Problem with $arg(\gamma (t))$

Question

I am see my notes about curves on complex spaces and I do not understand why it is so... I need help.

I do not understand what way take to do it, I need someone explain slowly please, thanks

Answer 1

Your notes are wrong, it's more like:

• Between $a$ and $t_1$, take $\arg(\gamma(t))$

As $\lim_{t\to t_1^-} \arg(\gamma(t)) = \pi$, you need that the next values be in the $]\pi, 3\pi]$ range, so

• Between $ t_1 $ and $t_2$, take $\arg( \gamma(t) )+2\pi$

As $\lim_{t\to t_2^-} \arg(\gamma(t))+2\pi = \pi$, you need that the next values be in the $]-\pi, \pi]$ range, so

• Between $ t_2 $ and $t_3$, take $\arg( \gamma(t) )$

Here, you miss a point, that I'll call $t_a$ at the intersection of your curve and the x axis in the negative part of the axe.

As $\lim_{t\to t_3^-} \arg(\gamma(t) ) = \pi$, you need the next value in the $]\pi,3\pi]$ range, so

• Between $ t_2 $ and $t_a$, take $\arg( \gamma(t) )+2pi$

As $\lim_{t\to t_a^-} \arg(\gamma(t) )+2\pi = 3\pi$, you need the next value in the $]3\pi, 5 \pi]$ range, so

• Between $ t_a $ and $t_4$, take $\arg( \gamma(t) )+ 4\pi$

And as $\lim_{t\to t_a^-} \arg(\gamma(t) )+4\pi = 5\pi$, you need the next value in the $]5\pi, 7 \pi]$ range, so

• Between $ t_4 $ and $b$, take $\arg( \gamma(t) )+ 6\pi$

Answer 2

If the curve $$\gamma:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\in{\mathbb R}^2\setminus\{{\bf 0}\}\qquad(a\leq t\leq b)$$ is continuously differentiable then you can define a continuous argument $t\mapsto\theta(t)$ $(a\leq t\leq b)$ along $\gamma$ as follows:

While the argument function ${\rm arg}$ is only defined up to multiples of $2\pi$ its gradient $\nabla{\rm arg}$ is well defined in the full punctured plane: $$\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2}, {x\over x^2+y^2}\right)\qquad\bigl((x,y)\ne(0,0)\bigr)\ .$$ It follows that $$\dot\theta(t)=\nabla {\rm arg}\bigl({\bf z}(t)\bigr)\cdot\dot{\bf z}(t)={x\dot y-y\dot x\over x^2+y^2}(t)$$ is well defined for all $t\in[a,b]$. Therefore you can choose an admissible value for $\theta(a)$ and then have $$\theta(t)=\theta(a)+\int_a^t {x(\tau)\dot y(\tau)-y(\tau)\dot x(\tau)\over x^2(\tau)+y^2(\tau)}\>d\tau\qquad(a\leq t\leq b)\ .$$ Written in terms of complex $z=x+iy$ instead of ${\bf z}=(x,y)$ this becomes $$\theta(t)=\theta(a)+{\rm Im}\int_a^t {\dot z(t)\over z(t)}\>dt\ .$$ The curve in your picture would then start with the argument $\theta(a)={3\pi \over 4}$, say, and end with the argument $\theta(b)={5\pi\over4}+4\pi$, because along $\gamma$ we see two surroundings of the origin before ending at $\gamma(b)$ in the third quadrant.