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:
功能的能力
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
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)= 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
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 Z1z2 + 3z + 1dz
in two di erent ways, the real integral Z 2013 + 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


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