If $z,s\in \mathbb{C}$ ( with norm less than $1$)  then show that $\left|\frac{1-s/z}{1/z-\overline{s}}\right|&lt;1$. I'm hoping for an algebraic proof but possibly also a proof by geometric means if possible ( even with a picture if possible).
12873
This is the question I'm stumbling with:   When $|\alpha| &lt; 1$ and $|\beta| &lt; 1$, show that:  $$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| &lt; 1$$    The chapter that contains this question contains (among others) the triangle inequalities:   $$\left||z_1| - |z_2|\right| \le |z_1 + z_2| \le |z_1| + |z_2| $$  I've tried to use the triangle inequalities to increase the dividend and/or decrease the divisor:  $$\left|\cfrac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| &lt; \cfrac{|\alpha| +|\beta|}{\left|1-|\bar{\alpha}\beta|\right|}$$  But it's not clear if or why that would be smaller than one.  I've also tried to multiply the equation by the conjugated divisor $\cfrac{1-\alpha\bar{\beta}}{1-\alpha\bar{\beta}}$, which gives a real divisor, but the equation does not appear solvable.  Any hint would be much appreciated.
107795
I have the following inequality on my class notes that I haven't been able to prove, I was even wondering if it is actually true:  $$ \forall a,b \in \mathbb{R}^{\ge0} \left( \left| \sqrt{a}-\sqrt{b} \right| \leq \sqrt{ \vert a -b \vert}\right)  $$  Thanks
104317
Let $z_1,\dots,z_n$ be non-zero complex numbers such that $\sum_{k=1}^n \frac{1}{z_k} = 0$. Prove that for any line $ax+by=0$ passing through the origin, $z_1,\dots,z_n$ cannot all lie in either of the half-spaces $\{ ax+by&lt;0\}$, $\{ ax+by &gt;0\}$. Any help would be greatly appreciated, thanks in advance!
48888
$$S_1=\left\{z:Im\left(\frac{z-z_1}{z-z_2}\right)=0, z_1,z_2 \in \Bbb C\right\}$$  $$S_2=\left\{z:Re\left(\frac{z-z_1}{z-z_2}\right)=0, z_1,z_2 \in \Bbb C\right\}$$  Can someone help me with the geometric description of the above two sets. I have no clue as to where I should start. Thank You!
7762
Let $|z_1|=\dots=|z_n|=1$ on the complex plane.  Prove that:  $$ \left(1+\frac{z_2}{z_1}\right) \left(1+\frac{z_3}{z_2}\right) \dots \left(1+\frac{z_n}{z_{n-1}}\right) \left(1+\frac{z_1}{z_n}\right) \in\mathbb{R} $$  I have tried induction and writing every "subexpression" as $(1+e^{i(\theta_n-\theta_{n-1})})$.  Any ideas?
101870
I'm studying some questions of A problem book of complex analysis, Daniel Alpay. The Question is:     Let $z_1$, $\ldots$, $z_m$ be complex different from $0$.   Show that $$\limsup\limits_{n\to\infty}\left|\sum\limits_{\ell=1}^m{z_\ell}^n\right|^{1/n}=\max\limits_{1\le\ell\le m}|z_\ell|$$   My doubt:  Let $M :=\max\limits_{1\le\ell\le m}|z_\ell|$ and $S_n:=\sum\limits_{\ell=1}^mz_\ell^n$  I prove that $\limsup _{n \rightarrow \infty}|S_n|^{1/n}\leq M$. But I don't saw anything to do for prove that $\limsup |S_n|^{1/n}\geq M$.  I suppose that using contradiction that $\limsup |S_n|^{1/n} = \tilde{M}&lt;M$ and take $z_l = M e^{i2\pi l}$, it'll be done.  Could Someone help me?
50174
Let $a,b,c\in\mathbb{R}^*_+$, $abc=1$.  How can i show that $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ ?  I got $\left(ab-b+1\right)\left(bc-c+1\right)\left(ca-a+1\right)\leq abc=1$, but i can't go any further ...  If i expand it :   $2-ab-\frac{b}a-c-ac-\frac{a}c-b+2a+\frac{2}a-bc-\frac{c}b-a+2b+\frac{2}b+2c+\frac{2}c-4$  Which gives using $c=\frac{1}{ab}$ :   $5ab-\frac{b}a-\frac{1}{ab}-a^2b+\frac{1}a+b+\frac{1}b-2$
94842
I am trying to prove that:  Let $f$ be analytic in the closed unit disc with $f(z)\leq 1$ for $|z|\leq1$.  For a given $0&lt;r&lt;1$ and $|z_1|,|z_2|\leq r$, $z_1\neq z_2$ then,   $$\left|\frac{f(z_2)-f(z_1)}{z_2-z_1}\right|\leq \frac{1}{(1-r)^2}.$$  My attempt is: part a  of the question asks to prove $$\left|\sum_k^\infty a_nz^n\right|\leq \frac{ |z^k|}{1-|z|}.$$   So, \begin{eqnarray*}  \left|\frac{f(z_2)-f(z_1)}{z_2-z_1}\right|&amp;=&amp; \frac{|\sum_1^\infty a_nz_2^n-\sum_1^\infty a_nz_1^n|}{|z_2-z_1|}\\  &amp;\leq&amp; \frac{\frac{ |z_1 |}{1-|z_1|}+\frac{ |z _2|}{1-|z_2|}}{|z_2-z_1|}\\  &amp;\leq&amp;\frac{1}{(1-r)^2} \frac{   |z_2| ({1-|z_1|})+  |z _1| (1-|z_2|) }{|z_2-z_1|}\\ \end{eqnarray*} But could not prove that the last term is controlled by 1.
