# Confused by conditions for smoothness for a complex curve. Tangent exists vs smoothness

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?

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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).

Alejandro Nasif Salum
September 20, 2018 22:37 PM