Basic questions in Complex Analysis (From Ahlfors)
Algebra of Complex numbers
- Prove the isomorphism between the system of matrices (wrt matrix multiplication and addition) of the following form and the field of complex numbers -
$$ \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} $$
Proof: Given a complex number $x=\alpha+i\beta$ we define the correspondence $\phi(x)=\phi(\alpha+i\beta)=$
$$ \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} $$
Let $y=\gamma+i\delta$. Thus, $\phi(x+y)=\phi(\alpha+\gamma+i(\beta+\delta))$ =
$$ \begin{pmatrix} \alpha+\gamma & \beta+\delta \\ -(\beta+\delta) & \alpha+\gamma \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} + \begin{pmatrix} \gamma & \delta \\ -\delta & \gamma \end{pmatrix} = \phi(x)+\phi(y) $$
We compute $\phi(x)\phi(y)=$ $$ \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} \begin{pmatrix} \gamma & \delta \\ -\delta & \gamma \end{pmatrix} = \begin{pmatrix} \alpha\gamma-\beta\delta & \alpha\delta+\beta\gamma \\ -(\alpha\delta+\beta\gamma) & \alpha\gamma-\beta\delta \end{pmatrix} = \phi(\alpha\gamma-\beta\delta+i(\alpha\delta+\gamma\beta)) = \phi(xy) $$
- Verify by calculation that the values of $\frac{z}{z^2+1}$ for $z=x+iy$ and $z=x-iy$ are conjucate.
Indeed - $$ \frac{\overline{x+iy}}{\overline{(x+iy)^2+1}} = \frac{\overline{x+iy}}{\overline{(x+iy)(x+iy)}+1} = \frac{{x-iy}}{\overline{(x+iy)}\space\overline{(x+iy)}+1} = \frac{{x-iy}}{(x-iy)(x-iy)+1} = \frac{{x-iy}}{(x-iy)^2+1} $$
- We Prove “Lagrange’s Identity” - $|\sum_{i=1}^{n}a_ib_i|^2=\sum_{i=1}^{n}|a_i|^2\sum_{i=1}^{n}|b_i|^2-\sum_{1\leq{i}<j\leq{n}}|a_i\bar{b_j}-a_j\bar{b_i}|^2$
We prove by induction on $n$. For the case of $n=1$ we have, by definition of the absolute value -
$$ |ab|=ab\bar{ab}=ab\bar{a}\bar{b}=a\bar{a}b\bar{b}=|a||b| \implies |ab|^2=|a|^2|b|^2 $$
We assume the statement is correct for $n\in{\mathbb{N}}$ and prove for $n+1$: $$ |\sum_{i=1}^{n+1}a_ib_i|^2=|\sum_{i=1}^{n}a_ib_i+a_{n+1}b_{n+1}|^2=|\sum_{i=1}^{n}a_ib_i|^2+|a_{n+1}b_{n+1}|^2+2Re(\overline{a_{n+1}b_{n+1}}\sum_{i=1}^{n}a_ib_i) = \\ \sum_{i=1}^{n}|a_i|^2\sum_{i=1}^{n}|b_i|^2-\sum_{1\leq{i}<j\leq{n}}|a_i\bar{b_j}-a_j\bar{b_i}|^2+|a_{n+1}b_{n+1}|^2+2Re(\overline{a_{n+1}b_{n+1}}\sum_{i=1}^{n}a_ib_i) = \\ \sum_{i=1}^{n}|a_i|^2\sum_{i=1}^{n}|b_i|^2-[\sum_{1\leq{i}<j\leq{n}}|a_i\overline{b_j}|^2+|a_j\overline{b_i}|^2-2Re(a_i\overline{b_j}\overline{a_j}b_i)] + |a_{n+1}b_{n+1}|^2+2Re(\overline{a_{n+1}b_{n+1}}\sum_{i=1}^{n}a_ib_i) $$ Next we note the following two statements -
$2Re(\overline{a_{n+1}b_{n+1}}\sum_{i=1}^{n}a_ib_i)=2Re(a_1\overline{b_{n+1}}\overline{a_{n+1}}b_1+\dots+a_n\overline{b_{n+1}}\overline{a_{n+1}}b_n)=\sum_{i=1}^n2Re(a_i\overline{b_{n+1}}\overline{a_{n+1}}b_i)$
Next, we add and subtract the following sums $\sum_i^n|a_ib_{n+1}|^2$ and $\sum_i^n|b_{i}a_{n+1}|^2$ to obtain the final result - $$ \sum_{i=1}^{n+1}|a_i|^2\sum_{i=1}^{n+1}|b_i|^2-[\sum_{1\leq{i}<j\leq{n+1}}|a_i\overline{b_j}|^2+|a_j\overline{b_i}|^2-2Re(a_i\overline{b_j}\overline{a_j}b_i)] = \\ |\sum_{i=1}^{n+1}a_ib_i|^2=\sum_{i=1}^{n+1}|a_i|^2\sum_{i=1}^{n+1}|b_i|^2-\sum_{1\leq{i}<j\leq{n+1}}|a_i\bar{b_j}-a_j\bar{b_i}|^2 $$ $\square$.
