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Let me explain my difficulty with this problem.

Q: If $z = x + iy \in \mathbb{C}$ such that $\arg \left[\frac{z-1}{z+1}\right] = 45$ i.e., $\pi/4$ then

  • (a) $x^2-y^2-2x-1 = 0$
  • (b) $x^2+y^2x-1 = 0$
  • (c) $x^2+y^2-2y-1 = 0$

My Approach-

I first simplified the complex number $\arg \left[\frac{z-1}{z+1}\right]$ by substituting $z = x + iy$ and obtained the complex number.

Then I used the formulae $\tan (\theta) = \Im (z)/\Re (z)$ but my doubt is whether we have to check quadrants for the obtained angle or not. I am confused as it is given argument instead of the principal argument. I always check quadrants for only principal argument but I am not sure about the argument.

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  • $\begingroup$ What is the question? If ... then what? $\endgroup$ – Robert Israel Dec 27 '17 at 4:56
  • $\begingroup$ Possible duplicate of Show that $\arg(z-1)=\arg(z+1) +\pi/4$. $\endgroup$ – Rohan Dec 27 '17 at 4:57
  • $\begingroup$ @Robert Israel Do we have to check the quadrants for this question.....that's my doubt $\endgroup$ – Hydrous Caperilla Dec 27 '17 at 4:57
  • $\begingroup$ What exactly is "this question"? $\endgroup$ – Robert Israel Dec 27 '17 at 4:58
  • $\begingroup$ @Robert Israel posted the question $\endgroup$ – Hydrous Caperilla Dec 27 '17 at 5:10
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It's not a matter of "principal argument" vs "argument". If $\pi/4$ is an argument of a point, that is by definition the principal argument.

For the argument to be $\pi/4$ your point must be in the first quadrant, but for $\tan(\theta) = \Im(z)/\Re(z) = 1$ it could be in either first or third quadrant. So if you wanted to check whether a point had argument $\pi/4$, you would need to check the quadrant.

However, that's not quite what's happening here. You are given that $(z+1)/(z-1)$ has argument $\pi/4$, and you want to check whether it satisfies a certain equation. If you find that all $z$ with $\Im((z+1)/(z-1))/\Re((z+1)/(z-1)) = 1$ satisfy that equation, then the answer is yes.

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  • $\begingroup$ Does argument of any complex number has to be it's principal argument or are there some restrictions.I studied that principal argument lies in the interval [-pi,pi].Does this help $\endgroup$ – Hydrous Caperilla Dec 28 '17 at 0:54
  • $\begingroup$ An argument of nonzero complex number $z$ is any $\theta$ such that $z = |z| e^{i\theta}$. If $\theta$ is an argument of $z$, so is $\theta + 2 n \pi$ for any integer $n$. $\endgroup$ – Robert Israel Dec 28 '17 at 2:21
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On substituting $z = x+iy$ you do get:

$$w = \left(\frac{x-1+iy}{x+1+ iy}\right) = \frac{(x-1+iy)(x+1-iy)}{(x+1)^2+(y^2)} = \frac{x^2+y^2-1}{(x+1)^2+(y^2)}+i \frac{2y}{(x+1)^2+(y^2)}$$

From here it is clear that since $\arg w = \tfrac{\pi}{4}$, we have $\tan (\arg w) = 1$

$$\frac{2y}{x^2+y^2-1}=1$$

But here, as you noted, we do need that $2y > 0$ and $x^2+y^2 - 1 > 0$, since $w $, belongs to first quadrant as $\arg w $ is acute.

Alternatively you can solve it using vectors. Two vectors one starting $-1$ and pointed towards $z$ and other starting at $1$ and pointed towards $z$. The angle between them needs to be $45^\circ$ and the angle which $z-1$ vector makes with $+x$ axis needs to be greater here.

You will get major arc of $x^2+y^2 -2y-1 = 0$ with ends $-1$ and $1$ as the answer in either way.

This link might be helpful: Desmos Graph. The desired curve is major arc of red circle with ends $-1,1$:

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