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**thespian** a) Find the mobius transformation which takes the points -1,1 and infinity to

0,2 and 2i.

I have done this bit, which gives me w=(2i+2ix)/(z-1+2i)

b)desribe and find an equation for the real axis under the transformation found in part a.(i'm stuck on this and part d)

Some ideas: we assume $\displaystyle x \in \mathbb{R}$ and we try to find the image of w(x). Now, we assume $\displaystyle a + bi \in Image(w(x))$, and we find the restrictions for x:

$\displaystyle a+bi=\frac{2ix+2i}{x-1+2i}=\frac{4(x+1)+2(x^2-1)i}{(x-1)^2+4} \Longrightarrow$$\displaystyle ax^2-2ax+5a+(bx^2-2bx+5b)i=4(x+1)+2(x^2-1)i$

Now compare real and imaginary parts in both sides and, surprisingly enough for me at least, we get that in order to get x's to solve the equations, it must be that $\displaystyle \,1- \sqrt{2} \leq a\,,\,b \leq 1 + \sqrt{2}$, so if I didn't make some mistake (and I wouldn't assume I didn't), the image of the real line under w is the above strip.

c) Find the length of the path y(t) =i+3e^2it [0,pi/4]

this gives me L=(pi/2)*root(2)

d) now find an upper bound for the function g(z) =(z^2)e^z where y(t) is the path in part c.

I don't get it: where in the equation $\displaystyle g(z)=z^2e^z$ does y(t) enter? I can't see it.

Tonio

Any ideas or help would be appreciated. This is a past exam question and we have not been given solutions, which sucks.