The complex function zsin (1 / z) has a limit when Z tends to 0. According to the theorem, 0 should be the removable pole of the function, which is expanded into a series with infinite negative terms 0 is its de - extreme or its essential extreme

The complex function zsin (1 / z) has a limit when Z tends to 0. According to the theorem, 0 should be the removable pole of the function, which is expanded into a series with infinite negative terms 0 is its de - extreme or its essential extreme

The limit of zsin (1 / z) in a complex variable is not zero. It is proved that
Let z = x + Yi
2. Z tends to zero in the direction of x = 0 and Y → 0
lim zsin(1/z)
=lim yisin(1/(yi))
=lim yi*(e^(1/y)-e^(-1/y))/2
It doesn't exist
That is to say, it's not going to the extreme

Finding the limit of complex variable function zre (z) / | Z |
Z tends to zero

When Z → 0, | Z | → 0, so
The modulus of is 0, so
→0

Let z = 0 be the zeros of order m of the function Z ^ 3-sin (Z ^ 3), then M =?
It's better to have a solution process to let me know how to do it? 、O(∩_ Thank you

z³-sinz³=z³-[z³-(z³)³/3!+(z³)^5/5!-...]=z^9/3!+z^15/5!-...
So m = 9

Transfer function of differential equation
How to write the transfer function from a differential equation, such as K-Y '- KY' '= my
The first k is wrong, f is the input, for example: Y (T) + μ y '(T) + KY' '(T) = f (T)

What about input?
Supplementary answer:
0 initial condition
Two sided Laplace transform
Y(s)+μ sY(s)+ks^2Y(s)=F(s)
Transfer function y (s) / F (s) = 1 / (KS ^ 2 + μ S + 1)
It's a second order system