Show that the equation \[ e^z=2z+1 \] has exactly one solution in \(D\).
Hint: Note that \[ e^z-1=\int _0^1 e^{tz}z\,dt. \] Conclude that \(|e^z-1|\le e-1\).
Question
Math 6210 • Real Analysis
Spring 2018 • Question 9
Question
Spring 2018 • Question 9
Show that the equation \[ e^z=2z+1 \] has exactly one solution in \(D\).
Hint: Note that \[ e^z-1=\int _0^1 e^{tz}z\,dt. \] Conclude that \(|e^z-1|\le e-1\).