Why is the range of argument z

1. May 27, 2012

Alshia

Why is the range of argument z more than -∏ and less than or equal to ∏?

2. May 27, 2012

DonAntonio

Choice. You may as well choose any other range of length $\,2\pi\,$ for it.

DonAntonio

3. May 27, 2012

algebrat

Why DonAntonio is correct: (a longer answer)

Depending on the problem you are faced with, it may be more convenient to use rectangular or polar coordinates, that is, either $z=x+iy=(x,y)$, or $z=re^{i\theta}=(r,\theta)$. But with the polar coordinates, you have an issue, that can be resolved by using a covering space which can be visualized like a corkscrew but $r$ goes to infinity (corkscrew is just $r=1$. All angles of $\theta$ and all $r>0$ are represented exactly once on this covering space. Then, to work back on the wonderful complex plane ℂ, we project, so for instance the infinitely long corkscrew ($r=1$) projects to the unit circle on the plane. So usually we don't want more than one angle representing the same point on the plane, so we slice the covering space so that $\theta$ changes by less than $2\pi$. Thus for instance you can take any $\theta_0$, and do work with our polar coordinates for $\theta_0<\theta<\theta_0+2\pi$. This means that we had to remove some ray from the complex plane. Depending on where we want to work on the complex plane, we remove a ray far enough away from where we want to work. We can never make more than a full rotation in the complex plane without some trouble being caused.