Locus of complex number in complex plane

Question

I have the following complex number:

$G = \xi + i\eta$

$\xi = 1-\sigma(1-\cos\phi_m)$

$\eta = -\sigma\sin\phi_m$

how can I find the locus of this complex number? I am told without proof that it is a circle of radius $\sigma$ and centre $1-\sigma$ but cannot even begin to think where this came from.

Any help would be appreciated.

Answer 1

$$\begin{align} G & = 1-\sigma + \sigma \cos\phi_m -i\sigma \sin \phi_m\\ & = 1-\sigma + \sigma(\cos\phi_m - i \sin\phi_m)\\ & = 1-\sigma + \sigma e^{-i\phi_m} \end{align}$$

And this is the equation of a circle wih radius $\sigma$ and center $1-\sigma$.

Answer 2

Think of $<\xi,\eta>$ as a vector in $\mathcal{R}^2$. We can write it as:

$ \begin{align} \left<\xi,\eta\right> &= \left<1-\sigma+\sigma\cos\phi_m,-\sigma\sin\phi_m\right> \\ &= \left<1-\sigma,0\right> + \sigma\left<\cos\phi_m,-\sin\phi_m\right> \end{align} $

The part that doesn't depend on $\phi_m$ gives us a center of $\left<1-\sigma,0\right>$, which corresponds to the real number $1-\sigma$. The part with the sine and cosine of $\phi_m$, gives us a circle traced out in a clockwise direction, with a radius of $\sigma$.