Aleph Null

  Theorem : If all zeros of a polynomial $P(z)$ lie in a half-plane, then all zeros pf the derivative $P'(z)$ lie in the same half-plane. Proof: Let $P(z)$ be any polynomial with degree $n$. Then $$P(z)=a(z-a_1)(z-a_2)\cdots(z-a_n)$$ where $a_1,a_2,\dots,a_n$ are the zeros of $P(z)$. Hence $$\frac{P'(z)}{P(z)}=\frac1{z-a_1}+\frac1{z-a_2}+\cdots+\frac1{z-a_n}$$Suppose the half plane $H$ defined as the part of the plane where $$IM\frac{z-a}{b}<0$$. Suppose $z$ If $a_k$ is in $H$ and $z$ is no, we have then $$Im\frac{z-a_k}{b}=Im\frac{z-a}{b}-Im\frac{a_k-a}{b}>0$$But the imaginary parts of reciprocal numbers have opposite signs. Therefore, under the same assumption, $Im\ b(z-a_k)^{-1}<0$.Now  this is true for all $k$ we conclude that $$Im\frac{bP'(z)}{P(z)}=\sum_{k=1}^{n} Im\frac{b}{z-a_k}<0$$ and consequently $P'(z)\neq 0$. Hence $z$ is not a root of $P'(z)$ concluding that all roots of $P'(z)$ lie in $H$. $\blacksquare$

Let \(f:U\to \mathbb{C}\) function which is differentiable and \(U\) is open in \(\mathbb{C}\). Suppose \(f'(z_0) \) exists where \(z_0=a+ib\in \)  \(U\subset \mathbb{C}\). \(f(z)=u+iv\) where \(u:U\to \mathbb{R}\) and \(v:U\to \mathbb{R}\). First take \(h=t\in \mathbb{R}\).$$f'(z_0)  = \lim_{t\to 0}\frac{f(a+t+ib)-f(a+ib)}{t}$$Breaking it we get $$ \lim_{t\to 0} \frac{u(a+t,b)-u(a,b)}{t}+i\lim_{t\to 0}\frac{v(a+t,b)-v(a,b)}{t}= \left.\frac{\partial u}{\partial x}\right|_{z_0}+i\left. \frac{\partial v}{\partial x}\right|_{z_0} \label{cd1}$$ Now take \(h=it\), \(t\in \mathbb{R}\) $$f'(z_0) = \lim_{t\to 0}\frac{f(a+ib+it)-f(a+ib)}{it}$$ Breaking it down we get $$ \lim_{t\to 0} \frac{u(a,b+t)-u(a,b)}{it}+i\lim_{t\to 0}\frac{v(a,b+t)-v(a,b)}{it}= \left. \frac{\partial v}{\partial y}\right|_{z_0} -i\left. \frac{\partial u}{\partial y}\right|_{z_0}\label{cd2}$$ Equating the two equations we get \(f\) is complex differentiable at \(z_0\) and $$\boxed{\left.\frac{\partial u}{\par...