Simple equation involving complex norm So I've just started complex analysis and was given the following problem:
$$|e^{i\theta}-1|=2$$
One only need recall Euler's identity to know that the solution is $\theta=\pi$. However, I was supposed to verify my solution geometrically and had a question.
The LHS of the norm is simply the unit circle in the complex plane. Since the angle $\pi$ provides a unit vector pointing toward $-x$, I assumed that the unit circle had a phase shift of $1$ unit to the left, meaning the sum of two unit vectors pointing left from the origin would have distance $2$ from the origin, thus satisfying the equation.
However, I believe my professor said that the circle was centred at $(1,0)$--not $(-1,0)$--and I just don't see how the problem works if this is true.
Could someone clear things up a bit?
 A: Geometrically, $\,|z-1|=2\,$ is the locus of points at distance $\,2\,$ from fixed point $\,+1\,$ i.e. the circle of radius $\,2\,$ centered at $\,1\,$, call it $\,\Gamma\,$.
On the other hand, $\,z = e^{i \theta} \mid \theta \in \mathbb{R}\,$ is the unit circle, which is internally tangent to $\,\Gamma\,$ at point $\,z=-1\,$, which is the only common point between the two circles.
Therefore, the unique solution is $\,z=-1\,$, which corresponds to $\,\theta = \pi\,$.
A: $\mid e^{i\theta}-1\mid=2\Rightarrow (\cos\theta-1)^2+\sin^2\theta=4\Rightarrow -2\cos\theta=2\Rightarrow \cos\theta=-1\Rightarrow \theta\in\{(2n+1)\pi:n\in\mathbb{Z}\}$
Hence $\mid e^{i\theta}-1\mid=2$ do ot represent a circle at all. But $\mid re^{i\theta}-1\mid=2$ does, where $r>0$.
After your comment:
In general $\mid re^{i\theta}-c\mid=r$ represents a circle centered at $c\in\mathbb{C}$. Here you have $\mid re^{i\theta}-c\mid=s$, where $r\neq s$. So this will not represent a circle. In particular in your case $r=1, s=2$.
Also $\mid re^{i\theta}-c\mid=\mid se^{i\theta}-d\mid$. For fixed $r,s,c,d$ you will get for which value of $\theta$ for which the those two circles intersect or touch. In your case $r=1,c=0,s=2,d=1$. So got $(-1,0)$, where those two circle touch.
