Whoops, I made a typo!: the original statement was i(1+ix)/(1-ix) so I just multiplied it in. So it should equal (i-x)/(1-ix). Sorry about that!
Also for the previous question, when you divided the fraction by costheta, do we need to say smth like costheta can't equal 0? or is that not required?
With the double angle formula, technically speaking you should, and then check the solutions for \(\cos x = 0\), i.e. \( x=2n\pi \pm \frac\pi2\) separately. There is another method that completely avoids this issue, but it's a bit harder to see. You can request to see that method if you wish in a later post.
\[ \text{Looking at that question now, the question would be split into multiple parts}\\ \text{should it appear in an exam.}\\ \text{The nature of the question looks very assignment-like.} \]
\[ \text{An assumption is made that }x\text{ is real.} \]
\[ \text{The initial expression given is not very manageable, so we're tempted to simplify it.}\\ \text{This should be done in the classic way of division by complex numbers.}\\ \begin{align*} \frac{i(1+ix)}{1-ix} &= \frac{i(1+ix)^2}{(1+ix)(1-ix)}\\ &= \frac{i(1+2ix-x^2)}{1^2+x^2}\\ &= \frac{i(1+2ix-x^2)}{1+x^2}\\ &= \frac{-2x}{1+x^2} + \frac{1-x^2}{1+x^2}. \end{align*} \]
\[ \text{Personally I feel there should be more cases - the case }x\leq 1\text{ looks a bit too powerful}\\ \text{at first glance. But I could be wrong.}\\ \text{For now, here's the solution to the second case:}\]
It might be fine if it is replaced by \( 0 < x \leq 1\) though, but then the bit about \(-\infty\) would be bad.
\[ \text{Suppose now that }x > 1. \\ \text{Then, note that }-\frac{2x}{1+x^2} < 0\\ \textbf{and }\frac{1-x^2}{1+x^2} < 0.\\ \text{Therefore, the complex number will be in the third quadrant.}\]
\[ \text{Upon drawing the diagram, we see that the related angle }\theta\text{ satisfies}\\ \tan \theta = \frac{x^2-1}{2x}.\\ \text{Therefore the argument will be}\\ \arg z = -\pi + \tan^{-1} \left( \frac{x^2-1}{2x}\right) \]
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\[ \text{So we've boiled the problem down to proving that}\\ \boxed{-\pi + \tan^{-1} \frac{x^2-1}{2x} = -\frac{3\pi}{2} + 2\tan^{-1}x}. \]
Of course, this may not be the recommended approach. But if the source of your question hasn't given any hints, anything goes. Here, I will now treat this problem similarly to a harder 3U question. On a separate sheet of paper, I've done some working backwards to make sure that this proof works. But I only present my answer with the working in a forwards, logical manner. (The intuition is that proving this statement is equivalent to proving that \( \frac\pi2 - 2\tan^{-1}x = \tan^{-1} \frac{1-x^2}{2x} \) for \(x > 1\).)
\[ \text{Observe that}\\\begin{align*} \tan \left( \frac\pi2 - 2\tan^{-1}x \right) &= \cot \left(2\tan^{-1}x \right)\\ &= \frac{1}{\tan (2\tan^{-1}x)}\\ &= \frac{1}{\left( \frac{2\tan(\tan^{-1}x)}{1-\tan^2(\tan^{-1}x)} \right)}\\ &= \frac{1}{\left( \frac{2x}{1-x^2} \right)}\\ &= \frac{1-x^2}{2x}\end{align*}\]
\[ \text{Hence taking inverse tan on both sides,}\\ \begin{align*} \frac\pi2 - 2\tan^{-1}x &= \tan^{-1} \frac{1-x^2}{2x}\\ \therefore \frac\pi2 - \tan^{-1} \frac{1-x^2}{2x} &= 2\tan^{-1}x\\ \frac\pi2 + \tan^{-1} \frac{x^2-1}{2x} &= 2\tan^{-1}x\\ \therefore -\pi + \tan^{-1} \frac{x^2-1}{2x} &= -\frac{3\pi}{2} + 2\tan^{-1}x \end{align*} \]
There's one huge subtlety here - I did not justify why we could cancel out the tan-inverse with the tan on the LHS. In general, \( \tan^{-1}(\tan x) = x \) only when \( -\frac\pi2 < x < \frac\pi2\).
Indeed, this situation holds here. Note that if \(x > 1\), then \( \frac\pi4 < \tan^{-1}x < \frac\pi2\). This can then be rearranged to give \( -\frac\pi2 < \frac\pi2 - 2\tan^{-1}x < 0 \), which satisfies the criteria that \( -\frac\pi2 < \frac\pi2 - 2\tan^{-1}x < \frac\pi2 \) regardless, so the cancellation is justified here.