\[e=\sum_{n=0}^{\infty}\frac{1}{n!}\quad(n=0,1,2,3,\cdots)\]
\[\begin{align} (1+\frac{1}{n})^n&=\sum_{m=0}^{n}{n \choose m}\cdot 1^{m}\cdot(\frac{1}{n})^{n-m}\\ &=\sum_{m=0}^{n}{n \choose m}\cdot(\frac{1}{n})^{n-m}\\ &={n \choose n}+{n \choose n-1}(\frac{1}{n})+{n \choose n-2}(\frac{1}{n})^2+\cdots+{n \choose 0}(\frac{1}{n})^n\\ &=1+n(\frac{1}{n})+\frac{n(n-1)}{2!}(\frac{1}{n})^2+\cdots+\frac{n!}{n!}(\frac{1}{n})^n\\ &=1+1+\frac{1}{2!}(1-\frac{1}{n})+\cdots+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{n-1}{n})\\ &\leq \sum_{k=0}^{n}\frac{1}{k!}\leq e \end{align}\] Next if \(n\ge m\), \[(1+\frac{1}{n})^n=1+1+\frac{1}{2!}(1-\frac{1}{n})+\cdots+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{n-1}{n})\\ \ge 1+1+\frac{1}{2!}(1-\frac{1}{n})+\cdots+\frac{1}{m!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{m-1}{n})\\ \ge 1+1+\frac{1}{2!}+\cdots+\frac{1}{m!}\\ =\sum_{k=0}^{m}\frac{1}{k!}\] let \(n\to\infty\), keep \(m\) fixed, we get \[\lim_{n\to\infty}\text{inf}(1+\frac{1}{n})^n\ge \sum_{k=0}^{m}\frac{1}{k!}\] when \(m\to\infty\) \[\lim_{n\to\infty}\text{inf}(1+\frac{1}{n})^n\ge e\] Then \[\lim_{n\to\infty}(1+\frac{1}{n})^n=e\]
For fixed rational number \(z\), \[\begin{align} \lim_{n\to\infty}(1+\frac{z}{n})^n&=\Bigl[\lim_{n\to\infty}(1+\frac{1}{n/z})^{n/z}\Bigr]^{z}\\ &=e^z \end{align}\]
The Ratio Test for series convergent: The series \(\sum a_n\) converges if \[\lim_{n\to\infty}\text{sup}\Biggl|\frac{a_{n+1}}{a_n}\Biggr|<1\] and diverges if \[\Biggl|\frac{a_{n+1}}{a_n}\Biggr|\ge1\] for all \(n\ge n_0\) where \(n_0\) is some fixed integer. Let \[\Biggl|\frac{a_{n+1}}{a_n}\Biggr|<\beta<1\] \[|a_{n+1}|<|a_n|\beta\] Then \[|a_{n}|<|a_N|\beta^{n-N}=(|a_N|\beta^{-N})\beta^{n}\quad(n>N)\] Because \(\sum\beta^n\) converges, then \(\sum a_{n}\) converges.
Ratio Test shows the series \[E(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}\] converges for every complex \(z\) we define the multiplication \[E(z)E(w)=\sum_{n=0}^{\infty}\frac{z^n}{n!}\sum_{m=0}^{\infty}\frac{w^m}{m!}=\sum_{m=0}^{\infty}\sum_{k=0}^{n}\frac{z^kw^{n-k}}{k!(n-k)!}\\ =\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}{n \choose k}z^kw^{n-k}=\sum_{n=0}^{\infty}\frac{(z+w)^n}{n!}=E(z+w)\] \[E(0)=\sum_{n=0}^{\infty}\frac{0^n}{n!}=1\] \[E(1)=\sum_{n=0}^{\infty}\frac{1^n}{n!}=e\] \[E(p)=E(1+\cdots+1)=E(1)E(1)\cdots E(1)=e^p\quad(p>0,p\text{ is rational})\] \(E(x)\) is called exponential function.
