UC知道复数和微积分的题目。。。救命啊
来源：百度知道 编辑：UC知道 时间：2024/07/02 08:40:31
1.
express the roots of the equation z^11=1 in any form.
Sketch the locus of P given by z+z*+1<=0 on an Argand diagram and determine the number of roots of z^11=1 that are located in the locus of P.
z and z* are conjugate complex number.(共轭)
2.
solve the equation (z-4i)^3=8i^3.
In Argand diagram, the points P1, P2, P3 represent the roots of the equation and S denotes the region for which -PI<=arg(z-4i)<=-PI/2.
Determine whether the points P1, P2, P3 lie in S.
(PI=3.1415926..大家懂的吧)
3.
（1）
Sketch， on the same diagram, the graphs of x^2+y^2=9 and y=e^(x*x/4). (上面的power读成四分之一x的平方，懂吧？？）
（2）
The finite region in the first quadrant bounded by the curves x^2+y^2=9, y=e^(x*x/4), the x-axis and the y-axis id denoted by R.
(a) Shade the region R.
(b) Find the volume of the solid of revolution formed when R id totated through 2*PI radians about the x-axis.
express the roots of the equation z^11=1 in any form.
Sketch the locus of P given by z+z*+1<=0 on an Argand diagram and determine the number of roots of z^11=1 that are located in the locus of P.
z and z* are conjugate complex number.(共轭)
2.
solve the equation (z-4i)^3=8i^3.
In Argand diagram, the points P1, P2, P3 represent the roots of the equation and S denotes the region for which -PI<=arg(z-4i)<=-PI/2.
Determine whether the points P1, P2, P3 lie in S.
(PI=3.1415926..大家懂的吧)
3.
（1）
Sketch， on the same diagram, the graphs of x^2+y^2=9 and y=e^(x*x/4). (上面的power读成四分之一x的平方，懂吧？？）
（2）
The finite region in the first quadrant bounded by the curves x^2+y^2=9, y=e^(x*x/4), the x-axis and the y-axis id denoted by R.
(a) Shade the region R.
(b) Find the volume of the solid of revolution formed when R id totated through 2*PI radians about the x-axis.
建议您分开提问，太多了
1.
设z=a+bi则z*=z-bi
z+z*+1=2a+1<=0
a<=-0.5
z^11=1的解，就是把复数坐标系（Argand diagram ）的单位圆11等分的点，共11个，其中一个在(1,0)
找出其横坐标在-0.5左边的点
有4个解
-cos(3/11*pi)+i*sin(3/11*pi)
-cos(1/11*pi)+i*sin(1/11*pi)
-cos(1/11*pi)-i*sin(1/11*pi)
-cos(3/11*pi)-i*sin(3/11*pi)
2.
我用相量表示复数，即它的模和相角
8i^3=(2^3)[270度]
则(8i^3)^1/3有3个分别为 2[90度]、2[90-120度=-30度]、2[90-240度=-150度]
其中只有第3个，-150度在要求的S域中
表示为 三角 2*(cos(-5/6*pi)+i*sin(-5/6*pi))
原方程只有一个解满足要求： z=2*(cos(-5/6*pi)+i*sin(-5/6*pi)) +4i
3.
如图
http://i.namipan.com/files/3a790ade3ad329094d9707e4bb8ccc64b272cb13d13800006515/0/untitled.jpg
两条曲线围成的阴影面积要求个积分，然后再旋转一周积分
我积分忘了，不能帮你解决了，抱歉
中文的我会做 ，英文懒得看..