Question

all of q1 please, a complex analysis question for complex numbers etc.

1. (a) Define the principal branch of Log(2). Find Log(1 + V3i). [6 marks] (b) Find all solutions to ex-1 = -ie3. (6 marks) (c) Find all solutions to 25 = 1+i. (8 marks) (d) Describe the image of the circle |z| = 5 under the mapping f(x) = Log(2). [6 marks]

Answer 1

(a) Prisciple branch of _Log(z) z can be coritten in polar form as follous 2 = 8.610 Log (2) = log (8.60) logy + loge = logotie tet If z= xriy then Z- s. 82742 - tan I

Log (1 + isi) – log($1 +(372) + i tan (1) og J7 +3 + it 2 log St & in = log 2 + 1/3 20. 3010 + 11-047 rad b) ez-, -ie he know, i = cos(-71) + i sint 12). Here whee n = 1, 2, 3 .. fece 8 = 1 2- E - int 3 = e .es -13n7

2-1 = - i 307 2 2 - I - i 3n1 where n=1,2,3. we knovo, iti = 12(t+ita) 12 (603 + i sof). here, 8=82; 0 2 2 Iti = √2.e 10 228.000 5 510 8.e = 12e 14 150 ina = 210 50 2 1 027/20

Parameterizing the boundaries as z(t) usually x(t) and y(t).

P
10 zaio til Yo e 1920 d) the giver The given circle, 12425 Radius of the giver circle is s. Boundaries of circle are x(+) 5 cost y(t) = 5 sint 0 <tL2R. where 2 atly

Given function f(x) = logle) f(x(i)) = log (scost tissist) - log5 + log (cost + i sint) = logs + log (1. et = 0.698 + it where os+220. This is function f(e) at the Boundary. 1095/0.698/ The above graph is image of circle 12/25 under mapping of f(2) = log(z).