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Proofs in Complex Analysis

More proofs in analysis:

Question (Composition of differentiable functions): Let $f,g$ be differentiable functions such that $g:X\rightarrow{Y}$ and $f:Y\rightarrow{Z}$ such that the image of $g$ is contained within the domain of $f$. Than, $f\circ{g}$ is a differentiable function and its derivative is given by $f(g(x))’=f’(g(x))g’(x)$.

Answer: Let $t\in{X}$, implying by our assumption $g(t)\in{Y}$. Than differentiability of $f$ we may write - $$ f(g(t)+h)-f(g(t))=f’(g(t))h+o(h) $$

We define $h(u)=g(t+u)-g(t)$ implying -

$$ f(g(t+u))-f(g(t)) = \ f(g(t)+h(u))-f(g(t)) = \ f’(g(t))h(u)+o(h(u)) = \ f’(g(t))(g(t+u)-g(t))+o(h(u)) $$

Since $h(u)=g(t+u)-g(t)=g’(t)u+o(u)$,

we have $h(u)=O(u)+o(u)=O(u)$, implying $o(h(u))=o(u).$

$$ f’(g(t))(g’(t)u+o(u))+o(u) = \ f’(g(t))g’(t)u+f’(g(t))o(u)+o(u) = \ f’(g(t))g’(t)u+o(u)+o(u) = \ f’(g(t))g’(t)u+o(u) $$

Concluding the proof.

Question: Prove that a convergent sequence is bounded

Answer: Let $\set{a_i}$ be a convergent sequence, ie having a finite limit $A$. Let $\epsilon>0$, by definition, there exists $N\in{\mathbb{N}}$ such that for $n>N$ we have - $|a_n-A|<\epsilon$ implying $|a_n|<|A|+\epsilon$ for all $n>N$. Implying that the tail end of the terms are bounded.

On the other hand, the finite set if terms $\set{a_i}_{i\leq{N}}$ is also bounded as a finite set (simply take its maximum). Together, we have that the sequence is bounded.

Question: (Cesaro) If $\lim_{n\rightarrow\infty}z_n=A$ prove that $$ \lim_{n\rightarrow\infty}\frac{1}{n}(z_1+…+z_n)=A $$

Answer:

We define $S_n=\frac{1}{n}(z_1+…+z_n)$. We compute $$ |S_n-A| = |\frac{1}{n}((z_1-A)+…+(z_n-A))| $$

Let $\epsilon>0$. From the convergence of $\set{z_i}$, there exists $N\in{\mathbb{N}}$ such that for $n>N$ we have -

$$ |S_n-A| = |\frac{1}{n}((z_1-A)+…+(z_n-A))|\leq\frac{1}{n}(|z_1-A|+..+|z_n-A|)<\\ \frac{1}{n}(|z_1-A|+..+|z_N-A|)+(1-\frac{N}{n})\epsilon $$

We let $n\rightarrow\infty$ to conclude $\limsup_n|S_n-A|\leq\epsilon$, Since, $\epsilon$ is arbitrarily small we conclude $\lim{S_n}=A$.

Question: Investigate the convergence of the sequence $\set{nz^n}^{\infty}_{n=1}$.

Answer: Let $z\in{\mathbb{C}}$ such that $z\neq{0}$ .We compute -

$$ |\frac{(n+1)z^{n+1}}{nz^n}|=(1+\frac{1}{n})|z| $$

As $n\rightarrow\infty$ we have

$$ |\frac{(n+1)z^{n+1}}{nz^n}|\rightarrow{|z|} $$

Case $|z|<1$:

Thus, for sufficiently large $n$ we may pick $|z|<q<1$ such that- $$ |a_n|=|nz^n|<n|z|^n<nq^n\rightarrow{0} $$

Since $nq^n\rightarrow{0}$ for any $0<q<1$.

Side quest Pf: $nq^n\rightarrow{0}$ for any $0<q<1$.

Recall: $a_{n+1}/a_{n}=(1+\frac{1}{n})q\rightarrow{q}<1$. Implying there exists $N$ such that for $n\geq{N}$, the sequence is decreasing and as it is a sequence of positive numbers it is bounded. Moreover there is $q<s<1$ such that for all $n\geq{N}$ we have

$$ \frac{|a_{n+p}|}{|a_n|}=\frac{|a_{n+p}|}{|a_{n+p-1}|}…\frac{|a_{n+1}|}{|a_n|}<s^p $$

Implying for $p\geq{1}$ we have, $a_{N+p}<a_Ns^p=Nq^Ns^p<Ns^{N+p}$. Than, from our previous analysis, since $s<1$, we may take $n$ to be arbitrarily large and make $|Ns^n|$ arbitrarily small. Implying $|a_n|<|Ns^{n}|$. Now let $\epsilon$, take $n$ such that $n>\log_s{\epsilon/N}$ to obtain $|a_n|<|Ns^{n}|<\epsilon$.

