x}(\cos(\pi+x)+i\sin(\pi+x)) $$ and the argument is $\pi+x$ (reduced to whatever interval you choose as the principal one).

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Convention usually says that $$ -\pi\lt\arg(z)\le\pi\tag{1} $$ $1+i\tan(x)$ is always in quadrants $1$ or $3$. Thus, $$ -\frac\pi2\le\arg(1+i\tan(x))\le\frac\pi2\tag{2} $$ and $$ \begin{align} \arg(1+i\tan(x)) &=\tan^{-1}(\tan(x))\\[6pt] &\equiv x\pmod\pi\tag{3} \end{align} $$ There is only one value that satisfies $(2)$ and $(3)$. That is, $$ x-\pi\left\lfloor\frac x\pi+\frac12\right\rfloor\tag{4} $$ Except when $x=\frac\pi2\pmod\pi$ where $\tan(x)$ is undefined (or $\infty$).

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  • $\begingroup$ Thanks for the answer, it was helpful. :) I have two more questions, though: 1) So the textbook answer was wrong? According to the book answer I wrote in the first post, the principal value is $x-\pi$, which, I think, doesn't agree with your answer. 2) Your answer is in principal value, could it be written in a simpler way if we didn't care about principal value? $\endgroup$ – Richard Smith Mar 24 '16 at 17:56
  • $\begingroup$ It depends on where $x$ lives. If $\frac\pi2\lt x\lt \frac{3\pi}2$, then the answers agree. $\endgroup$ – robjohn Mar 24 '16 at 19:58
  • $\begingroup$ Note that $\left\lfloor\frac x\pi+\frac12\right\rfloor$ is an integer, so $(4)$ represents an integer multiple of $\pi$ added to or subtracted from $x$. $\endgroup$ – robjohn Mar 24 '16 at 20:04

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