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.

  • $\begingroup$ What is the question? If ... then what? $\endgroup$ Dec 27 '17 at 4:56
  • $\begingroup$ Possible duplicate of Show that $\arg(z-1)=\arg(z+1) +\pi/4$. $\endgroup$
    – user371838
    Dec 27 '17 at 4:57
  • $\begingroup$ @Robert Israel Do we have to check the quadrants for this question.....that's my doubt $\endgroup$ Dec 27 '17 at 4:57
  • $\begingroup$ What exactly is "this question"? $\endgroup$ Dec 27 '17 at 4:58
  • $\begingroup$ @Robert Israel posted the question $\endgroup$ Dec 27 '17 at 5:10

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.

  • $\begingroup$ 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 $\endgroup$ 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 $n$. $\endgroup$ Dec 28 '17 at 2:21

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 $\tan (\arg w) = 1$


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 $-1,1$:


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.