Question for Ahlfors experts (re: conformal mappings) This comes from the part in Ahlfors where he introduces conformal mappings (page 73-4). I'm having trouble following one of the steps he takes.
We have an arc $\gamma$ in a region $\Omega $ defined by the equation z=z(t), a continuous function f, and an image arc $\gamma'$ defined by w=w(t)=f(z(t)). Ahlfors is showing that if the partial derivatives of f are continuous, then the fact that $f'$ preserves angles between $\gamma$ and $\gamma'$ implies that f is analytic. 
The argument is: by assuming that the partial derivatives $\partial{f}/\partial{x}$ and  $\partial{f}/\partial{y}$ are continuous, we can write
$$w'(t_0)=\frac{\partial{f}}{\partial{x}}x'(t_0)+\frac{\partial{f}}{\partial{y}}y'(t_0)$$
He then rewrites this in terms of $z'$:
$$w'(t_0)=\frac{1}{2}(\frac{\partial{f}}{\partial{x}} - \imath\frac{\partial{f}}{\partial{y}})z'(t_0) + \frac{1}{2}(\frac{\partial{f}}{\partial{x}} + \imath\frac{\partial{f}}{\partial{y}})\overline{z'(t_0)}$$
and then 
$$\frac{w'(t_0)}{z'(t_0)}=\frac{1}{2}(\frac{\partial{f}}{\partial{x}} - \imath\frac{\partial{f}}{\partial{y}}) + \frac{1}{2}(\frac{\partial{f}}{\partial{x}} + \imath\frac{\partial{f}}{\partial{y}})\frac{\overline{z'(t_0)}}{z'(t_0)}$$
If we suppose that angles are preserved, then $arg[w'(t_0)/z'(t_0)]$ is independent of $arg[z'(t_0)]$. Therefore the RHS of the last equation has to have a constant argument. 
He writes, "as arg $z'(t_0)$ is allowed to vary, the point represented by (the RHS) describes a circle having the radius $\frac{1}{2}|(\frac{\partial{f}}{\partial{x}} + \imath\frac{\partial{f}}{\partial{y}})|$." He concludes that the argument cannot be constant on the circle unless the radius=0, implying that
$$(*)        \frac{\partial{f}}{\partial{x}}=-\imath\frac{\partial{f}}{\partial{y}},$$
which is the complex form of the Cauchy-Riemann equations.
MY QUESTION: How do we see that the RHS of the last equation is a circle? I am assuming that he is using the form, $z = a+re^{it}$, with $a$ being the center and $r$ the radius. I see that $\frac{\overline{z'(t_0)}}{z'(t_0)}$ can be written in the form $e^{it}$. The conclusion in (*) suggests that the part corresponding to the radius is $\frac{1}{2}(\frac{\partial{f}}{\partial{x}} + \imath\frac{\partial{f}}{\partial{y}})$. But of course the radius needs to be a real number, and this expression is complex. Plus, Ahlfors wrote that the radius is the modulus, $\frac{1}{2}|(\frac{\partial{f}}{\partial{x}} + \imath\frac{\partial{f}}{\partial{y}})|$. Is it simply a question of rearranging the equation? I may be being dense here... Any ideas/hints what I'm missing? 
 A: As far as I can tell, you have a set of the form
$$ \eta + \overline{\eta} e^{it} $$
where $\eta$ is a fixed complex number and $t$ varies in $\mathbb{R}$. Did I understand correctly?
This is always a circle. Indeed, we can rewrite it as
$$ \eta \left(1 + \frac{\overline{\eta}}{\eta} e^{it}\right) $$
which is just
$$ \eta \left(1 + e^{i(t+\alpha)}\right) $$
where $\alpha$ is just the argument of the modulus-$1$ complex number $\frac{\overline{\eta}}{\eta}$.
As $t$ varies, the stuff in the parentheses describes a circle. Multiplication by the fixed complex number $\eta$ sends circles to circles.
Of course, you could always brutally compute with real and imaginary parts to show this as well.
==== Edit
Stupid of me to skim and not notice that the two factors are not really conjugate in your question, but it really doesn't make a difference. You can easily show that $\alpha + \beta e^{it}$ is a circle for any complex numbers $\alpha$ and $\beta$. Indeed, the first term $\alpha$ is irrelevant as it's just a translation, and the $\beta e^{it}$ term traces out the circle
$$ |\beta|e^{i(t+\arg(\beta))} $$
which is just a circle of radius $\beta$. (This is what I should have written originally, but coffee is useful.)