For $|z|>1$: The necessary conditions for convergence do not apply bc the terms do not tend to zero, and therefore the sequence diverges.

Question: Show that the sum of an absolute convergent series does not change if the terms are rearranged.

Proof:

Let $S=\sum_{i=0}^\infty{a_i}$ be an absolutly convergent sequence.

Let $\pi:\N\rightarrow\N$ be a permutation of $\N$.

Since the original sequence is absolutly convergent, let $\epsilon>0$, we find $n_0$ such the for $n>n_0$ we have $|a_n|+….+|a_{n+p}|<\epsilon$ for any $p\geq0$.

This implies there are only finitely many terms in the original series such that the Cauchy criteria does not apply.

For those finite set $\set{1,…,n_0}$, the permutation $\pi$ must place all of them in possibly large set therefore there exists $n_\pi$ such that all the indicies appears among them.

Thus, we may find the largest index $n_\pi$ such that for any $n>n_\pi$ we have $|a_{\pi(n)}+…+a_{\pi(n+p)}|<\epsilon$ for any $p>0$. (since for each $\pi(n)$ we have $\pi(n)>n_0$)

Implying the rearranged series converges from the Cauchy criteria. Moreover, for any $n>n_\pi$ we have

We have - $$ |S-\sum_{i=0}^n{a_{\pi(i)}}|=|\sum_{i=n+1}^\infty{a_{\pi(i)}}|\leq \\ \sum_{i=n+1}^\infty{|a_{\pi(i)}|} \leq \sum_{i=n_0+1}^\infty{|a_{i}}|=\epsilon $$

Concluding the pf.

Question:

Discuss the uniform convergence of the series: $$ \sum_{n=1}^{\infty}\frac{x}{n(1+nx^2)} $$

Proof: Let $x\in{\R}$ such that $|x|<1$. We have - $$ |a_n|=|\frac{x}{n(1+nx^2)}|=\frac{|x|}{n(1+nx^2)}\leq\frac{1}{n^2} $$

if $|x|=1$ we have - $$ |a_n|=\frac{1}{n(1+n)}\leq\frac{1}{n^2} $$

The latter are terms of a convergent series implying that the original series is converges uniformly inside the unit ball due to the Wierstrass M-test.

if $|x|>1$ we have $$ |a_n|=|\frac{x}{n(1+nx^2)}|=\frac{|x|}{|n(1+nx^2)|}\leq\frac{|x|}{|n(1+n)|}<\frac{M}{|n(1+n)|} $$ For some $M>|x|$, implying the convergence is only point-wise.

Question: For real $y$, show that every remainder in the series expansion for $\cos{y}$ has the same sign of the last omitted term in the series.

Answer: Indeed the terms of the series are given by $1-\frac{x^2}{2}+…+\frac{(-1^n)x^{2n}}{2n!}$. Since the sign of the terms alternate the remainder has the same sign of the last omitted term in the series. Indeed, by Taylor’s formula the remainder is the following - $$ Re(y)=\frac{f^{(2n+1)}(\epsilon)}{(2n+1)!}=\frac{(-1)^n\sin{\epsilon}}{(2n+1)!} $$

For some $0<\epsilon<y$. This term indeed has the same sign has the $n$th terms in the series. The same is true for $\sin$.

Proofs in Complex Analysis:

Question: If $g(z)$ and $f(z)$ are analytic functions show the $f(g(z))$ is analytic.

Answer: Since $f,g$ are both analytic there exists $u_f,u_g$ and $v_f,v_g$ differentiable such that - $$ f(s+it) = u_f(s,t)+iv_f(s,t) \ g(x+iy) = u_g(x,y)+iv_g(x,y) $$

$f,g$ satisfy the Cauchy-Riemann equations respectively.

