Here's the question you clicked on:
.Sam.
Prove that where w = f (z) is analytic and one-to-one.
\[\frac{\partial^2 \Phi}{\partial x^{2}}+\frac{\partial^2 \Phi}{\partial y^2}=|f'(z)|^2(\frac{\partial^2 \Phi}{\partial u^2}+\frac{\partial^2 \Phi}{\partial v^2})\]
wahh... I'm ready to faint !! LOL , I don't know :P
unfortunately--google books is missing a rather needed page, but yah. http://books.google.com/books?id=1DItN16-t-MC&pg=PA366&lpg=PA366&dq=%22%E2%88%82%5E2%CE%A6/%E2%88%82x%5E2%2B%E2%88%82%5E2%CE%A6/%E2%88%82y%5E2%3D%7Cf%E2%80%B2(z)%7C%5E2(%E2%88%82%5E2%CE%A6/%E2%88%82u%5E2%2B%E2%88%82%5E2%CE%A6/%E2%88%82v%5E2)%22&source=bl&ots=bLDvOGPmRR&sig=5rgnEEIWehIwfj-0OaDZE1Kp4VE&hl=en&sa=X&ei=Glx6T5KCJsfi2AXa3vycAw&ved=0CCEQ6AEwAA#v=onepage&q&f=false
WTH! I'm just a 15 year old self studying calc two. I DON GET THOSE FUNNY PHI AND PARTIAL DERIVATIVE SIGNS! :(
That is, if that is even phi :S
I thought you know because you'd say ERF(), sorry
I only know what the erf is because I wanted to figure out the empirical rule by indefinite integration, and later I realized that to be impossible.
Cause wolfram alpha said so :)
Good luck though, try reddit's r/cheatatmathhomework
They can do group theory there, i'm not sure why.