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Let $z$ be a complex number with positive real part. Is it possible that $e^{2z}+\frac{z+1}{z-1} = 0$?

xen
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2 Answers2

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Define: $$f:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto(z-1)e^{2z}+(z+1),$$ $$g_1:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto|f(x+iy)|^2,$$ $$g_2:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto g(x,y)e^{-2x},$$ $$h:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto(x-1)e^{2x}+1.$$ Now prove this: $$(\forall x>0)\frac{\partial g_1}{\partial x}>0\Rightarrow(\forall x>0)g_1(x,y)>0,$$ $$(\forall x>0)\frac{\partial g_2}{\partial x}>0\Rightarrow(\forall x>0)\frac{\partial g_1}{\partial x}>0,$$ $$(\forall x>0)\frac{\partial g_2}{\partial x}\geqslant 2x(h(x)+h(-x)),$$ $$(\forall x\neq 0)h(x)+h(-x)>0.$$

donaastor
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  • How did you come up with this wonderful, but imho unintuitive proof? E. g., what did you make to choose $g_1, g_2$ and $h$ as you have done? – user7427029 Feb 25 '21 at 12:28
  • @user7427029 Needless to say, your question is very in place! I do agree that's it's unintuitive, but I can't say that it's wonderful. Just look at it :D I never decline to use other tools to aid my imagination, such as computer graphics, so I obviously used computer to plot $f$. Then I made guesses about what holds and what doesn't and the rest of the work is turning all intuition and ideas into rigorous proof. That proof is sometimes wonderful, but sometimes its terrible (like now), so I pointed out only major steps in the proof and left for the reader to make its own intuitive path to them. – donaastor Feb 25 '21 at 17:15
  • @user7427029 More concretely, I was just making a ton of guesses and these particular functions seemed the most satisfying. Just plot them using this tool for example: https://samuelj.li/complex-function-plotter/. Just after I realized their usefulness, I started to think about how could I have come up with that intuitively myself. So, this method is new to me too, but I don't consider it wonderful, whatsoever, I feel that there is some much more elegant way to pack all those characteristic properties of the function $f$ that my method barely exploits. – donaastor Feb 25 '21 at 17:18
  • @user7427029 That's the story for $g_1$ and $g_2$, but for $h$, I was simply trying to reduce the "dimension", namely the number of variables. I wanted to lose $y$ and to keep working with $x$ only. So, for the third line in proof, you'd fix $x$ as constant to prove the inequality as for $y$ only, and then solve the last line as an inequality as for $x$ only. – donaastor Feb 25 '21 at 17:22
  • When you plot $g_2$ using the tool I suggested, you will see that the lines of absolute value (click HELP to find "abs" function under "MISC") are going "always up" in the sense that they aren't going up then down then up and so on. That means that the absolute value will only increase as you go further from the imaginary axis. – donaastor Feb 25 '21 at 17:24
  • Thanks a lot for your detailed explanations! I did not know of a useful tool like this one before. Despite the process of reasoning was merely "only" trial-and-error, it surely needed much time and good intuition to get your result. – user7427029 Feb 25 '21 at 18:04
  • This "intuition" that you call is mostly made of knowledge of generally interesting properties of the subfunctions of $f$ that you can learn in standard courses like complex analysis. The idea to prove that the absolute value is strictly greater than zero is, I think, very standard. I was just trying to fit those generally interestion properties in, while playing with absolute value and real analysis. It is always a good strategy to spend time tidying up your head after learning a new course with a whole lot of new definitons, theorems and, most importantly, ideas. I hope it can help you :) – donaastor Feb 25 '21 at 18:19
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Near $z=1$, the second term dominates the first; far from $z=1$, the first dominates the second. By numerical methods, the curve in the right half plane where $\left|e^{2z}\right| = \left|(z+1)/(z-1)\right|$ looks like this:

enter image description here

and nowhere on that curve (except the origin) is there a zero of your function.

Robert Israel
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