Going back to the question of $\overline{a}\overline{b}=$ $$ (\alpha-i\beta)(\gamma-i\delta) = (\alpha\gamma-\beta\delta)-i(\alpha\delta+\gamma\beta)=\overline{ab} $$
- A proof of Cauchy’s inequality - $$ |\sum_{i=1}^na_i\overline{b_i}|^2\leq\sum_{i=1}^n|a_i|^2\sum_{i=1}^n|b_i|^2 $$ This line of proof is taken from Ahlfors (p. 11) with details introduced by myself: Let $\lambda\in\mathbb{C}$. We write - $$ 0\leq\sum_{i=1}^n|a_i-\lambda\overline{b_i}|^2=\sum_{i=1}^n|a_i|^2+|\lambda|^2\sum_{i=1}^n|b_i|^2-\sum_{i=1}^n2Re(a_i\overline{\lambda}b_i) $$ We plug in $\lambda=\frac{\sum_{i=1}^na_i\overline{b_i}}{\sum_{i=1}^n|b_i|^2}$ - $$ \sum_{i=1}^n|a_i|^2+\frac{|\sum_{i=1}^na_i\overline{b_i}|}{(\sum_{i=1}^n|b_i|^2)^2}^2\sum_{i=1}^n|b_i|^2-2Re(\overline{\lambda}\sum_{i=1}^na_ib_i) = \\ \sum_{i=1}^n|a_i|^2+\frac{|\sum_{i=1}^na_i\overline{b_i}|}{(\sum_{i=1}^n|b_i|^2)^2}^2\sum_{i=1}^n|b_i|^2-2Re({\frac{\overline{\sum_{i=1}^na_i\overline{b_i}}}{\overline{\sum_{i=1}^n|b_i|^2}}}\sum_{i=1}^na_i\overline{b_i}) = \\ \sum_{i=1}^n|a_i|^2+\frac{|\sum_{i=1}^na_i\overline{b_i}|}{\sum_{i=1}^n|b_i|^2}^2-2Re(\frac{{|\sum_{i=1}^na_i\overline{b_i}|^2}}{\overline{\sum_{i=1}^n|b_i|^2}}) = \\ \sum_{i=1}^n|a_i|^2+\frac{|\sum_{i=1}^na_i\overline{b_i}|}{\sum_{i=1}^n|b_i|^2}^2-2\frac{{|\sum_{i=1}^na_i\overline{b_i}|^2}}{\sum_{i=1}^n|b_i|^2} = \\ \sum_{i=1}^n|a_i|^2-\frac{{|\sum_{i=1}^na_i\overline{b_i}|^2}}{\sum_{i=1}^n|b_i|^2} = \\ \sum_{i=1}^n|a_i|^2-\frac{{|\sum_{i=1}^na_i\overline{b_i}|^2}}{\sum_{i=1}^n|b_i|^2} \geq{0} $$
Implying the inequality. $\square$.
