0
$\begingroup$

A simple question, but I like to be clinical with my choice of words:

I have a complex number, $z=-i$.

If I were to calculate the argument of this complex number, $arg(z) = tan^{-1}( \frac{-1}{0}) \approx -\frac{π}{2}rad$.

I know that something is tending towards infinity. What is this something? The argument? Is it just values that go into calculating the argument? Or something else?

$\endgroup$
  • $\begingroup$ You're dividing by zero, you cant do that, also the answer is $-\frac\pi2$ or $\frac{3\pi}2$ not $\frac\pi2$ $\endgroup$ – Alice Ryhl Sep 15 '14 at 19:36
  • $\begingroup$ I understand that you can't divide by zero (one could use a very large number to approximately calculate $arg(z)$), but how would you describe this? Is it the argument that is tending towards infinity? $\endgroup$ – Jack G Sep 15 '14 at 19:39
  • $\begingroup$ arg is only defined as arctangent of $\Re(z)>0$ $\endgroup$ – Alice Ryhl Sep 15 '14 at 19:41
1
$\begingroup$

This really isn't a question specific to arguments and complex numbers.

Essentially the arctan function horizontally asymptotes to $pi/2$ and $-pi/2$, as seen by the graph:

enter image description here

So there actually isn't a value of $x$ (infinity is not a number!) that gives an output of $-pi/2$.

Rather, the limit as $x$ approaches negative infinity for $arctan(x$) yields the $-pi/2$ you are seeking.

In turn $x$ can be made to approach infinity by minimizing the denominator for the opposite/adjacent triangle that defines the tangent. This means a complex number with a minimized real part and a negative imaginary part will yield $-pi/2$. So what is the limit of $x$ (the real part), when it approaches $0$? Well it is $0$ itself.

So the complex number with an argument of $-pi/2$ can be concluded to be of the form $0-ki$ (where $k$ is a positive real number). And do note, we have taken limits to find this.

In order to really make this whole system of complex arguments be less tedious, one can simply define such problematic cases to make them conform to our desires. So always simply state that a complex number only with imaginary parts either has an argument of $3pi/2$ or $pi/2$ - this, we can say is the "definition" and so questions regarding it cease (although is good to have used limits to rigourously justify our definition ;) )!

$\endgroup$
  • 1
    $\begingroup$ Excellent. That's the clarification I needed. $\endgroup$ – Jack G Sep 15 '14 at 19:53
1
$\begingroup$

The problem is that $\arg(z)=\arctan\left(\frac{\Im(z)}{\Re(z)}\right)$ is only true for $\Re(z)>0$

Here's the full definition

$$\arg(x+iy)=\left\{ \begin{array}{lr} \arctan\frac yx&\quad x>0\\ \pi+\arctan\frac yx&\quad y\ge0,x<0\\ -\pi+\arctan\frac yx&\quad y<0,x<0\\ \frac\pi2&\quad y>0,x=0\\ -\frac\pi2&\quad y<0,x=0\\ \color{gray}{\text{undefined}}&\quad y=0,x=0\\ \end{array} \right.$$

Where $x,y\in\mathbb R$

$\endgroup$

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.