New PDF release: Algebraic Methods in the Global Theory of Complex Spaces

By Constantin Banica, Octavian Stanasila

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But that Evidently, then, the set of values of log(i2) is not the same as the set of values of 2logi. That is, log(i 2 ) :t: 2Iogi. 7. To solve the equation logz = in 12, write exp(logz) = exp(in / 2), or z = e;1ri 2 = i. 10. Since ln(x 2 + y2) is the real component of any (analytic) branch of 2logz, it is harmonic in every domain that does not contain the origin. This can be verified directly by writing u(x,y) ln(x 2 + y2) and showing that u'"(x,y) + u,,(x,y) O. = 1. = Suppose that Rez1 > Oand Rez2 > O.

N = 0,±1,±2, ... ). )m (n = 0,±1,±2, ... ). 41 On the other hand, (n = 0,±1,±2, ... ). 1ogi. 2 (b) Note that log(i2) = log(-1) = In 1 + (n + 2nn)i = (2n + l)m (n = 0,±1,±2, ... ) 2logi =2[1n1+{; +2nn)] =(4n+ l)m (n = 0,±1,±2, ... ). but that Evidently, then, the set of values of log(i2) is not the same as the set of values of 2logi. That is, log(i 2 ) :t: 2Iogi. 7. To solve the equation logz = in 12, write exp(logz) = exp(in / 2), or z = e;1ri 2 = i. 10. Since ln(x 2 + y2) is the real component of any (analytic) branch of 2logz, it is harmonic in every domain that does not contain the origin.

2 2 Thus _1 2 - 1 1z I= 1z 1-11 <.! - 3· 2 62 Also, the length of C is ~ (4n) =n. _,~ML= n. 2 Cz 2. =i and L =n, we find that -1 3 Toe path C is as shown in the figure below. Toe midpoint of C is clearly the closest point on C to the origin. The distance of that midpoint from the origin is clearly -fi , the length of C 2 being -vi. y o Hence if z is any point on C, 1zl Consequently, by taking M 3. ~ ~. _l=-1 4 Toe contour C is the closed triangular path shown below. y -4 To find an upper bound for lfc

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