Complex numbers and calculus problems... Help

Let z=a+bi, then z*=z-bi

z+z*+1=2a+1<=0

The solution of a<=z 11=1 is to divide the unit circle of the argand diagram into 11 equal points, a total of 11, one of which is at (1,0).

Find the point whose abscissa is to the left.

There are 4 solutions. 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)

I use phasor to represent complex numbers, i.e., its modulus and phase angle.

8i 3 = (2 3) [270 degrees].

Then (8i, 3), 1, 3, there are 3, respectively, 2[90 degrees, [90-120 degrees=-30 degrees, [90-240 degrees=-150 degrees].

Only the 3rd of these, -150 degrees is in the required s domain.

It is expressed as trigonometric 2*(cos(-5 6*pi)+i*sin(-5 6*pi)).

Only one solution of the original equation satisfies the requirement: z=2*(cos(-5 6*pi)+i*sin(-5 6*pi)) 4i

I forgot my points, I can't help you solve it, sorry.

It is advisable to ask questions separately, there are too many.

I will do it in Chinese, and I am too lazy to read English.

With the NL formula.

One of the original functions of e (-it) is f(t)=ie (-it), substituting t=+ and 0 respectively.

f(+ f(0), since when t +, -it- so ie (-it) 0, or f(+ = 0.).And f(0)=i, so the result is 0-i=-i.

Define the function f(s) = [0, ]e (-st)dt, where s cThen just calculate the integral (which is a function of s) and let s=i be the desired integral.

Since the integrand e(-st) can be seen as 1*e(-st), then define the function f(t)=1, where t 0, then f(s)= [0, ]f(t)e (-st)dt

This is the Laplace transformation of the function f(t), look up the table to get f(s)=1 s, substitute s=i into it, and get the original formula =-i

There is another way to calculate it, which is to find the original function directly, using the nl formula.

Since one of the original functions of e (-it) is f(t)=ie (-it), substituting t=+ and 0 respectively.

f(+ f(0), since when t +, -it- so ie (-it) 0, or f(+ = 0.).f(0)=i, so the result is 0-i=-i, which is the same as the result of the first method.

Hopefully, the idea of this question will help you.

With the NL formula.

One of the original functions of e (-it) is f(t)=ie (-it), substituting t=+ and 0 respectively.

f(+ f(0), since when t +, -it- so ie (-it) 0, or f(+ = 0.).And f(0)=i, so the result is 0-i=-i.

Take a closer look at complex functions, I've learned. In fact, complex transformation is equivalent to the basic operation of complex numbers plus calculus, which includes the limit, continuity, derivative, pole number and integration of complex numbers. The general idea is still about the same, such as the derivable continuous.

However, there are still many differences between complex and real fields. For example, sin x is no longer a bounded function in a complex number field, but can take all the numbers in the complex number field. complex number field.

You take a closer look at the complex.

Functions, I've learned. Complex transformation.

In fact, it is equivalent to a plural number.

The basic arithmetic plus calculus, inside from complex numbers.

There are limits, continuities, derivatives, poles, and even integrals. The general idea is still about the same, such as the derivable continuous. But in the plural.

There are still many differences between the field and the field of real numbers. For example, sin x is in the plural.

The domain is no longer bounded functions, but can be used for plurals.

All the number of domains. Plural.

domain.

It's a very important formula, it's in the book, just memorize it. It is proved as follows, so that z=z0+e(i), then z is in the circumference |z-z0|=1 on dz=ie (i )d , so integral = ie (i )d e (in )=i e [i(1-n)]d , if n = 1, integral = i d = 2 i, when n≠1, integral = i [cos(n-1) -isin(n-1) ]d = 0

1) and 2) are not different.

Cause. 1 primitive function = x -4x + c1

2 primitive function = (x-2) +c2=x -4x+4+c2c1=4+c2