- #1
chaoseverlasting
- 1,050
- 3
[SOLVED] Complex Numbers
A very happy new year to all at PF.
Kreyszig, P.665 section 12.3,
A function f(z) is said to have the limit l as z approaches a point z_0 if f is defined in the neighborhood of z_0 (except perhaps at z_0 itself) and if the values of f are "close" to l for all z "close" to z_0; that is, in precise terms, for every positive real \epsilon we can find a positive real \delta such that for all z not equal to z_0 in the disk |z-z_0|<\delta, we have
|f(z)-l|<\epsilon...(2)
that is, for every z not equal to z_0 in that \delta disk, the value of f lies in the disk (2).
I think this means that if you were to plot z and f(z), for all values of z near the point z_0 (within the \delta disk), but not necessarily at z_0 itself (as the function may not exist at z_0), the value of f(z) would be very close to l (but not necessarily l, as f(z_0) may not exist) and that this value of f(z) would be inside the \epsilon disk.
Is that right?
A very happy new year to all at PF.
Homework Statement
Kreyszig, P.665 section 12.3,
A function f(z) is said to have the limit l as z approaches a point z_0 if f is defined in the neighborhood of z_0 (except perhaps at z_0 itself) and if the values of f are "close" to l for all z "close" to z_0; that is, in precise terms, for every positive real \epsilon we can find a positive real \delta such that for all z not equal to z_0 in the disk |z-z_0|<\delta, we have
|f(z)-l|<\epsilon...(2)
that is, for every z not equal to z_0 in that \delta disk, the value of f lies in the disk (2).
I think this means that if you were to plot z and f(z), for all values of z near the point z_0 (within the \delta disk), but not necessarily at z_0 itself (as the function may not exist at z_0), the value of f(z) would be very close to l (but not necessarily l, as f(z_0) may not exist) and that this value of f(z) would be inside the \epsilon disk.
Is that right?