Homework 4
Hint for IV.6.3 --Steve Bell
We assume $ (f)''=f $ on $ \mathbb C $.
Notice that
$ (e^z f)''=e^zf +2e^zf'+e^zf''=2(e^zf + e^zf')=2(e^zf)'. $
Let $ g=(e^zf)'. $ Then $ g'=2g $ and now you can use the theorem from class that concerns solutions of this first order complex ODE. By the way, you will also need to use the fact that if two analytic functions on the complex plane have the same derivative, then they must differ by a constant.
Could you just substitute the f in the form they provided into the ODE to prove that you can assume the solution is of that form? --Adrian Delancy
Adrian, if you plug that form into the ODE, you are only verifying that those ARE solutions to the equation. The harder and more interesting part is to show that ONLY functions of that form are solutions. If you follow my hint above, you'll get an expression for g, and then you'll need to antidifferentiate and then mess around with a little linear algebra to see that f has to be of the form mentioned in the problem. --Steve Bell
Question on IV.5.2 (And subsequently IV.5.3)
If anyone could, I'd like some clarification. I'm confused on what the question is asking. Is it asking for the curve of the values of z that make the conditions true? --Whoskins
