By A. I. Markushevich

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**Sample text**

35. of the arc PLQ, not a single one? We shall now show that the images fill the whole ray. Let us begin with the point P' (the origin); it is the image of the point P since z' = z - a vanishes when z = a. We shall z- b take an arbitrary point z' on the ray P'E (Fig. 35) differing from the point P' (i. e. z' =F 0). It is evident that z' cannot be a positive real number since the ray P'L: does not coincide with the positive real axis. Considering z to be unknown, we solve the equation z' = z- a z'b - a = - - b for z and find zz' - z'b = z - a whence z = , 1.

45. scientists N. E. Zhukovsky and S. A. Chaplygin for which reason they are called Zhukovsky-Chaplygin sections. By changing the angle of inclination cp of the tangent line to the circle at point 1 (Fig. 42) and the radius of the smaller circle, various sections can be obtained. In particular, if the angle cp is right, i. e. if the larger circle is constructed on the interval from - 1 to + 1 as on the diameter, the corresponding section is symmetric with respect to the real axis (Fig. 45). Such a section is sometimes called Zhukovsky's vane.

So we wish to prove that for every point z on PLM there exists a corresponding point z' which the points 0 and c' = ::=~ = ~ =- ~ on the circle P'I:M' for are the ends of the diameter. It is, evidently, sufficient to prove that from each point z' = z - a (provided that z lies on PLM) the segment P'E' can be z- b seen at right angles, i. e. the angle E' R'P' is the rightangle *. But the angle E' R' P' is formed by the vectors E'R' and P'R' representing the numbers z' - c' and z'; it is equal to the angle S'P'R' (the vectors P'S' and E'R' being equal), reckoned in the direction from P'S' to P' R'.