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Respected everyone.

I have got stuck in the following problem on the determination of principal argument of a non-zero complex number. Can you please help me to find out the answer ? I am not sure about the title either. Please feel free to edit according to the question suitability if necessary.

Let me give in details what is the situation here.

We know that if $z=a+ib$ be a non-zero complex number then either Case 1: $z$ is lying on real or imaginary axis only ( so that either $a$ or $b$ but not both will be zero) Or Case 2 : $z$ is lying in one of the four quadrant ( so that here both $a, b$ will be non-zero) .

For case 1, we can readily find the principal argument $\arg(z)$. If $z=a+i0$ then $\arg(z)$ is $0$ if $a>0$ and $\pi$ if $a<0$. Whereas if $z=0+ib$ then $\arg(z)=\frac{\pi}{2}$ if $b>0$ or $-\frac{\pi}{2}$ if $b<0$. Am I right till here ?

For Case 2: Let $z=a+ib$ where both $a, b$ are non-zero. We define $\alpha:=\tan^{-1}(\frac{b}{a})$.

Now here in wikipedia the complete list has been given how to compute the principal argument of a complex number which I found that the list includes those complex numbers which are on real or imaginary axis.

I am trying to separate the cases as I started this question above. My question is the following : Instead of defining $\alpha$ above, if we define $\alpha:=\tan^{-1}\left(\frac{|a|}{|b|}\right)$ can we say that

$\bullet~ \arg(z)=\alpha$ if $z$ is in 1st Quadrant

$\bullet~ \arg(z)=\pi-\alpha$ if $z$ is in 2nd Quadrant

$\bullet~ \arg(z)=-(\pi-\alpha)$ if $z$ is in 3rd Quadrant

$\bullet~ \arg(z)=-\alpha$ if $z$ is in 4th Quadrant

I am not sure if I am right or not. If the above are wrong, what should be the correct ones ? Please help me. Thank you in advance.

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  • $\begingroup$ Please define domain, range, and rule of assignment for your $\tan^{-1}$ $\endgroup$ – MPW Sep 3 '16 at 3:59
  • $\begingroup$ @MPW Alright. Actually I meant the usual $\arctan$ function. In my schooldays I was taught using the symbol $tan^{-1}$ only. Will it be sufficient ? please let me know. $\endgroup$ – Anjan3 Sep 3 '16 at 4:05
  • $\begingroup$ $\tan$ is many-to-one, so you must define a specific inverse (it has many). That's a huge part of what the question is about. $\endgroup$ – MPW Sep 3 '16 at 4:08
  • $\begingroup$ Ohh ok ok. en.wikipedia.org/wiki/Inverse_trigonometric_functions Can this be ok ? Right now maximum I can say domain is $[0, \infty)$. $\endgroup$ – Anjan3 Sep 3 '16 at 4:09
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Let us define the complex exponential function by $$\sum_{k = 0}^\infty \frac{z^k}{k!}$$ for $z \in \mathbb{C}$ and the trigonometric functions by $$\sin(z) := \frac{e^{iz} - e^{-iz}}{2i}$$ and $$\cos(z) := \frac{e^{iz} + e^{-iz}}{2}$$ Thus we have $$e^{iz} = \cos(z) + i\sin(z)$$ for any $z \in \mathbb{C}$. Consider the following lemma.

Lemma. The function $f:[0,2\pi) \rightarrow \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1\}$, defined by $f(t) := e^{it}$ is bijective.

For $t \in \mathbb{R}$, $t \neq (n + 1/2)\pi$ define $$\tan t := \frac{\sin t}{\cos t}$$ Let $z \in \mathbb{C}$ with $z \neq 0$, then $\frac{z}{\left| z \right|}$ has modulus $1$ an thus by the lemma, there exists $\varphi \in [0,2\pi)$ such that $$\frac{z}{\left| z \right|} = e^{i\varphi}$$ We call $$z = \left| z \right|e^{i\varphi}$$ the polar representation of $z$ and $\varphi$ is called an argument of $z$ since it is only unique modulo $2\pi$.We arbitrarily define the $\varphi$ satisfying $-\pi < \varphi \leqslant \pi$ the principal value of the argument of $z \neq 0$. Let $z := x + iy$ where $x > 0$. Then $$z = \left| z \right| e^{i\varphi} = \left| z \right| \left(\cos \varphi + i\sin \varphi\right)$$ implies $$x = \left| z \right|\cos \varphi$$ and $$y = \left| z \right|\sin \varphi$$ so $$\varphi = \arctan(y/x)$$ where we choose $\arctan$ as the inverse of $\tan$ which is defined on $(-\pi/2,\pi/2)$. Now we have $$ \begin{aligned} \arg(x + iy) = \begin{cases} \arctan(y/x) &x > 0\\ \arctan(y/x) + \pi &x < 0, y \geqslant 0\\ \arctan(y/x) - \pi &x < 0, y < 0\\ \pi/2 &x = 0, y>0\\ -\pi/2 &x = 0, y < 0 \end{cases} \end{aligned}$$

the so defined function $\arg z$ is continuous on $\mathbb{C} \setminus (-\infty,0]$. If you know german, you can find more informations here on pages 81 - 85.

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  • $\begingroup$ So in this case can I now say as deduction : Let $\alpha:= \arctan(\frac{|y|}{|x|})$. Then if $x>0, y>0$ implies $\arg(z)=\alpha$, when $x>0, y<0$ implies $\arg(z)=-\alpha$, when $x<0, y>0$ implies $\arg(z)=\pi-\alpha$ and when $x<0, y<0$ implies $\arg(z)=-(\pi-\alpha)$ ? Please correct me if wrong. $\endgroup$ – Anjan3 Sep 6 '16 at 8:23

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