## Complex lines in C^2

Problem: (a) Suppose ${l_1}$ and ${l_2}$ are two complex lines in ${\mathbb{C}^2}$ which are not orthogonal. Prove that the projection of circles ${\Gamma\subset l_1}$ to ${l_2}$ are also circles.

(b) Suppose ${l_1}$ is a 2 dimensional real plane in ${\mathbb{C}^2}$ and ${l_2}$ is a complex line. If for any circle ${\Gamma\subset l_1}$, its projection to ${l_2}$ is a circle on ${l_2}$, then ${l_1}$ must be a complex or anticomplex line. Proof: (a) Under some translation and rotation which preserves the projection and circles, we can assume

$\displaystyle l_1=\{(z,wz)|z\in \mathbb{C}\}$

$\displaystyle l_2=\{(\xi,0)|\xi\in \mathbb{C}\}$

where ${w}$ is a fixed constant in ${\mathbb{C}}$. Here we have used the property that ${l_2}$ is not orthogonal to ${l_1}$. Suppose ${\Gamma}$ is circle in ${l_1}$ with center ${(z_0, wz_0)}$ with radius ${r}$, then points on ${\Gamma}$ satisfy

$\displaystyle r=|(z-z_0,w(z-z_0))|=|z-z_0|\sqrt{1+|w|^2}$

which means ${|z-z_0|=const}$. This exactly means the projection on ${l_2}$ is a circle.

(b) Any 2 dimensional real space through origin can be respresented by

$\displaystyle l_1=\{\alpha\zeta+\beta\bar{\zeta}|\zeta\in \mathbb{C}\}$

where ${\alpha=(\alpha_1,\alpha_2)\in\mathbb{C}^2,\beta=(\beta_1,\beta_2)\in\mathbb{C}^2}$. WLOG, assume ${l_2}$ is the same as (a).

If ${|\alpha_1|\neq |\beta_1|}$, then let ${z=\alpha_1\zeta+\beta_1\bar{\zeta}}$,

$\displaystyle l_1=\left\{\left(z,\frac{(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}}{|\alpha_1|^2-|\beta_1|^2}\right)\right\}$

By projecting to ${l_2}$, we get ${|z|=const}$ must imply ${(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}}$ has constant norm also. One can verify that this is only true when ${\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2=0}$ or ${\alpha_1\beta_2-\alpha_2\beta_1=0}$, which means ${l_1}$ must be an anti-complex line or complex line.

If ${|\alpha_2|\neq |\beta_2|}$, we can porject circles on ${l_1}$ to ${l_3=\{(0,\zeta)\}}$, which are also circles by assumption. So we can apply all the above proof.

Next suppose ${|\alpha_1|=|\beta_1|\neq 0}$ and ${|\alpha_2|= |\beta_2|\neq 0}$, first note that

$\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}\equiv const$

In fact, for any ${\zeta}$, there exists ${c_1>0}$ such that

$\displaystyle |\alpha_1\zeta+\beta_1\bar{\zeta}|+|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1$

then ${\zeta/c_1}$ will be in a circle in ${l_1}$ whose center is origin and radius 1. So ${|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1c_0}$ for some constant ${c_0}$. Then

$\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{c_1c_0}{c_1-c_1-c_0}=const$

Choosing ${\zeta}$ in particular, this constant must be ${|\alpha_1|/|\alpha_2|=|\beta_1|/|\beta_2|}$. From

$\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{|\alpha_1|}{|\alpha_2|}$

Squaring it and write it as equality, we get

$\displaystyle [|\alpha_2|^2\alpha_1\bar{\beta}_1-|\alpha_1|^2\alpha_2\bar{\beta}_2]\zeta^2=[|\alpha_1|^2\bar{\alpha_2}\beta_2-|\alpha_2|^2\bar{\alpha_1}\beta_1]\bar{\zeta}^2$

Then

$\displaystyle |\alpha_2|^2\alpha_1\bar{\beta}_1=|\alpha_1|^2\alpha_2\bar{\beta}_2$

$\displaystyle |\alpha_1|^2\bar{\alpha_2}\beta_2=|\alpha_2|^2\bar{\alpha_1}\beta_1$

This two imply ${\alpha_1/\beta_1=\alpha_2/\beta_2=\eta^{-1}}$. Hence

$\displaystyle l_1=\{(\alpha_1,\alpha_2)(\zeta+\eta\bar{\zeta})|\zeta\in\mathbb{C}\}$

However, ${|\zeta|=|\eta\bar{\eta}|}$ implies that

$\displaystyle \text{Arg}\,(\zeta+\eta\bar{\zeta})=\frac{1}{2}\text{Arg}\,\eta$

then ${l_1}$ is in fact only one dimesional in this case. Contradiction. $\Box$

Remark: Shabat, Introduction to complex analysis Part II. Functions of several variables.