Let $z\in{\mathbb{C}}$ we compute -

$$ u_{f\circ{g}} = u_f(u_g(x,y),v_g(x,y)) \implies \ d_xu_{f\circ{g}} = d_su_fd_xu_g+d_tu_fd_xv_g = \ $$

Using the Cauchy-Riemann equations for both $f$ and $g$ in the last equality:

$$ d_su_f=d_tv_f \ -d_tu_f=d_sv_f $$

And - $$ d_xu_g=d_yv_g \ -d_yu_g=d_xv_g $$

We obtain - $$ d_su_fd_xu_g+d_tu_fd_xv_g = \ d_tv_fd_yv_g+d_sv_fd_yu_g = d_yv_{f\circ{g}} $$

Similarly, computing $d_yu_{f\circ{g}} = d_su_fd_yu_g + d_tu_fd_yv_g$ and applying Cauchy-Riemann equations yields $d_yu_{f\circ{g}}=-d_xv_{f\circ{g}}$.

Question: Let $f$ be analytic function. If $|f|$ is constant then $f$ is constant.

Answer:

Since $|f|$ is constant so is $|f|^2=u^2+v^2$ and thus we may deduce - $d|f|^2=(d_x|f|^2,d_y|f|^2)=0$

$$ d_x|f|^2=2ud_xu+2vd_xv = 0 \\ d_y|f|^2=2ud_yu+2vd_yv = 0 $$ Using the Cauchy-Riemann equations for $f$ we obtain - $$ 2ud_xu+2vd_xv=0 \\ -2ud_xv+2vd_xu=0 $$

We obtain two linear equations in $d_xu,d_xv$ corresponding to the following matrix -

$$ \begin{pmatrix} u & v \\ v & -u \end{pmatrix} $$

Which has a determinant, $-|f|^2=const$, by assumption.

If $|f|=0$ we have $u,v=0$. Otherwise, the matrix is invertible and $d_xu=0,d_xv=0$.

We repeat, in the same manner for $d_yu=0,d_yv=0$

Implying that $f$ is constant.

Question: Show that $3<\pi<2\sqrt{3}$:

We first find independently $\cos\frac{\pi}{3}$ and $\sin\frac{\pi}{3}$. Indeed since $\exp{\pi i}=-1$ we have $(\exp{\frac{\pi}{3}i})^3=-1$ implying $\exp{\frac{\pi}{3}i}$ is a solution for $z^3=-1$.

Thus we reduce this question to finding solutions to $z^3+1=0$.

Indeed, we have for any $a^3+b^3=0$ the following decomposition - $(a+b)(a^2-ab+b^2)$

This can be viewed by observing that $a=-b$ is a solution to the qubic equations and solving for the quadratics $(a+b)(Aa^2+Bab+Cb^2)$

We multiply and obtain $Aa^3+Ba^2b+Cab^2+Aa^2b+Bab^2+Cb^3$

From the original equation $a^3+b^3=0$, we must have $A=C=1, B=-1$.

Resulting in the decomposition mentioned. We than have $z^3+1=(z+1)(z^2-z+1)$

Giving a single real root $z=-1$ and two imaginary roots: $\frac{1+i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}$

Since $\sin y>0$ for $y<\pi$ (from continuity of the polynomial everywhere and the intermediate value theorem) we arrive at $\exp{\frac{\pi}{3}i}=\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}$ implying $\cos{\frac{\pi}{3}}=\frac{1}{2}$ and $\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}$.

We return to the original inequlities now with this information.

We have $1-\frac{\pi^2}{18}<\cos{\frac{\pi}{3}}=\frac{1}{2}$ Implying - $1<\frac{\pi^2}{9}$ giving $3<\pi$.

On the other hand $\frac{\sqrt{3}}{2}=\sin{\frac{\pi}{3}}>\frac{\pi}{3}-\frac{\pi^3}{162}$. We study the polynomial $f(y)=\frac{y}{3}-\frac{y^3}{162}$ and find a (crude) bound such that the above inequality is satisfied. We have $Df=\frac{1}{3}-\frac{3y^2}{162}\implies$ f has extremum at $y^2=\frac{162}{9}=18\implies y=\pm3\sqrt{2}$.

We note that $f$ increases for $|{y}|<{3\sqrt{2}}$

We would like to show that $\pi<2\sqrt{3}$. Indeed, compute $f(2\sqrt{3})=\frac{2\sqrt{3}}{3}-\frac{(2\sqrt{3})^3}{162}=\frac{252}{162\sqrt{3}}=\frac{84*3}{162\sqrt{3}}=\frac{84\sqrt{3}}{162}>\frac{81\sqrt{3}}{162}=\frac{\sqrt{3}}{2}$. However, since $f$ is increasing in this range and is continous everywhere we deduce that $y<2\sqrt{3}$ to obtain $f(y)<\frac{\sqrt{3}}{2}$ implying $\pi<2\sqrt{3}$.