Continuity of $f(z)=\overline{z}/z$ I have the following problem:
Is the function $f:\mathbb{C}\to\mathbb{C}$ defined by $f(z)=\dfrac{\overline z}{z}$ continuous over $\mathbb{C}$?
I don't understand. I think $f$ isn't even a function (what is $f(0)$?)
What would you answer?
 A: Well, $\overline{z}$ is a continuous function and $z$ is a continuous function, so it follows that $$f(z)=\frac{\overline{z}}{z}$$ is continuous everywhere in $\mathbb{C}-\{0\}$. We only need to consider continuity at zero. Lets analysis limits approaching this point on the real and imaginary axis. $$\lim_{x\to 0} \frac{\overline{x+0i}}{x+0i} = 1\ \ \ \ \ \ \ \lim_{y\to 0} \frac{\overline{0+yi}}{0+yi} = -1$$ So, the function cannot be continuous at zero, but is continuous everywhere else. 
A: You're right; we need to define $f(0)$ for this to make sense.  But if the intended domain for $f$ is $\mathbb{C} \setminus \{0\}$, then it's continuous.
Writing $z = a + bi$, we have $$f(z) = \frac{a - bi}{a + bi} = \frac{a^2 - b^2 - 2abi}{a^2 + b^2} = \frac{a^2 - b^2}{a^2 + b^2} + \frac{2ab}{a^2 + b^2} i$$
To show this is continuous (still assuming $z \ne 0$), look at the real and imaginary parts.
A: Just to firm up what Dan Fischer said, you can look at it in polar coordinates.  Then z*/z = Reit/Re-it = e2it.  This is clearly continuous in t.  The R doesn't matter since it divides out.  So z*/z maps every z into the unit circle.  
Nonetheless, it will not map into the unit circle or anything else at z = 0 since it isn't defined there, and cannot be defined in any continuous way. 
