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Tutorial including questions from complex analysis
Typology: Exercises
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(a)
∫ C 0
dz z − z 0 = 2πi; (b)
∫ C 0
(z − z 0 )n−^1 dz = 0 (n = ± 1 , ± 2 , · · ·).
f (z) dz = 0
when the contour C is the circle |z| = 1, in either direction, and when
(a) f (z) = z
2 z − 3
; (b) f (z) = ze−z^ ; (c) f (z) = 1 z^2 + 2z + 2
(d) f (z) = sech z; (e) f (z) = tan z; (f) f (z) = log(z + 2).
f (z) =
reiθ/^2
( r > 0 ,
−π 2 < θ <^
3 π 2
)
of the multiple-valued function z^1 /^2. Show that ∫ C
f (z)dz = 0
by evaluating separately the integrals of f (z) over the semicircle and the two radii which make up C. Why does the Cauchy-Goursat theorem not apply here?
(a)
∫
C
e−z^ dz z − (πi/2) ; (b)
∫
C
cos z z(z^2 + 8)
dz;
(c)
∫ C
z dz 2 z + 1
(d)
∫
C
cosh z z^4
dz;
(e)
∫
C
tan(z/2) (z − x 0 )^2 dz^ (−^2 < x^0 <^ 2).
g(w) =
∫
C
2 z^2 − z − 2 z − w dz^ (|w| 6^ = 3), then g(2) = 8πi. What is the value of g(w) when |w| > 3?
1
C
f ′(z) dz z − z 0 =
∫
C
f (z) dz (z − z 0 )^2.
g(z) = 1 2 πi
∫
C
f (s) ds s − z is analytic at each point z interior to C and that
g′(z) = 1 2 πi
∫ C
f (s) ds (s − z)^2 at such a point.
z cosh(z^2 ) =
∑^ ∞ n=
z^4 n+ (2n)! (|z|^ <^ ∞).
∑^ ∞ n=
(z − 1)n n! (|z^ −^1 |^ <^ ∞) for the function f (z) = ez^ by (a) using f (n)(1) (n = 0, 1 , 2 , · · ·); (b) writing ez^ = ez−^1 e.
4 z +
∑^ ∞ n=
zn 4 n+^.
f (z) = (^) z (^2) (1^1 − z) ,
and specify the regions in which those expansions are valid.
z (z − 1)(z − 3)
∑^ ∞ n=
(z − 1)n 2 n+^
2(z − 1)
f (z) =
z(1 + z^2 ) in certain domains, and specify those domains.