A mapping of the interior of unit circle to the exterior of an ellipse Doing the usual $W=u+iv$ and $z=x+iy$, I can see that $w=\dfrac{1}{2}\left[z e^{-a}+\dfrac{e^a}{z}\right]$ yields the equation $\dfrac{u^2}{\cosh^2 a}  + \dfrac{v^2}{\sinh^2 a}$  which is the equation of an ellipse .... what I do not understand is that why is the interior of the unit circle mapped to the exterior of an ellipse with major and minor axis indicated by the above $u$ ,$v$ equation , and more importantly what I must do to show this..........
 A: An information which is missing in the announcement is that $a>0$ in order for the described scenario to prevail. Note first that if $\alpha>0$ and $0<|\beta|<\alpha$ then 
 $$ \gamma_{\alpha,\beta}(t) = \alpha \cos t + i \beta \sin t , \; \; t\in [0,2\pi]$$
parametrizes an ellipse with major and minor axes $\alpha$ and $|\beta|$, respectively. For positive $\beta$ the ellipse is traversed counter-clockwise, and clockwise when $\beta$ is negative. $\gamma_{\alpha,\beta}$ will traverse the boundary of the filled ellipse given by:
$$ E_{\alpha,\beta} = \left\{ (u,v) : \frac{u^2}{\alpha^2}+\frac{v^2}{\beta^2}\leq 1\right\}$$
The exterior of the ellipse $E^c_{\alpha,\beta}$ is given by replacing $\leq$ by $>$.
When $0<\alpha<\alpha'$ and $0<|\beta| < |\beta'|$ then $E_{\alpha,\beta}$ is disjoint from $E^c_{\alpha',\beta'}$. In particular, $\partial E_{\alpha',\beta'}$ is lying in the exterior of the filled ellipse $E_{\alpha,\beta'}$. 
Now, in the present case setting $z=re^{it}$, $0<r<1$, $t\in [0,2\pi]$ we have:
  $$ f(z) = \frac{1}{2} \left( \frac{z}{e^a} + \frac{e^a}{z} \right) =
\frac12\left(\frac{r}{e^a} + \frac{e^a}{r}\right) \cos t  + \frac12\left(\frac{r}{e^a}-\frac{e^a}{r} \right) \sin t = \alpha \cos t + i \beta \sin t 
  $$
with $$ \alpha= \alpha_{r} =\frac12\left(\frac{r}{e^a} + \frac{e^a}{r}\right) , \; \; \beta=\beta_{r} = \frac12\left(\frac{r}{e^a}-\frac{e^a}{r} \right)$$
Note here, that since $a>0$  and $0<r<1$ we have $\alpha_r>0$ and $-\alpha_r<\beta_r <0$. Both $\alpha_r$ and $|\beta_r|=-\beta_r$ are strictly decreasing functions of $0<r\leq 1$. So turning this around, when $r$ decreases, $t\in [0,2\pi] \rightarrow f(re^{it})$ describes an increasing family of ellipses, i.e. these ellipses are all in the exterior of the ellipse
$E_1 = E_{\alpha_1,\beta_1}$ given by $r=1$. Btw, the map $\{0<|z|<1\} \mapsto f(z)\in E_1^c$ is a bijection, injective since the ellipses are disjoint and surjective since both $\alpha_r$ and $-\beta_r$ goes to infinity as $r\rightarrow 0$ (you may also solve an equation to see this).
A: Consider the transform
$$w=\dfrac{1}{2}\left[z e^{-a}+\dfrac{e^a}{z}\right]$$
Indeed, it's true that the interior of the unit circle is mapped exterior to the ellipse. This is a consequence of the fact that the terms in the square bracket is are a form of inversion, that is, for any value of $|z|\lt 1$ the bracketed term is greater than $|w|$.
For the unit circle $z=e^{i\theta}$ and thus
$$w=\cosh(i\theta-a)$$
is the ellipse. Similarly, for any point within the unit circle, say $Z=re^{i\theta},~r\lt1$, we can say that
$$
\begin{align}
W=
&=\cosh\big(\ln(Ze^{-a})\big)\\
&=\cosh\big(i\theta-(a+|\ln(r)|)\\
&=\cos\theta\cosh(a+|\ln(r)|)-\sin\theta\sinh(a+|\ln(r)|)
\end{align}
$$
I'll leave it as exercise for the reader to show that $W\gt w$.
A: A nice way to look at this is to define $Q(z):=(z+1/z)/2$. Your mapping is then $Q(ze^{-a})$. The mapping $Q$ is a two-to-one mapping since $Q(1/z)=Q(z)$ with fixed points $1$ and $-1$ where its derivative is zero. The mapping takes the orthogonal family of circles centered at the origin and straight lines passing through the origin onto the orthogonal family of confocal ellipses and hyperbolas where the foci are $1$ and $-1$. A circle with radius $r$ is mapped to the same ellipse as the circle with radius $1/r$. Points $z$ such that $|z|>r$ or $|z|<1/r$ get mapped outside the ellipse. Points $z$ such that $1/r<|z|<r$ get mapped inside the ellipse. A borderline case is the unit circle where $|z|=1$ and these get mapped two-two-one onto the the line segment $[-1,1]$. The mapping is conformal everywhere except at $1$ and $-1$ where it doubles angles.
