Skip to main content
added 9 characters in body
Source Link
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = \pm 1$$\sqrt{u(z)} \sqrt{u(a-z)} = u(a/2) = \pm 1$: if it is $1$, then $f(z) = \sqrt{u(z)}$ is a solution. If it is $-1$, then $f(z) = (z - a/2) \sqrt{u(z)}$ is a solution.

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = \pm 1$: if it is $1$, then $f(z) = \sqrt{u(z)}$ is a solution. If it is $-1$, then $f(z) = (z - a/2) \sqrt{u(z)}$ is a solution.

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = u(a/2) = \pm 1$: if it is $1$, then $f(z) = \sqrt{u(z)}$ is a solution. If it is $-1$, then $f(z) = (z - a/2) \sqrt{u(z)}$ is a solution.

added 75 characters in body; deleted 2 characters in body
Source Link
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = \pm 1$: if it is $1$, then $f = \sqrt{u}$$f(z) = \sqrt{u(z)}$ is a solution. If it is $-1$, then $f(z) = (z - a/2) \sqrt{u(z)}$ is a solution.

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = \pm 1$: if it is $1$, then $f = \sqrt{u}$ is a solution.

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = \pm 1$: if it is $1$, then $f(z) = \sqrt{u(z)}$ is a solution. If it is $-1$, then $f(z) = (z - a/2) \sqrt{u(z)}$ is a solution.

Source Link
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

A necessary condition for a nonzero solution is $u(z) u(a-z) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(a-z)} = \pm 1$: if it is $1$, then $f = \sqrt{u}$ is a solution.