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 $\frac{-2\pi}{3}$
Best Answer
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).