A different perspective:
Let $u,v\in{\mathbb{C^n}}$. Define $\braket{u,v}=\sum_{i=0}^na_i\overline{b_i}$. Implying - $|u|^2=\sum_{i=0}^na_i\overline{a_i}=\sum_{i=0}^n|a_i|^2$
Implying we are to prove the Cauchy-Schwarz inequality - $|\braket{u,v}|^2\leq{|u|^2|v|^2}$. Thus we can use the fact that $\frac{\braket{u,v}}{|v|^2}$ represents the ortogonal projection of $u$ onto $v$ to minimize $\lambda$ in the expression $|a-\lambda \bar{b}|^2$.
We choose $\lambda=\frac{\braket{a,b}}{|b|^2}$ to minimize the orthogonal projection and attain an expression with $\braket{a,b}$ -
$$ 0\leq|a-\lambda b|^2=|a-\frac{\braket{a,b}}{|b|^2}b|^2 = \\ \braket{a-\frac{\braket{a,b}}{|b|^2}b,a-\frac{\braket{a,b}}{|b|^2}b} = |a|^2+\frac{|\braket{a,b}|^2}{|b|^4}|b|^2-2\frac{|\braket{a,b}|^2}{|b|^2} = \\ |a|^2+\frac{|\braket{a,b}|^2}{|b|^2}-2\frac{|\braket{a,b}|^2}{|b|^2}=|a|^2-\frac{|\braket{a,b}|^2}{|b|^2} $$
Concluding the proof. $\square$
Analytical Geometry in the Complex plane
A line in the complex plan is define by $f(t)=a+bt$ for $t\in{\mathbb{R}}$.
Theorem: Two lines $f(t)=a+bt$ and $g(t)=a’+b’t$ are identical $\iff$ $a-a’$ and $b’$ are both multiple of $b$.
$\implies$ Assume $g=f$. If $a=a’$ we are done, otherwise there exists $t_0$ such that $g(0)=f(t)$ implying $a’=a+bt_0$ implying $a-a’=-bt_0$.
Similarly, there exists $t_1$ such that $g(1)=f(t_1)$ implying, $a’+b’=a+bt_1 \implies b’=b(t_1-t_0)$
$\impliedby$ Assume $a-a’$ and $b’$ are both multiples of $b$. Let $t\in{\mathbb{R}}$. Than, by assumption there exists constants $q,c\in{\mathbb{R}}$ such that $f(t)=a+bt=(a’+qb)+cb’t=a’+b’c(q+t)=g(c(q+t))$. $\square$
Question: Prove that the two diagonals of a parallelogram bisect each other.
Answer: Let $a,b\in{\mathbb{C}}$. We assume $|a|\neq0$, $|b|\neq0$
We consider the following real function - $g(t)=|(a+b)t|^2-|(a(1-t)+bt)|^2$
Than $g(0)=-|a|^2$ while $g(1)=|a+b|^2-|b|^2\geq{0}$ than by the intermediate value theorem there exists a point $p\in{[0,1]}$ such that $g(p)=0$ implying the result.
While the above result only shows existance we may also prove the following - Define $f(t)=(a+b)t$ and $g(t)=a(1-t)+bt$. Then -
$f(1/2)=\frac{a+b}{2}=g(1/2)$ concluding the pf.
Stereographic Projections
We present the following mapping from the sphere $S$ in $\mathbb{R}^3$ to the complex plane defined by $z=\frac{x_1+ix_2}{1-x_3}$ for $x_3\neq{1}$ Alternatively $(0,0,1)\notin{D}$.
We note the following -
$$ |z|^2=\frac{|x_1+ix_2|^2}{|1-x_3|^2}=\frac{x_1^2+x_2^2}{(1-x_3)^2}=\\ \frac{1-x_3^2}{(1-x^3)^2}=\frac{1+x_3}{1-x_3} $$
Indeed $|z^2|+1=\frac{2}{1-x_3}$ and $|z^2|-1=\frac{2x3}{1-x_3}$ implying $\frac{|z|^2-1}{|z|^2+1}=x_3$
Implying the circle $x_3=0$ is assigned to the circle $|z|^2=1$ in the plane and the $x_3<0$ hemisphere is assigned to $|z|^2<1$ and the opposite is also true.
