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# is this function bounded watch

1. im considering the boundary of the set

B={z:im(z)<=0 and 2<=z<3} in the complex plane

is the the function

f(z)=1/(sinh(z+2)) bounded on the boundary of B?

im saying no,as the point z=-2 is on the boundary so f is not continuous on the boundary of B

in particular 1/f(z) tends to 0 as z tends to -2 so f(z) tends to infinity.

is this enough to show its not bounded ?

is there a way to show |f(x)|>M for any M directly from f(z)?

thanks
2. (Original post by mathz)
im considering the boundary of the set

B={z:im(z)<=0 and 2<=z<3} in the complex plane

is the the function

f(z)=1/(sinh(z+2)) bounded on the boundary of B?

im saying no,as the point z=-2 is on the boundary so f is not continuous on the boundary of B

in particular 1/f(z) tends to 0 as z tends to -2 so f(z) tends to infinity.

is this enough to show its not bounded ?

is there a way to show |f(x)|>M for any M directly from f(z)?

thanks
Please check your definition of B. It doesn't make sense as it stands. And if you meant B={z:im(z)<=0 and 2<=|z|<3} then f(z) isn't even defined at z=-2, since you'd have 1/0.
Edit: Or are such singularities permitted - looks like you're heading into complex analysis.
3. (Original post by ghostwalker)
Please check your definition of B. It doesn't make sense as it stands. And if you meant B={z:im(z)<=0 and 2<=|z|<3} then f(z) isn't even defined at z=-2, since you'd have 1/0.
Edit: Or are such singularities permitted - looks like you're heading into complex analysis.
sorry,my mistake

B={z:im(z)<=0 and 2<=|z-1|<3}

and yes it is complex analysis. the question just says is the function bounded on the boundary of B.

im guessing just saying its not continuous or it tends to infinity at z=-2 will suffice.
4. (Original post by mathz)
sorry,my mistake

B={z:im(z)<=0 and 2<=|z-1|<3}

and yes it is complex analysis. the question just says is the function bounded on the boundary of B.

im guessing just saying its not continuous or it tends to infinity at z=-2 will suffice.
OK. Yes, you can say that as z tends to -2, then f(z) tends to infinity....(1)

Saying it's not continuous doesn't imply it's not bounded.

I find it an odd question as f(-2) isn't actually defined, since you'd have 1 divided by 0.

Setitng that aside, (1) still holds.

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