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Unformatted text preview: Math 122A Midterm Winter 2009 1311111; Your Perm Number Name Solution Total Score: 50
Your Scores: 1. o 2‘ . 3 4. 5 o. Your Total ., . . 1 (10 13138) Find all four roots of —1 + ﬁi Locate these poinis on the complex plane Soiution:
m1+ x/Ei = 2e?“ (~1+\/§2‘)ﬁ are We’(2'7"+2k”)/4,wit11k20,1,2 ,3‘
Fork=0 Cg—fee—f(ooss+zsm5) \/:(§+2'§),
k=101= We“; =\/:Z(§ 7%)={V§(—§+i‘—/23),
wwopwfe“ H) T)m{‘/§(‘g_+i§), w303m {Viao M/ﬁ(%~<§) 2. (10 pts) a) Describe the image under the mapping w z 23 of" the region
{I31 < 1/2, 0 < Arg(z) < 7r/4}. Then Sketch it on the complex plane;
1:) Sketch the following subset of the compiex plane: {2 2 12 m 22'} 2 Isl} Solution: 8.) In exponentia} form, 2: = rem, then 23 2 T3639
The domain has 0 < r < ":"1 O < 6%, so the image has 1 3 W 1 3:
O<T<(§) mg, O<g<ﬁr~ b) Let’s look at the equivalent inequality: 12: m 2i2 2 z3 This is equivalent to ISC + (y  2W 21$ + yil2 3 (10 pts) Using Cauchy—Riemann equations, determine where the derivative of
f(3) = {232 — My + 1)2 exists. Then compute f’(2:) at those points Solution: We have u. = 332, v = —(y + 1F. Take partial derivative gives
um = ‘22:, my 2 0, v; =2 0, v1, = —2(y +1) These are a1} continuous, so derivative of f(z) exists when Cauchy—Riemann equations n1. = vy, try = —vm are satisﬁed. Namely when 29: = —2(y —i— 1), or a: = —y m 1“
Therefore f’(z) exists when x = my — 17 And f’(z) um um + tum = 2m at these
points. 4. (10 pts) Let ﬁe) be an analytic function in a connected open set A, and the
argument of _f(z) is a constant. Show that f is a constant Proof: Write Hz) = u(:r:, y)+z'v($, y). ~We have tan(arg(f(3))) 2 11/11. If u = 0 at
a point, then u _=. O on .4 since argument of Hz) is constant. In this case Hz) equals
to some pure imaginary constant Now we can assume it % 0 at all points Since argument of ﬂz) is constant,
11/11. M c, where c is some constant Since a % 0, we have “u = cu. Hence Um m cum, vy 2 any On the other hand} by Cauchy—Riemann equations, um = vy, uy 2 ——v1.
Combine these we have vi = wcumuy, of, = cuzuy Therefore "03 + v: = 0, which gives Um = Dy 2 0. By CauchyRiemann equations, or w my m 0. We have ﬁe) is a
constant 5‘ (10 pts) Show that u(:1:, y) m $(1 ~ y) is harmonic and ﬁnd a harmonic conju~
gate of n. Solution: We have 15m x 1 — y, um m O, 11.1, m ——$, uyy m 0 Hence um + uyy “m” 0
and u, is harmonic Let v(:I:, y) be a harmonic congngate of u Then vy = um mm 11; and Um m ~uy m 51:,
integrate um = :1: with respect to 2: gives 1 v = y + my). This gives Dy 2 <p’(y)i Combining by above, we have g0’(y) = 1 — y) 80 (My) =
 §y2 + c. We have u = ”@4122 w éyg ~+~ y is a harmonic conjugate of us Bonus Problem: (5 pts) Suppose that f : (C me (C is continuous and that
f(z) m f(2z) for all z E (C. Show that f is constant on {C Proof: Since f(z) m “23), we have f(
any fixed 23 E (C, n. a positive integer, z) m f(z) for all z E (C. This gives, for [Gil— 1 1 1 fizo) ” “520) = “320) = = f(—30)‘ Since f(z) is continuous, 21m nizo m n0) nmoo 272 Therefore f(3()) = f(0) for any 25 E (C and f(z) is a constant on (C. ...
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 Fall '07
 RickRugangYe

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