0

I'm new on stack Exchange

I have a problem I think I can solve using Rouché's theorem But I have no idea how to start.

How should i show that $$f(z) = 5\sin(z) - e^z$$ has exactly 1 zero in the square with the origin as center and sides having length $\pi$?

Using Rouché's theorem, I should probably say that $$|5\sin(z)|<|e^z|$$ But I cant understand why.

Im so sorry for my bad english and probably stupid question But i cant figure it out myself

Thank you in advance :)

Jean Marie
  • 81,803
Math lover
  • 1
  • Please use MathJax to write mathematical equations: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – jvdhooft Jun 04 '17 at 20:47
  • Condition $|5\sin(z)|<|e^z|$ must be fulfilled on the sides of the square. Take the different sides one by one: if it is the east side with $z=\pi/2+it$ (with $-\pi/2 \leq t \leq +\pi/2$), you have to show that $|5\sin(\pi/2+it)|<|e^{\pi/2+it}|$, etc... – Jean Marie Jun 04 '17 at 21:05
  • Thanks! I've now found for $$z = \pi/2 +iy$$ that $$|5sin(z)| > |e^z|$$ by using $$sin(z) = sin(x)cosh(y)+i cos(x)sinh(y)$$ now for the case $$z = x + i \pi/2$$ I'm stuck with $$sin(x)cosh(\pi/2)+i cos(x)sinh(\pi/2)$$ Any tips on how to find that this is > 1? – Math lover Jun 05 '17 at 14:38
  • Using Rouché might prove difficult as there is a root at $z=-3.15016179728$ only slightly outside the boundary. The two roots inside are at $z=0.263264797883$ and $z=1.60871868731$. The next pair of roots is complex at $z=4.38522616338\pm i·3.46816444659$. – Lutz Lehmann Jun 09 '17 at 08:48

0 Answers0