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QUESTION
 (More difficult)
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\item[(a)]
Let $w=\tanh z$. Show that the points of the $z$-plane on the line
$x=a(\neq 0)$ correspond to the points of the circle with centre
$(\coth 2a,0)$ and the radius cosech $2a$ of the $w$-plane,
Indicate in a diagram of the complex plane the way these circles
vary with $a$.

\item[(b)]
Find all solutions of $\tanh z=2$.

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ANSWER
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\item[(a)]
$w=$tanh$z=\frac{e^z-e^{-z}}{e^z+e^{-z}}$. Let $u=e^z$. Then the
line $x=a$ in the $u$-plane is mapped to points of the form
$e^a(\cos y+i\sin y)$, that is to a circle center 0 radius $e^a$,
and $u^2$ lies on a circle center 0 radius $e^2a$. Now
$w=\frac{u-\frac{1}{u}}{u+\frac{1}{u}}=\frac{u^2-1}{u^2+1}$. This
gives $u^2=\frac{1+w}{1-w}$. Thus $|\frac{1+w}{1-w}|=e^{2a}$,
Hence $|1+w|^2=e^{4a}|1-w|^2$ and so
$(1+w)(1+\overline{w})=e^{4a}(1-w)(1-\overline{w})$. Put $w=X+iY$.
Then we find $Y^2+Y^2-2X(\frac{e^{4a}-1}{e^{4a}+1})+1=0$ which
after a bit of work we see is a circle centre (coth $2a$,0),
radius cosech $2a$.

\item[(b)]
Go back to the equation $u^2=\frac{1+w}{1-w}$. Here $w=2$, so
$u^2=-3$ and $u=\sqrt{3}i$. Thus
$z=\log(\sqrt{3}i)=\log\sqrt{3}+i\frac{\pi}{2}+2\pi n i=\frac{\log
3}{2}+i(\frac{\pi}{2}+2\pi n)$

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