Continue solutions of ODEs around the origin

[psie]

Homework Statement

Solve the following equations when $z$ is real and positive. How will the solutions change if continued one revolution around the origin in the positive direction? a) $2zx'-x=0$ and b) $z^2x'+x=0$.

Relevant Equations

By "continued one revolution around the origin", they basically mean evaluate $x(ze^{2\pi i})$.

What confuses me is that my solution differs from that given in the answers at the back of the book.

Solving the ODEs is fairly simple. They are both separable. After rearrangement and simplification, you arrive at $x(z)=Cz^{1/2}$ for a) and $x(z)=De^{1/z}$ for b). In both solutions, $C$ and $D$ are positive constants.

Then I'm asked to evaluate $x(ze^{2\pi i})$ and a TA claims that $x(ze^{2\pi i})=C(ze^{2\pi i})^{1/2}=Cz^{1/2}e^{\pi i}=-Cz^{1/2}$ for a) and that $x(ze^{2\pi i})=De^{1/(ze^{2\pi i})}=De^{1/z}$ for b). This is also the answer given in the back of the book. I agree for b), but for a), why isn't it true that $x(ze^{2\pi i})=C(ze^{2\pi i})^{1/2}=C(z\cdot1)^{1/2}=Cz^{1/2}$?

Edit: This problem appears in a section on differential equations with singular points, i.e. where the matrix $A(z)$ in the corresponding system $x'(z)=A(z)x(z)$ has a singularity.

 

[pasmith]

Which branch of the square root are you using? In the principal branch, \arg z \in (-\pi, \pi] and \arg z^{1/2} = \frac12 \arg z \in (-\pi/2, \pi/2] so that \operatorname{Re}(z^{1/2}) \geq 0. However, in the branch in which \arg z \in (\pi, 3\pi] then \arg z^{1/2} = \frac12 \arg z \in (\pi/2, 3\pi/2] so that \operatorname{Re}(z^{1/2}) \leq 0. e^{0i} is not in this branch, but e^{2\pi i} is.

In both cases, \sqrt{re^{i\theta}} = r^{1/2}e^{i\theta/2} with r^{1/2} \geq 0.

 

[psie]

Right, a branch...I forgot! I guess the book uses the latter branch, i.e. for which $\arg z \in (\pi, 3\pi]$, since they input a real positive number and get a negative positive number. It makes more sense now. Thank you.