For fixed rational number \(z\), \[\lim_{n\to\infty}(1+\frac{z}{n})^n=\sum_{n=0}^{\infty}\frac{z^n}{n!}\]
Since \(exp(z)\ne0\) \[\begin{align} exp'(z)&=\lim_{h\to0}\frac{exp(z+h)-exp(z)}{h}\\ &=exp(z)\lim_{h\to0}\frac{exp(h)-1}{h}\\ &=exp(z)\lim_{h\to0}\frac{\sum_{n=0}^{\infty}\frac{h^n}{n!}-1}{h}\\ &=exp(z)\lim_{h\to0}\frac{\sum_{n=1}^{\infty}\frac{h^n}{n!}}{h}\\ &=exp(z)\lim_{h\to0}\sum_{n=1}^{\infty}\frac{h^{n-1}}{n!}\\ &=exp(z)\cdot1\\ &=exp(z) \end{align}\]
For complex numbers \[E(-it)=\sum_{n=0}^{\infty}\frac{(-it)^n}{n!}\] is the complex conjugate of \[E(it)=\sum_{n=0}^{\infty}\frac{(it)^n}{n!}\] Thus \[|e^{it}|^2=e^{it}\overline{e^{it}}=e^{it}e^{-it}=e^{0}=1\] then \(|e^{it}|=1\quad (t \text{ is real})\) so \(e^{it}\) lies on the unit circle. We define \(\cos t, \sin t\) to be the real and imaginary parts of \(e^{it}\) \[\cos t=\text{Re }[e^{it}],\quad \sin t=\text{Im }[e^{it}]\quad(t\text{ real})\] and \[e^{it}=\cos t+i\sin t\] Since \[\cos' t+i\sin' t=(e^{it})'=ie^{it}=-\sin t+i\cos t\] Then \[\cos' t=-\sin t,\quad \sin' t=\cos t\] and \[\cos t=\text{Re }[e^{it}]=\text{Re }\Biggl[\sum_{n=0}^{\infty}\frac{(it)^n}{n!}\Biggr]\\ =1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\cdots\] Since \(\cos0=1\), we define the smallest positive number \(t_0\) for which \(\cos t_0=0\) and we define \[t_0=\frac{\pi}{2}\] Since \(|e^{it}|=1\) and \(e^{it}=\cos t+i\sin t\) then \(\sin t_0=\pm1\) since \(\sin' t=\cos t\ge0 \quad(t\in [0,t_0])\) and \(\sin0=0\) then \(\sin t_0>0\) hence \(\sin t_0=1\) and \[e^{\frac{\pi i}{2}}=i\] It follows that \[e^{\pi i}=i^2=-1\quad e^{2\pi i}=1\] and \[e^{z+2\pi i}=e^{z}e^{2\pi i}=e^{z}\]
If \(z=x+iy, \quad(x,y\text{ are real})\) then \[e^z=e^{x+iy}=e^xe^{iy}\] and \(|e^z|^2=e^xe^{iy}\overline{e^xe^{iy}}=e^{2x}\) then \[|e^z|=e^{x}\] If \(e^z=e^xe^{iy}=1\) then \(|e^z|=e^x=1\) and \(x=0\) and \(e^{iy}=1\) since \(e^z=e^{z+2\pi i}\) then \[e^{iy}=e^{yi+2\pi i}=e^{(y+2\pi)i}=1\] Suppose \(0<y<2\pi\) Since \(0<\frac{y}{4}<\frac{\pi}{2}\) and \(e^{iy/4}=\cos(y/4)+i\sin(y/4)\) we have \(\cos(y/4)>0,\quad\sin(y/4)>0\) also \[e^{iy}=(e^{iy/4})^4=(\cos(y/4)+i\sin(y/4))^4\\ =\cos^4(y/4)+\sin^4(y/4)-6\cos^2(y/4)\sin^2(y/4)+4i\cos(y/4)\sin(y/4)(\cos^2(y/4)-\sin^2(y/4))\] Since \(e^{iy}=1\), then \[cos^2(y/4)=\sin^2(y/4)=\frac{1}{2}\] then \[e^{iy}=(e^{iy/4})^4=(\cos(y/4)+i\sin(y/4))^4\\ =\cos^4(y/4)+\sin^4(y/4)-6\cos^2(y/4)\sin^2(y/4)\\ =(\frac{1}{2})^2+(\frac{1}{2})^2-6\cdot\frac{1}{2}\cdot\frac{1}{2}\\ =-1\ne1\] then \(y\notin(0,2\pi)\) and \(\frac{y}{2\pi}\) must be an integer.
The symbol \[f:X\to Y\] means that \(f\) is a function (mapping) of the set \(X\) into the set \(Y\). If \(f(X)=Y\), \(f\) is said to map \(X\) onto \(Y\).
Since \(|e^{it}|=1\quad (t \text{ is real})\) so the mapping \(f: t\to e^{it}\) maps the real axis into the unit circle. For \(w=u+vi\quad(u,v\text{ are real})\) that \(|w|=1\), (a) suppose that \(u\ge0, v\ge0\), since that \(u\leq1\) then there exists a \(t, 0\leq t\leq \pi/2\) such that \(\cos t=u\) then \(\sin^2t=1-u^2=v^2\), and since \(\sin t\ge0\) if \(0\leq t\leq\pi/2\) we have \(\sin t=v\), thus \(w=e^{it}\). (b) If \(u<0, v\ge0\) then \(-iw=-ui+v=e^{it}\quad(t\text{ is real})\) and \(w=\frac{e^{it}}{-i}=\frac{e^{it}}{-e^{\pi i/2}}=e^{i(t+\pi/2)}\). (c) If \(v<0\), then \(-w=e^{it}\quad(t\text{ is real})\) hence \(w=\frac{e^{it}}{-1}=\frac{e^{it}}{-e^{i\pi}}=e^{i(t+\pi)}\). Then the mapping \(f: t\to e^{it}\) maps the real axis onto the unit circle.
If \(w\) is a complex number and \(w\ne0\), since \(w=\frac{w}{|w|}|w|\), there is a real \(x\) such that \(|w|=e^x\), since \(\Biggl|\frac{w}{|w|}\Biggr|=1\), then \(\frac{w}{|w|}=e^{iy}\quad(y\text{ is real})\) for some real \(y\). Hence \(w=e^{iy}e^x=e^{x+iy}\)