## 本次英国作业案例主要为Complex Analysis 的数学代写限时测试

2.以下是关于复杂差异的以下哪一项陈述：

f（z）= z + jzj2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ž1个
z2 + 3z + 1
dz

Z 2
0
1个
3 + 2吨
dt

4页
7
2个
3
2页

（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）= jzj2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 + 1n）= 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

（i）
（ii）
（iii）
（iv）

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 Z1z2 + 3z + 1dz
in two di erent ways, the real integral Z 2013 + 2 cos tdt 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