Recent content by mxmt

A guess is that the line segment of z=(i,-i) is mapped onto the normal branch cut of the logarithm (-inf,0). Therefore, f(1)=exp(iPi/2) because this is where i is located in the complex plane.

There was a typo in my first post: Of course this should be w=log(\frac{z+i}{z-i}) Anybody who understands it now?

Homework Statement Define an analytic branch f(z) of w, such that f(z)=0 for the limit of z->\infty Now what is f(1)? Homework Equations w=\frac{z+i}{z-i} The Attempt at a Solution The branch cut of the logarithm is: (-\infty,0) All branches of the logarithm are...

If you get infinity in the numerator as well as in the denumerator you should check which one goes to infinity fastest. In this case, p^2 approaches infinity much slower then e^(2p), so your fraction will become zero for p=infinity.

Never mind, I figured it out.

The problem Find Res(f,z1) With: f(z)=\frac{z}{(z^2+2aiz-1)^2} The attempt at a solution The singularities are at A=i(-a+\sqrt{a^2-1}) and at B=i(-a-\sqrt{a^2-1}) With the normal equation (take limit z->A of \frac{d}{dz}((z-A)^2 f(z)) for finding the residue of a pole of order 2, my attempt...