1. The problem statement, all variables and given/known data Let f : I → C be a smooth complex valued function and t0 ∈ I fixed. (i) Show that the initial value problem z'(t) = f(t)z(t) z(t0) = z0 ∈ C has the unique solution z(t) = z0exp(∫f(s)ds) (where the integral runs from t0 to t. Hint : for uniqueness let w(t) be another solution of the same initial value problem and contemplate the expression w(t)/z(t). (ii) Show that if f : I → iR is imaginary valued, the length of the solution z(t) is constant, i.e. |z(t)| = |z0|. Does the converse also hold, i.e. if you know that the length of the solution to the ODE z' = fz is preserved, then f necessarily has to be imaginary valued. (iii) Show that if z1(t), z2(t) solve the ODE z'(t) = f(t)z(t), then z1(t) = cz2(t) for some c ∈ C (this is another way to state the uniqueness property). 2. Relevant equations Not entirely sure what equations I should put in here but I suppose I could put z = x + iy 3. The attempt at a solution I'm currently working on part i) but will continue to post here as I work further. So I almost did separation of variables here but then realized I could just directly integrate (rather than get dz/z on the LHS and f(t)dt on the RHS) but doing the direct integration I see that if I could just integrate out Z(s) to get Z(t0) in the front but that's obviously not possible as the integral of two functions requires the use of integration by parts. I integrated by parts using u = z(t) and v = f(t) to find ∫f(t)z(t)dt = z(t)f(t) - ∫f(s)z'(s)ds and so I tried continuing on but am not quite sure it's going to be fruitful. I tried a few things but none really seem right. Can anyone tell me where I'm going wrong?