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The inequality given is as stated bellow:

$$\frac{\pi}{2}< \arg\left(\frac{z-i}{z+i}\right) \le \frac{3\pi}{4}$$

There are not any other constraints except $z\ne-i$. I tried letting $z=x+yi$, which leads to: $\frac{\pi}{2}<\arctan\big(\frac{2x}{x²+y²-1}\big)\le \frac{3\pi}{4}$, which I tried solving as: $\frac{2x}{x²+y²-1}\le-1$ for the right part but I got stuck. Is this the correct away to approach this problem?

Arturo Magidin
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kesetovic
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  • Hint: Draw a picture of ${w : \pi/2 < \operatorname{arg} w \le 3\pi/4}$. Examine how the linear fractional transformation $w = (z-i)/(z+i)$ transforms it. – GEdgar Dec 08 '23 at 17:47
  • @GEdgar Something tells me that I may be over my head. That is, it may be that my analysis is either incomplete or much less elegant than it needs to be. If there is a better way, then I will delete my answer. What do you think? – user2661923 Dec 08 '23 at 21:23

1 Answers1

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Yes, it is the correct way. The only mistake that you made is that you focused on the tangent, rather than focusing explicitly on both the cosine and sine.

$\pi/2 < \theta \leq 3\pi/4 \implies $

  • $\cos(\theta) < 0.$
  • $\sin(\theta) > 0.$
  • $\sin(\theta) \geq | ~\cos(\theta) ~|.$

The above constraints can be somewhat streamlined by

  • $~\dfrac{1}{\sqrt{2}} \leq \sin(\theta) < 1.$
  • $~\dfrac{-1}{\sqrt{2}} \leq \cos(\theta) < 0.$

However (strangely enough), the first set of constraints above will be easier to deal with, for this problem, because it will allow you to ignore the denominator (explained below).

Starting from scratch:

$$\frac{x + i(y-1)}{x + i(y+1)} \times \frac{x - i(y+1)}{x - i(y+1)} = $$

$$\frac{[ ~x^2 + (y^2 - 1) ~] + i[-2x]}{\text{some positive real number - who cares}}. \tag1 $$

Note:
If $\displaystyle ~z = \frac{x + iy}{s} ~: ~x,y \in \Bbb{R}, ~s \in \Bbb{R^+},~$ then the standard approach is to re-express $~z~$ as

$$z = \frac{\sqrt{x^2 + y^2}}{s} \times \frac{x + iy}{\sqrt{x^2 + y^2}},$$

so that the argument to $~z~$ is the angle $~\theta~$ that has $~\displaystyle \cos(\theta) = \frac{x}{\sqrt{x^2 + y^2}}, ~\sin(\theta) = \frac{y}{\sqrt{x^2 + y^2}}.$

In this problem, since it is sufficient to interrogate the signs of $~\cos(\theta),~$ and $~\sin(\theta),~$ and the relative sizes of $~\cos(\theta)~$ and $~\sin(\theta),~$ the denominator in (1) above can be ignored.

To satisfy the original constraints, you need:

  • $[-2x] > 0 \implies x < 0.$

  • $[x^2 + y^2 - 1] < 0 \implies $
    $y^2 < 1 - x^2.$

  • $~(-2x) \geq 1 - (x^2 + y^2) \implies $
    $y^2 \geq 1 - x^2 + 2x.$

Since $~x^2 + y^2 < 1,~$ you must have that
$~-1 < x < 0.$

Then, $~y~$ can be positive or negative, but you must have:

  • $y^2 < 1 \implies |y| < 1.$

  • $1 - x^2 + 2x \leq y^2 < 1 - x^2.$

user2661923
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