Find a Mobius transformation that maps the right half-plane to the upper half-plane carrying the point 7+5i to 3i.
Can I get some help please?
This method is messy, but at least it ought to work. The idea is that you should have a library of known Möbius transformations that you combine in order to get the desired result.
The transformation $\displaystyle f(z) = \frac{z-1}{z+1}$ maps the right half-plane to the unit circle, and it takes 7+5i to $\displaystyle \frac{73+10i}{89}$ (that's where the numbers start to get messy).
The transformation $\displaystyle g(z) = \frac{z-i}{z+i}$ maps the upper half-plane to the unit circle, and it takes 3i to 1/2. Its inverse transformation, from the unit disk to the upper half-plane, is given by $\displaystyle g^{-1}(z) = \frac{i(1+z)}{1-z}$.
For any point a in the unit disk, the transformation $\displaystyle h_a(z) = \frac{z-a}{1-\overline{a}z}$ maps the unit disk to itself, and takes a to the origin. Its inverse transformation is given by $\displaystyle h_a^{-1}(z) = \frac{z+a}{1+\overline{a}z}$, and it obviously takes the origin to $\displaystyle a$.
Thus the composition of mappings $\displaystyle g^{-1}\circ h_b^{-1}\circ h_a\circ f$, where $\displaystyle a = \tfrac{73+10i}{89}$ and $\displaystyle b=1/2$, will be a Möbius transformation that maps the right half-plane to the upper half-plane and carries the point 7+5i to 3i.
Having written all that, I have a better idea. You can think of the right half-plane as consisting of all those points that are closer to 7+5i than they are to –7+5i. In other words, they are the points for which $\displaystyle |z-7-5i| < |z+7-5i|$. It follows that the Möbius transformation $\displaystyle k(z) = \frac{z-7-5i}{z+7-5i}$ maps the right half-plane to the unit disk, and it takes 7+5i to the origin. So you can use that map to replace the composite map $\displaystyle h_a\circ f$ in the above construction by the single map k, and that will avoid all the messy arithmetic.