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I tried solving the pair of equations \begin{align} e^x\cos y&=1+2x\\ e^x\sin y&= 2y \end{align} but I got stuck.

dancingmusic
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2 Answers2

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Hint: one solution is quite easy. It also happens to be the only one.

Robert Israel
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let $f(z) = e^{z}$ and $g(z) = -(1+2z)$ then on the boundary $e^{i\theta}$,

$$ |f(z)| = |e^{\cos \theta}| < \sqrt{5 + 4 \cos(\theta)} \tag{1}$$ Since $g(z)$ has one root on $|z|<1$, by Rouché's Theorem $f(z)+g(z)$ has one root on $|z|< 1 $. In particular, $z=0$ satisfies the equation and it is only the root of it as well.

To show that $(1)$ holds, one can simply show that $h(x) = e^x - \sqrt{5 + 4x}$ is smooth function on $[1,1]$ has $h(x)$ has minimum value of $3-e$ on $x=1$.

S L
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  • How do you get $g(z)$? – seldon May 11 '14 at 09:23
  • @mattecapu i just constructed :) – S L May 11 '14 at 15:17
  • $\lvert f(z)\rvert < \lvert g(z)\rvert$ on the boundary is not the condition you need. You could choose $g(z) \equiv 10$ and have that inequality. You need to bound $\lvert f(z)-g(z)\rvert$. – Daniel Fischer May 11 '14 at 15:37
  • @DanielFischer hmm ... here in my book it is stated as "if $f(z)$ and $g(z)$ are analytic function on some region and $|f(z)|<|g(z)|$ on the boundary curve, then $f(z)$ and $f(z)+g(z)$ has same number of zeroes on the region R". Did I make any mistake somewhere? – S L May 11 '14 at 15:49
  • Perhaps I could modify the $f(z)$ a bit ... – S L May 11 '14 at 15:50
  • It says "$f(z)$ and $f(z)+g(z)$" have the same number of zeros, not "$f(z)$ and $g(z)$". But the condition should be $\lvert g(z)\rvert < \lvert f(z)\rvert$ on the boundary then, since $g = (f+g)-f$. – Daniel Fischer May 11 '14 at 15:53
  • @DanielFischer does my new modification work?? are there still flaws? – S L May 11 '14 at 16:13
  • You need a strict inequality in $(1)$. (You have the strict inequality in fact, but you wrote a weak inequality.) – Daniel Fischer May 11 '14 at 16:20
  • @DanielFischer I hope it works now :S – S L May 11 '14 at 16:32