We derive the following from our initial mapping -
$$ z(1-x_3)=x_1+ix_2 \implies z(1-\frac{|z|^2-1}{|z|^2+1})=x_1+ix_2 \implies \\ \frac{2z}{|z|^2+1}=x_1+ix_2 $$
Together with our initial condition - $1-(\frac{|z|^2-1}{|z|^2+1})^2=x_1^2+x_2^2$
We can obtain expressions for $x_1$ and $x_2$.
If we treat the $x_1,x_2$-axis in $\mathbb{R}^3$ as the complex plane than for any point $z$ in the plane defined by $z=x+iy$ there exists a point on the sphere that lies on the line between the point $z=(x,y,0)$ and the pole $(0,0,1)$. This vector is thus - $A-N=(x,y,0)-(0,0,1)=(x,y,-1)$.
Quesiton: What is the slope of $(x,y,-1)$ in the z-plane?
Answer: Indeed, the slope in the z-plane is $\frac{\Delta{z}}{|z|}=\frac{1}{|z|}=\frac{1}{\sqrt{x^2+y^2}}$
We can project this point to the sphere $S$ to obtain a point $(x_1,x_2,x_3)$ on the sphere using the expressions we derived previously (which depend on $|z|^2$). Where the line from $(0,0,1)$ to $(x,y,0)$ in the z-plane is thus $x_3=1-\frac{1}{|z|}t$
Quesiton: At what point on the sphere $S$ does the line given previously meet the sphere?
By our previous definition, to assign $x,y$ to $x_1,x_2$-plane coordinates we can use the following mapping - $$ x+iy=\frac{x_1+ix_2}{1-x_3} $$
Implying $x_1=x(1-x_3)$ and $x_2=y(1-x_3)$, we thus get - $$ (x(1-x_3))^2+(y(1-x_3))^2+x_3^2 = 1 \\ (x(1-(1-\frac{1}{|z|}t)))^2+(y(1-(1-\frac{1}{|z|}t)))^2+(1-\frac{1}{|z|}t)^2 = 1 \\ x^2(\frac{1}{|z|}t)^2 +y^2(\frac{1}{|z|}t)^2 + (1-\frac{1}{|z|}t)^2 = 1 \\ t^2 + (1-\frac{1}{|z|}t)^2 = 1 \\ t^2 + 1 - \frac{2t}{|z|} + \frac{t^2}{|z|^2} = 1 \\ t^2|z|^2-2t|z|+t^2 = 0 \\ t^2(|z|+1)-2t|z| = 0 $$
Implying $t=0$ or $t=\frac{2|z|}{1+|z|^2}$. We plug back in to the line equation $x_3=1-\frac{1}{|z|}t$ to get $x_3=1-\frac{2}{1+|z|^2}=\frac{|z|^2-1}{|z|^2+1}$
We get finally - $(\frac{2x}{1+|z|^2}, \frac{2y}{1+|z|^2}, \frac{|z|^2-1}{|z|^2+1})=(x_1,x_2,x_3)$. Implying the straight line from the pole through $(x,y,0)$ passes through $(x_1,x_2,x_3)$. $\square$
This implies that the transformation defined above corresponds to projecting points from the sphere onto the complex plane along lines passing through the pole $(0,0,1)$.
Quesiton: What is the equation describing a circle in the sphere $S$ passing through $(0,0,1)$ and $(x_1,x_2,x_3)$?
Question: Show that two points on the sphere $z,z’$ are diametrically opposite points iff $z\bar{z}’=-1$
Answer: Indeed we use the equation of the distance derived with $d(z,z’)=2R=2$: $$ 2=\frac{2|z-z’|}{\sqrt{(1+|z|^2)(1+|z’|^2)}} \\ \sqrt{(1+|z|^2)(1+|z’|^2)}=|z-z’| \\ (1+|z|^2)(1+|z’|^2) = |z-z’|^2 \\ 1+|z|^2+|z’|^2+|z|^2|z’|^2=|z|^2+|z’|^2-z\bar{z}’-\bar{z}z’ \\ 1+|z|^2|z’|^2+z\bar{z}’+\bar{z}z’=0 \\ (1+\bar{z}z’)(1+z\bar{z}’)=0 \implies |1+z\bar{z}|^2=0 \implies z\bar{z}’=-1 $$