By Leif Mejlbro

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**Extra resources for Complex Functions Examples c-3 - Elementary Analytic Functions and Harmonic Functions**

**Example text**

1) If a ∈ R+ , then Log(a + iy) is continuous in y ∈ R, thus lim {Log(a + iy) − Log(a − iy)} = 0 y→0+ for a ∈ R+ . com 33 Complex Functions Examples c-3 The exponential function and the logarithm function 2) If a ∈ R− , then Log(a + iy) tends towards ln |a| + iπ for y → 0+, and Log(a − iy) tends towards ln |a| − iπ for y → 0+, hence lim { Log(a + iy) − Log(a − iy)} = 2πi y→0+ for a ∈ R− . 3) If a = 0, then lim {Log(iy) − Log(−iy)} = y→0+ π π i − − i = πi, 2 2 for a = 0. 6 Given the function f (z) = Log z 2 + 2 , where Log denotes the principal branch of the logarithm, and denote by A the largest domain in C, in which f is analytic.

If we put w = eiz , 2 3 then the latter equation is written 0 = 2 cos z − 10 1 10 1 =w+ − = 3 w 3 w w2 − 10 w+1 3 = 1 w w− 1 3 {w − 3}, hence iz = ± log 3 = ± ln 3 + 2pπi, p ∈ Z. Summing up the equation has the solutions ⎧ π ⎪ + p1 π, ⎪ ⎪ 2 ⎪ ⎪ ⎨ p1 , p2 , p3 ∈ Z. 13 Find all solutions of the equation sin z = i, in the form x + iy. We get by using the deﬁnition of sin z, i = sin z = 1 1 eiz − e−iz = e2iz − 1 . 2i 2i eiz If this is multiplied by 2i eiz = 0, then e2iz − 1 = −2 eiz , hence by and addition of 2eiz + 2 and by interchanging the two sides of the equation 2 = eiz 2 + 2 eiz + 1 = eiz + 1 thus iz e √ = −1 ± 2 = 2 , √ √ 2 − 1 > 0, −( 2 + 1) < 0.

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