The transformation $T$ from the $z$-plane where $z=x+iy$, to the $w$-plane where $w=u+iv$ is given by
$$w=\frac{-iz+i}{z+1}$$
- The transformation $T$ maps the points on a circle $\lvert\,z\,\rvert=1$ , to points on a line l in the $w$-plane. Find the equation of l
I managed to do this part, the equation of the line is $v=0$, i.e. the $u$ axis
- Shade on an Argand diagram the region $R$ of the $w$-plane which is the image of $\lvert\,z\,\rvert\,\leq1$
I thought the shaded region should be everything above the $u$ axis, but the answer given in the book is everything below the $u$ axis.
What I tried to arrive at my answer is just subbing a point inside the unit circle, say $z=0.75$ which gives $w = 0+0.01428i$ which is above the $u$ axis, so my reasoning was that a point inside the unit circle ($z$-plane) corresponds to a point above the $u$-axis ($w$-plane). Any thoughts?
Thanks in advance.
While the question has been essentially answered in a comment, here is a longer answer.
If one knows that Möbius transformations map circles to generalized circles, then it's enough to determine the transforms of $\,3\,$ points on the unit circle. Choosing those to be $1 \mapsto 0\,$, $\,i \mapsto 1\,$ and $\,-i \mapsto -1\,$, it follows that the image of the unit circle is the real axis.
Otherwise, the same result can be derived by rewriting $\displaystyle\,w=\frac{-iz+i}{z+1} \iff z = \frac{1+iw}{-1+iw}\,$, then $|z|=1 \iff |1+iw|=|1-iw|\,$. Squaring the latter equality and using that $\,|a|^2=a \bar a\,$:
$$\require{cancel} \begin{align} (1+iw)(1-i \bar w) = (1-iw)(1+i \bar w) \quad &\iff \quad \bcancel{1} - i \bar w+iw+ \cancel{w \bar w} = \bcancel{1} + i \bar w -i w + \cancel{w \bar w} \\[5px] &\iff \quad w - \bar w= 0 \end{align} $$
Therefore the equation of l is $\,w-\bar w = 0\,$, which represents the real axis.
By continuity, the interior of the unit circle will map to one of the half-planes having the real axis as boundary. To determine which one, it is enough (as noted in a comment already) to choose an arbitrary point strictly inside the unit circle and determine its image. Choosing the origin for example $\,0 \mapsto i\,$, so the interior of the unit disc maps to the upper half-plane above the real axis.