Feng Fuji s detailed explanation of the after class answers of complex variable functions and integr

Solution: (sinz) 2dz=(1-cos2z) 2dz, primitive=[z 2-(1 4)sin2z]丨(z=- i, i)= i-(1 2)sin2 i=( -sNH)i.

2e (-3z)+3cos2z]dz=d[(-2 3)e (-3z)+(3 2)sin2z)]], the original formula = [(-2 3)e (-3z)+(3 2)sin2z)]丨(z=0,i)=(2 3)[1-e (-3i)]+3 2)sin2i=(2 3)(1-cos3)+[2 3)sin3+(3 2)sinh2]i. FYI.

Complex functions are usually used as curve integrals, so the ones discussed below are also curve integrals.

1) This is a formal transformation.

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At the end of the second line of the above equation, we can see that both the real and imaginary parts of the integration result are the second type of curve integrals about the real and imaginary parts of the function, if there is a parametric equation for curve c.

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Then the above equation can be reduced to a definite integral.

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Of course, x(t) and y(t) are required to satisfy the first-order derivability.

In addition, of course, the integration of the second type of curve can be reduced to the integration of the first curve, which will not be discussed in depth.

If you want to ask what is the meaning of the integral, about the second type of curve integral, it can be understood as the work done by the variable force on the object that is doing the curvilinear motion.

The integration of the second type curve into a definite integral is to multiply the variable force by the path derivative to obtain the power, and then the power is integrated with time to obtain the work done by the variable force.

The integral of the real variable function is like this, and the integral of the complex variable function can also be understood in this way.

2) Turn leftTurn right.

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Here zk can be seen as a small segment of the curve c, then f(zk) is the "complex density" of a point on the curve, so the result of the integration can be regarded as the "complex mass" of the entire curve

3) If the integral is a plane integral or multiple integral, then it is usually about the integration of real variables, which can be regarded as the separate integration of the real and imaginary parts.

The distance between -1 and 2 is 3, and the distance between 1 and 2 is 1, so in |z-2|<5 has two singularities in the integrand-1 and 1, where.

1 is the first-order pole and -1 is the second-order pole, according to the remainder theorem:

Or use the Cauchy integral theorem of the complex line to deduce the compound closed-circuit theorem, make two curves c1 that only encompasses 1 but does not contain 1 and c2 that contains only -1 and does not contain 1, and uses the Cauchy integral formula and the higher-order derivative formula:

<> textbooks generally have similar example questions, <>

2) For the small question, let f(z)=z (z -1). In the 丨z丨=2 domain, f(z) has two first-order poles z1=1 and z2=-1.

res[f(z),z1]=lim(z→z1)[(z-z1)f(z)]=lim(z→1)[(z-1)f(z)]=1/2。Similarly, res[f(z),z2]=lim(z z2)[(z-z2)f(z)]=lim(z-1)[(z+1)f(z)]=1 2.

By Cauchy's integral theorem, the original formula = (2 i) = 2 i.

4) For small questions, let f(z)=cosz (z -4). In the "c:x +y =4x" domain, f(z) has a first-order pole z1=2.

res[f(z),z1]=lim(z→z1)[(z-z1)f(z)]=lim(z→2)[(z-2)f(z)]=cos2)/4。

From the Cauchy integral theorem, the original formula = (2 i)res[f(z),z1]=(cos2) i 2.

FYI.