Mixing
which kind of singularity has $f(z) = frac{1}{z} + e^{frac{1}{z}}$?
$begingroup$
I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.
I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.
complex-analysis
asked Jan 23 at 13:22
Simon GreenSimon Green
845
$endgroup$
add a comment |
$begingroup$
I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.
I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.
complex-analysis
asked Jan 23 at 13:22
Simon GreenSimon Green
845
$endgroup$
add a comment |
$begingroup$
I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.
I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.
complex-analysis
asked Jan 23 at 13:22
Simon GreenSimon Green
845
$endgroup$
I wondering which kind of sigularity point has $f(z) = frac{1}{z} + e^{frac{1}{z}}$.
I know that $e^{frac{1}{z}}$ has essential singularity in 0, and $1/z$ has a pole in same point, but I can't find an easy way to found out $f$'s singularity kind in 0.
complex-analysis
complex-analysis
asked Jan 23 at 13:22
Simon GreenSimon Green
845
asked Jan 23 at 13:22
Simon GreenSimon Green
845
asked Jan 23 at 13:22
Simon GreenSimon Green
845
asked Jan 23 at 13:22
Simon GreenSimon Green
845
asked Jan 23 at 13:22
Simon GreenSimon Green
845
845
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
Another possible answer to my question:
The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
implies:
$frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.
answered Jan 28 at 13:12
Simon GreenSimon Green
845
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
$endgroup$
Since the principal part of the Laurent series at $z= 0$ has infinitely many (non-zero) members, the singularity is essential.
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
edited Jan 23 at 15:05
David C. Ullrich
61.2k43994
61.2k43994
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
answered Jan 23 at 13:27
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
$begingroup$
Another possible answer to my question:
The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
implies:
$frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.
answered Jan 28 at 13:12
Simon GreenSimon Green
845
$endgroup$
add a comment |
$begingroup$
Another possible answer to my question:
The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
implies:
$frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.
answered Jan 28 at 13:12
Simon GreenSimon Green
845
$endgroup$
add a comment |
$begingroup$
Another possible answer to my question:
The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
implies:
$frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.
answered Jan 28 at 13:12
Simon GreenSimon Green
845
$endgroup$
Another possible answer to my question:
The fact that the meromorphic functions are field, and $frac{1}{z}$ meromorphic,
implies:
$frac{1}{z} + e^{frac{1}{z}}$ meromorphic $Rightarrow ((frac{1}{z} + e^{frac{1}{z}}) - frac{1}{z})$ meromorphic $Rightarrow e^{frac{1}{z}}$ meromorphic, contradiction.
answered Jan 28 at 13:12
Simon GreenSimon Green
845
answered Jan 28 at 13:12
Simon GreenSimon Green
845
answered Jan 28 at 13:12
Simon GreenSimon Green
845
answered Jan 28 at 13:12
Simon GreenSimon Green
845
845
add a comment |
add a comment |
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