**Problem:** (a) Suppose and are two complex lines in which are not orthogonal. Prove that the projection of circles to are also circles.

(b) Suppose is a 2 dimensional real plane in and is a complex line. If for any circle , its projection to is a circle on , then must be a complex or anticomplex line. *Proof:* (a) Under some translation and rotation which preserves the projection and circles, we can assume

where is a fixed constant in . Here we have used the property that is not orthogonal to . Suppose is circle in with center with radius , then points on satisfy

which means . This exactly means the projection on is a circle.

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

where . WLOG, assume is the same as (a).

If , then let ,

By projecting to , we get must imply has constant norm also. One can verify that this is only true when or , which means must be an anti-complex line or complex line.

If , we can porject circles on to , which are also circles by assumption. So we can apply all the above proof.

Next suppose and , first note that

In fact, for any , there exists such that

then will be in a circle in whose center is origin and radius 1. So for some constant . Then

Choosing in particular, this constant must be . From

Squaring it and write it as equality, we get

Then

This two imply . Hence

However, implies that

then is in fact only one dimesional in this case. Contradiction.

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

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