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The problem is: Find all $z \in \mathbb{C}$ such that $ 0< \arg(\frac{z-i}{z+i}) < \frac{\pi}{4} $.

If $z= x+iy$, then $\operatorname{Re}(\frac{z-i}{z+i}) = \frac{x^2+y^2-1}{x^2 +(y+1)^2}$ and $\operatorname{Im}(\frac{z-i}{z+i}) = \frac{-2x}{x^2 +(y+1)^2}$. Also, we have: $$ 0< \frac{-2x}{x^2+y^2-1} <1$$

due to the condition from beginning.

My question is: do we also need the conditions: $\operatorname{Re}(\frac{z-i}{z+i})>0$ and $\operatorname{Im}(\frac{z-i}{z+i})>0$? I know how to plot all regions, but I'm just wondering if these two conditions are necessary. Thanks in advance.

Ted Shifrin
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User154
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    Let $S={z=\xi+i\beta \in \mathbb C:0< Arg(z) < \frac{\pi}{4}}$. Then $S={ z=\xi+i\beta \in \mathbb C:\xi>0,\beta>0,\beta<\xi}$. Isn't it ? – Stéphane Jaouen Oct 25 '22 at 06:34
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    If you know about Möbius (linear fractional) transformations, a more conceptual way to attack the problem is to look at the preimage of the sector $0<\mathop{\rm Arg}(w)<\frac\pi4$ under the map $z\mapsto\frac{z-i}{z+i}$; the two boundary lines must go to circles-or-lines under the inverse Möbius transformation. – Greg Martin Oct 25 '22 at 07:16

3 Answers3

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As indicated by the comment of Stéphane Jaouen, a necessary and sufficient condition on $z = u + iv$ having an Argument $\theta$ in the range $0 < \theta < \dfrac{\pi}{4}$ is for the following condition to be met:

$$0 < v < u.$$

In the posted problem, you have (in effect)

$$u = \frac{x^2+y^2-1}{x^2 +(y+1)^2}, ~~v = \frac{-2x}{x^2 +(y+1)^2}. \tag2 $$

In (2) above, since the (shared) denominator is always positive, except for $(x,y) = (0,-1)$, you are looking for all values of $(x,y)$ except $(0,-1)$, such that

$$0 < -2x < x^2 + y^2 - 1.$$

So, you must have that

  • $x < 0$

  • $0 < x^2 +2x - 1 + y^2 \implies 2 < (x+1)^2 + y^2.$

user2661923
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2

Geometrical Approach:

As I have explained in my answer here, $\arg\left(\dfrac{z-z_1}{z-z_2}\right)$ refers to the counterclockwise angle (measured starting from the segment represented in the denominator) made at the vertex $z$ by segments joining $z,z_1$ and $z,z_2$.

If $\arg\left(\dfrac{z-z_1}{z-z_2}\right)$ is fixed at value $\alpha$, then the locus of $z$ is the major/minor arc of a circle with the segment joining $z_1,z_2$ as a chord, depending on whether $\alpha$ is acute/obtuse respectively, as shown: enter image description here enter image description here

So now you have to plot all those arcs where the chord subtends an angle less than $45^{\circ}$. (The chord will subtend $0^\circ$ at infinity)

The circle on which the chord joining $i$ and $-i$ subtends $45^\circ$ is $(x+1)^2+y^2=2$. (Why?) So you need this region:

enter image description here

If you ask why the circle has its major arc towards -X and not +X, then note that the measurement of the anticlockwise angle must start from the segment joining $z$ and $-i$.

insipidintegrator
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  • "the anticlockwise angle must start from the segment joining z and −i" : I was wrong and I couldn't understand my mistake; thank you :) – Stéphane Jaouen Oct 25 '22 at 08:16
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It is very helpful, to determine the pre-image of the boundaries first. I attack these questions this way. But, I always do mistakes when identifying the region by test points.

Pre-image of $u=v$ is the circle $$(x+1)^2+y^2=2\tag{*}$$ that User2661923 found, the pre-image of $u$-axis is $y$-axis.

Then by choosing test points we find the solution: Left of the $y$-axis and outside of the circle $(*). $

insipidintegrator
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Bob Dobbs
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