本次英国代写主要为Complex Analysis 的限时测试

2。
以下是关于复杂差异的以下哪一项陈述:
功能的能力
f(z)= z + jzj
2个
+ iz
是真的?
f(z)在0 2 C为全纯
f(z)在i 2 C时是可微且全纯的
f(z)在0 2 C时是可微的,但在0 2 C时不是全纯的
f(z)在i 2 C时是可微的,但在任何地方都不是全纯的
在C中
f(z)在C中的任何地方均不可微分
没有其他

3。
设等高线(t)= eit
; 0 t 2,对设备进行参数设置
圆圈。通过评估轮廓积分
ž

1个
z2 + 3z + 1
dz
有两种不同的方式,即真正的积分
Z 2
0
1个
3 + 2吨
dt
被视为等于以下哪个值?
4页
7
2个
3
2页
没有其他

请考虑以下语句:
(i)有一个全纯函数f:C! C(Re(f(z))= x2

(ii)如果g:C! C在C上是全纯的,然后f(z)= jzj
2个
g(z)是
在z = 0且g的每个零处均成立。
(iii)C上有一个全纯函数,使得f(
n + 1
n)= 0
对于n = 1; 2; :::并且f(0)= 1。
(iv)连续函数f:Cnf0g!如果Cnf0g C是全纯的
并且只有在
[R
对于Cnf0g中的每个闭合轮廓,f(z)dz = 0
下列哪一项是正确的:
(i)
(ii)
(iii)
(iv)
没有其

2.
Which one of the following statements about the complex di erentia-
bility of the function
f(z) = z + jzj
2
+ iz
is true?
 f(z) is holomorphic at 0 2 C
 f(z) is di erentiable and holomorphic at i 2 C
 f(z) is di erentiable at 0 2 C but is not holomorphic at 0 2 C
 f(z) is di erentiable at i 2 C but is not holomorphic anywhere
in C
 f(z) is not di erentiable anywhere in C
 None of the others

3.
Let be the contour (t) = eit
; 0  t  2, parametrising the unit
circle. By evaluating the contour integral
Z

1
z2 + 3z + 1
dz
in two di erent ways, the real integral
Z 2
0
1
3 + 2 cos t
dt
is seen to be equal to which one of the following values?
 4 p7
  p7
 2
  p3
 2 p5
 None of the others

Consider the following statements:
(i) There is a holomorphic function f : C ! C with Re(f(z)) = x2
.
(ii) If g : C ! C is holomorphic on C then f(z) = jzj
2
g(z) is di eren-
tiable at z = 0 and at each zero of g.
(iii) There exists a holomorphic function on C such that f(
n+1
n ) = 0
for n = 1; 2; : : : and f(0) = 1.
(iv) A continuous function f : Cnf0g ! C is holomorphic on Cnf0g if
and only if
R
f(z) dz = 0 for every closed contour in Cnf0g
Which one of the statements holds true:
 (i)
 (ii)
 (iii)
 (iv)
 None of the others


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