I would like to understand and answer the following question from Serge Lange's Introduction to Complex Analysis at a graduate level.

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I understand how is one supposed to use the hint to prove the desired result but I do not understand how one proves the hint. (I think once the hint is proven we have basically shown that about every point the function is locally constant, even about the poles, so the poles are removable singularities and hence the function is constant).

I tried setting up equations of the lines $$aw_1+bw_2+cw_3=\epsilon$$ in complex and vector forms but it was a dead end as I wasn't able to prove that such $a,b,c$ exist for all epsilon not equal to zero.

Does anybody have any ideas has what the construction of such proof would look like and moreover, how should one think about this problem intuitively?

  • $\begingroup$ You need to show that $S=\{b_1w_1+b_2w_2+b_3w_3: b_1,b_2,b_3\in\Bbb Z; b_1,b_2,b_3 \text { not all } 0\}$ has members arbitrarily close to $0$. Note $0\not\in S$ because $w_1, w_2,w_3$ are linearly independent over $\Bbb Q$, so, a fortiori, they are lin. indep. over $\Bbb Z.$ Also if one of the ratios $w_i/w_j$ (with $i\ne j$) is real, it is irrational and we can apply an old result to obtain $b_i,b_j\in \Bbb Z$ \ $\{0\}$ with $b_iw_i+b_jw_j$ arbitrarily close to $0$. I have not yet got an A for the case where $w_1/w_2,w_2/w_3,w_3/w_1\not\in \Bbb R$. $\endgroup$ Jul 16 at 15:37
  • $\begingroup$ The "old result" I mentioned is that if $r=w_i/w_j\in\Bbb R$ \ $\Bbb Q$ then there are members of $\{nr: n\in\Bbb Z^+\}$ that are arbitrarily close to integers. $\endgroup$ Jul 16 at 15:43
  • $\begingroup$ We might need to use the following: For $x\in\Bbb R$ let $[x]$ denote the largest integer not exceeding $x.$ If $r\in\Bbb R$ \ $\Bbb Q$ then $T=\{nr-[nr]: n\in\Bbb Z^+\}$ is dense in $[0,1].$ In particular $\inf \,T=0.$ $\endgroup$ Jul 16 at 15:52
  • $\begingroup$ @DanielWainfleet thank you for the comments. I am too still puzzled by this. To be honest this just looks like an awful question given the fact that there has been no discussion for such methods before this exercise. $\endgroup$ Jul 16 at 19:43
  • $\begingroup$ It's more like a Number Theory problem with an immediate corollary in Analysis. $\endgroup$ Jul 17 at 4:16

Case 1: $w_1/w_2 \not\in \Bbb R.$ Consider $\Bbb C$ as a 2-dimensional vector space over the field $\Bbb R.$ So $\{w_1,w_2\}$ is a vector-space basis for $\Bbb C.$ So there exist $r_1,r_2\in\Bbb R$ such that $r_1w_1+r_2w_2=w_3.$

Definition. For $x\in \Bbb R$ let $d(x,\Bbb Z)=\min \{|x-z|:z\in\Bbb Z\}.$

Lemma. For any finite $S\subset \Bbb R$ and any $\epsilon >0$ there exist infinitely many $n\in\Bbb N$ such that $\forall x\in S\,(d(nx,\Bbb Z)<\epsilon).$

We only need this lemma when $S$ has 1 or 2 members, and we do not need infinitely many $n\in \Bbb N$. Just one $n$ will be needed for a given $\epsilon$.

Return the the 1st paragraph. By the Lemma, with $S=\{w_1,w_2\}$, for any $\epsilon >0$ there exists $n_{\epsilon}\in\Bbb N$ such that $n_{\epsilon}r_j=m_j+\delta_j$ for $j\in \{1,2\}$, with $m_j\in\Bbb Z$ and $|\delta_j|<\epsilon.$ Hence $$|m_1w_1+m_2w_2-n_{\epsilon}w_3|=|n_{\epsilon}(r_1w_1+r_2w_2-w_3)-(w_1\delta_1+w_2\delta_2)|=$$ $$=|0-(w_1\delta_1+w_2\delta_2)|\le$$ $$\le (|w_1|+|w_2|)\epsilon$$ which can be arbitrarily small but not $0$, because if $m_1w_1+m_2w_2-n_{\epsilon}w_3=0$ then $w_1,w_2,w_3$ would be linearly dependent over $\Bbb Q.$

Case 2. $w_1/w_2\in \Bbb R.$ By the Lemma with $S=\{w_1/w_2\},$ for any $\epsilon>0$ there exists $n_{\epsilon}\in \Bbb N$ such that $n_{\epsilon}(w_1/w_2)=m+\delta$ with $m\in\Bbb Z$ and $|\delta|<\epsilon.$ Hence $$|n_{\epsilon}w_1-mw_2|=|\delta|\cdot |w_2|\le\epsilon |w_2|$$ which can be arbitrarily small but not $0$, because if $n_{\epsilon}w_1-mw_2=0$ then $w_1,w_2$ would be linearly dependent over $\Bbb Q.$

  • $\begingroup$ I can give a proof of the Lemma if needed. $\endgroup$ Jul 18 at 8:04
  • $\begingroup$ Hi thank your for the answer. A proof of the lemma would be great as I have not seen anything like this. $\endgroup$ Jul 18 at 17:17
  • $\begingroup$ I will add a proof of the lemma soon. $\endgroup$ Jul 30 at 1:53

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