by Ameet Sharma
Last Updated September 21, 2018 13:20 PM

Dealing with curves in the complex plane... curves of the form

$$z(t) = x(t) + iy(t)$$

I'm looking at page 2 of the notes here. The conditions for smooth curves.

https://sites.oxy.edu/ron/math/312/14/ws/17.pdf

In my mind, a parametrized curve is smooth at a particular t if the curve has a tangent line at that t. And for that I'd think having $z(t)$ be continuous and having $z'(t) \ne 0$ would be sufficient.

But the notes require the stronger condition of $z'(t)$ being continuous. I'm wondering why.

So I have 2 questions.

1) Is my idea that a tangent line only requires $z(t)$ be continuous and $z'(t)$ exists and is nonzero wrong (ie do I need the additional condition that $z'(t)$ is continuous?)

2) Is smoothness at a point the same as having a tangent line at a point or is it something else?

It depends on the definition. The existence of a tangent line of the curve parameterized by $z\colon I\subset \mathbb R \rightarrow \mathbb C$, at $t_0\in I$ is equivalent to the existence of $z'(t_0)$ and $z'(t_0)\neq0$. In that case, the tangent line has equation $$l(t)=z(t_0)+t\cdot z'(t_0).$$

With the definition you mention, you have to add the fact that $z'(t)$ is defined around $t_0$ and that it be a continuos function at $t_0$ to say the curve is *smooth* at that point. But really, that depends on the author.

For instance, if $z\colon \mathbb R\rightarrow \mathbb C$ is given by $$z(t)=\left\{\begin{matrix}t+i\cdot t^2\sin\left(\frac1t\right)& \text{if}&t\neq0\\0& \text{if}&t=0,\\\end{matrix}\right.$$ then it can be proven that $z'(t)$ exists in the whole domain and that $$z'(t)=\left\{\begin{matrix}1+i\cdot \Big( 2t\sin\left(\frac1t\right)-\cos\left(\frac1t\right)\Big)& \text{if}&t\neq0\\1& \text{if}&t=0,\\\end{matrix}\right.$$

But this function is not continuous at $t=0$ since $$\lim_{t\to0}z'(t)$$ does not exist (it does not exist the limit of the imaginary part).

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