By A. I. Markushevich
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Additional info for Complex Numbers and Conformal Mappings
See 19. increases, w en point z Z - SIn ex ". moves in the indicated direction, acquiring infinitely large values. But cp + (1 + P= 180°, whence p = 180° - (CJ. + cos ex sin
Approaching indefinitely the limiting position A' M'. In the process, the point B', which is the nearest to A' point of intersection of the secant and the curve, will approach A' indefinitely, since the distance A' B' = D 2 sin? ex tends to zero as ex tends to zero. It follows that A' M', which is the limiting position of the secant, is the tangent to the arc A'B'lB2". at the point A'. It can also be proved that A'M' is the tangent to the arc A' B;B(, ... at the same point A'. 34. Now let us turn to Zhukovsky's function z' = ~(z + +).
Now passing from secants to tangents, we shall make point Z1 approach indefinitely point zo along the curve L. Then the point Z'1 = will approach indefinitely the point Zo = z5 along the curve L'. Therefore, the secants will also approach indefinitely the tangents drawn at the points Zo and Zo and the angle between the secants will approach indefinitely the angle between the tangents. , = 90°; consequently, the tangent at the point Zo = i to any curve L drawn through that point and the tangent at the point z5 = i 2 = - 1 to the image on the curve L' are mutually perpendicular.