9279
Let's suppose that $|z|&lt;1$ and $|w|=1$. Show that the modulus of $\displaystyle \frac{z-w}{1-\bar{z}w}$ is always $1$. Some hint.
3540
I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$
54940
Let $U\subset \mathbb{C}$ be a nonempty connected open set such that for every $z\in U$, $\overline z\in U$.  Let $f$ be analytic on $U$. Suppose $f(x)\in\mathbb R$ for every $x\in U\cap\mathbb R$. Prove that $f(\overline{z})=\overline{f(z)}$ for any $z \in U$.    By definition, I know that $f$ analytic on $U$ means that for every $z_0 \in U$, there exists $r&gt;0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(z)=\sum_{n=0}^\infty a_n\left(z-z_0\right)^n$ on the disc $D(z_0,r)\subseteq U$. I see that $f$ takes points without imaginary components to other points without imaginary components. But I don't see how this implies a symmetry that $f(\overline{z})=\overline{f(z)}$ for $z \in \mathbb C \setminus \mathbb R$. It seems like a very strong conclusion and I'm not sure how I would prove it.  I have also shown in the preceding question that $U\cap\mathbb R$ contains an open interval, however I am not sure if that detail is meant to be helpful to this question.
68106
Prove   $$\bigcup _ { n \in \mathbb { N } } \left[ \frac { 1 } { n } , 1 \right] = ( 0,1 ]$$    I'm having some trouble starting this proof because I'm not really sure how to get started on this. Any help would be appreciated.
80304
We have been using this result without proof in my class, but I don't know how to prove it. Could someone point me in the right direction?  $$|1-e^{i\theta}|\le|\theta|$$  I believe this is true for all $\theta\in\mathbb{R}$. It is easy to show that the left side is bounded by 2 (triangle inequality), but I'm stuck otherwise.
82352
I am trying to prove that for $x \in \mathbb{R}^n$ $$\frac{1}{\sqrt{n}}||x||_1 \le ||x||_2 \le ||x||_1$$  So far I have successfully proven that $||x||_2 \le ||x||_1$, which was fairly easy.  I took the square of both norms and used that to show that $||x||_2^2 \le ||x||_1^2$.  The needed result then follows easily.  Where I am getting stuck is in proving that $\frac{1}{\sqrt{n}}||x||_1 \le ||x||_2$.  I took the square of both norms again, which results in $$(\frac{1}{\sqrt{n}}||x||_1)^2 = \frac{1}{n} \sum_{i=1}^n|x_i|^2+\frac{2}{n}\sum_{i \neq j}|x_i||x_j|$$ and $$(||x||_2)^2 = \sum_{i=1}^n|x_i|^2$$  My next step was to prove that $\frac{2}{n}\sum_{i \neq j}|x_i||x_j|$ must be less than the difference between $\frac{1}{n} \sum_{i=1}^n|x_i|$ and $\sum_{i=1}^n|x_i|$, meaning that $$\frac{2}{n}\sum_{i \neq j}|x_i||x_j| \le \frac{n-1}{n}\sum_{i=1}^n|x_i|$$ Which is where I am now stuck.  Is there some identity that I am missing here that will make this easier?
79847
If $z$ is any non-zero complex number, how to prove the following inequality? $$ | z-1 | \le | | z | -1 | + | z| \cdot | \arg z | $$  Hints please!
112065
Let $f\in C^{\infty}(\mathbb{R})$:  1)  $\exists$ $L&gt;0$ : $\forall$ $x\in \mathbb{R}$ and $\forall$ $n\ge 1$  $$|f^{(n)}(x)| \le L,$$  2) $$f\left(\frac{1}{n}\right)=0, \quad \mbox{$\forall$} n\ge 1.$$  With these conditions, prove that: $$f(x)\equiv 0, \quad  \mbox {in}\quad \mathbb{R}.$$
16666
Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.