10:47 AM
0

Aim A quick search for duplicates. Question If $|f(z) |\leq 1 + |z|$, show that $f(z) = az + b$ is a recently-posted question. I tried to search for its duplicates on Approach0 but I didn't manage to get something like Showing that $f$ is linear function if $\forall z \in \mathbb{C}$, $|f(z)| \l... 1 Let$f$be an entire function that satisfies$|f(z)| \leq 1 + |z|$for all$z\in \mathbb{C}$. Show that$f(z) = az +b$for fixed complex numbers$a,b$. The hint tells us to try and use Cauchy's Integral Formula on an arbitrary circle. This is my attempt: Consider an arbitrary large circle w... @WeiZhong Is there some simple way to find whether some question is in the index or not? Ok, so it is in the index, when I tried to copy a formula from that question, it is among the search result. So the question remains why it isn't among the results of the search linked in the post on meta. Interestingly, if I take the search you linked and omit all occurrences of \left... and \right.... then it appears near the top. I am not sure whether this should make much difference - I have asked about this in the searching chatroom. — Martin Sleziak 17 secs ago 11:44 AM Now I noticed that we have discussed something similar. 24 Approach0 Searching for a specific formula is often rather difficult. A tool which is very suitable for this is the Approach0 search engine. Some basic information can be found in Guide for New Users (which contains also animations of some examples). If you want to know more about the project, ... > There is also a distinction between |x| and \left|x\right|, Approach0 treats these two differently as a result of some technical considerations. > For example, compare the search for$\frac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$with and ... > ... without including \left and \right. Although it seems that previous discussion was mainly about absolute value. As far as I can tell from the source code, both "(" and "\left(" are _L_BRACKET. 6 hours later… 5:55 PM @MartinSleziak \left \right for "(" or ")" does not change anything, as you said. So if you try this one and this one you get the same result (2nd result is what you are looking for, and they have exact the same score which is 142.042) As you have pointed out, to find the linked post as mentioned on meta, you need to remove \left and \right around vertical bar |, it is exact the case you mentioned here, there is no easy way to know how to pair vertical bars, because left bar and right bar are the same. That is why I treat them differently. I confidently treat "(" and "\left(" as _L_BRACKET because ( should be the left one, while in vertical bar case, you cannot easily figure out. So normally, without specifying \left or \right, I will just treat a bar as a normal leaf token. So maybe I misunderstood something, but shouldn't then searching for "f\left(z\right)" and for "f(z)" return the same results? The post in that meta question is an example where the results are rather different. It should, so you see my two links return the same results. what matters is the \left \right surrounding bars, not those \left \right surrounding parenthesis. So what is actually the difference between the search that the OP posted on meta and the one I posted in chat? The OP posted this link. And in the comments I posted this link. The results are different. And the search queries are:$\left|f\left(z\right)\right|\le1+\left|z\right|$,$f\left(z\right)=az+b|f(z)|\le1+|z|$,$f(z)=az+b\$
Oh, I see.

Again, your query omits left right.

So only \left|...\right| matter.
Whether I omit \left and \right next to (...) does not matter.
Is that correct?

6:09 PM
@MartinSleziak Exactly. So you can only strip the \left and \right for the first keyword, it will still find you that post.

Thanks, I should have noticed this. (I somehow missed that it was not only about (....) but there is also absolute value.)
In any case, you're definitely in better position to explain why the results are different. Will you post an answer to the question on meta?