Mixing
the point $z =0$ is choose the correct option
$begingroup$
for the function $$f(z) = sinleft(frac{1}{cosfrac{1}{z}}right)$$, the point $z =0$ is
choose the correct option
$1)$ removable singularity
$2) $ pole
$3)$ an essential singularity
$4)$ a non-isolated singularity
My attempt : option 4) will correct i thinks its will have non-isolated singularity because limit of singularity are $dfrac{2}{pi}, dfrac{2}{3pi}, dfrac{2}{5pi}, ldots$.
is its true ?
complex-analysis
edited Jan 1 at 20:39
Dando18
4,66741235
asked Jan 1 at 20:02
jasminejasmine
1,650416
$endgroup$
add a comment |
$begingroup$
for the function $$f(z) = sinleft(frac{1}{cosfrac{1}{z}}right)$$, the point $z =0$ is
choose the correct option
$1)$ removable singularity
$2) $ pole
$3)$ an essential singularity
$4)$ a non-isolated singularity
My attempt : option 4) will correct i thinks its will have non-isolated singularity because limit of singularity are $dfrac{2}{pi}, dfrac{2}{3pi}, dfrac{2}{5pi}, ldots$.
is its true ?
complex-analysis
edited Jan 1 at 20:39
Dando18
4,66741235
asked Jan 1 at 20:02
jasminejasmine
1,650416
$endgroup$
1
$begingroup$
Correct answer. Indeed $z=0$ is not an isolated singularity.
$endgroup$
– Yiorgos S. Smyrlis
Jan 1 at 20:43
$begingroup$
@YiorgosS.Smyrlis thanks'
$endgroup$
– jasmine
Jan 1 at 21:08
add a comment |
$begingroup$
for the function $$f(z) = sinleft(frac{1}{cosfrac{1}{z}}right)$$, the point $z =0$ is
choose the correct option
$1)$ removable singularity
$2) $ pole
$3)$ an essential singularity
$4)$ a non-isolated singularity
My attempt : option 4) will correct i thinks its will have non-isolated singularity because limit of singularity are $dfrac{2}{pi}, dfrac{2}{3pi}, dfrac{2}{5pi}, ldots$.
is its true ?
complex-analysis
edited Jan 1 at 20:39
Dando18
4,66741235
asked Jan 1 at 20:02
jasminejasmine
1,650416
$endgroup$
for the function $$f(z) = sinleft(frac{1}{cosfrac{1}{z}}right)$$, the point $z =0$ is
choose the correct option
$1)$ removable singularity
$2) $ pole
$3)$ an essential singularity
$4)$ a non-isolated singularity
My attempt : option 4) will correct i thinks its will have non-isolated singularity because limit of singularity are $dfrac{2}{pi}, dfrac{2}{3pi}, dfrac{2}{5pi}, ldots$.
is its true ?
complex-analysis
complex-analysis
edited Jan 1 at 20:39
Dando18
4,66741235
asked Jan 1 at 20:02
jasminejasmine
1,650416
edited Jan 1 at 20:39
Dando18
4,66741235
asked Jan 1 at 20:02
jasminejasmine
1,650416
edited Jan 1 at 20:39
Dando18
4,66741235
edited Jan 1 at 20:39
Dando18
4,66741235
edited Jan 1 at 20:39
Dando18
4,66741235
4,66741235
asked Jan 1 at 20:02
jasminejasmine
1,650416
asked Jan 1 at 20:02
jasminejasmine
1,650416
asked Jan 1 at 20:02
jasminejasmine
1,650416
1,650416
1
$begingroup$
Correct answer. Indeed $z=0$ is not an isolated singularity.
$endgroup$
– Yiorgos S. Smyrlis
Jan 1 at 20:43
$begingroup$
@YiorgosS.Smyrlis thanks'
$endgroup$
– jasmine
Jan 1 at 21:08
add a comment |
1
$begingroup$
Correct answer. Indeed $z=0$ is not an isolated singularity.
$endgroup$
– Yiorgos S. Smyrlis
Jan 1 at 20:43
$begingroup$
@YiorgosS.Smyrlis thanks'
$endgroup$
– jasmine
Jan 1 at 21:08
1
1
$begingroup$
Correct answer. Indeed $z=0$ is not an isolated singularity.
$endgroup$
– Yiorgos S. Smyrlis
Jan 1 at 20:43
$begingroup$
Correct answer. Indeed $z=0$ is not an isolated singularity.
$endgroup$
– Yiorgos S. Smyrlis
Jan 1 at 20:43
$begingroup$
@YiorgosS.Smyrlis thanks'
$endgroup$
– jasmine
Jan 1 at 21:08
$begingroup$
@YiorgosS.Smyrlis thanks'
$endgroup$
– jasmine
Jan 1 at 21:08
add a comment |
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$begingroup$
Correct answer. Indeed $z=0$ is not an isolated singularity.
$endgroup$
– Yiorgos S. Smyrlis
Jan 1 at 20:43
$begingroup$
@YiorgosS.Smyrlis thanks'
$endgroup$
– jasmine
Jan 1 at 21:08