Determine the branch cuts of f(z)
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Consider the function f(z) = $({z^2} + 1)^{1/2}$, that is, the principal value of the function f(z) = $({z^2} + 1)^{1/2}$.
Determine the branch cuts of f(z); check the discontinuity of the branch along the branch cuts.
How do I determine those branch cuts?
complex-analysis
asked Nov 22 at 12:49
Peter van de Berg
198
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up vote
-3
down vote
favorite
Consider the function f(z) = $({z^2} + 1)^{1/2}$, that is, the principal value of the function f(z) = $({z^2} + 1)^{1/2}$.
Determine the branch cuts of f(z); check the discontinuity of the branch along the branch cuts.
How do I determine those branch cuts?
complex-analysis
asked Nov 22 at 12:49
Peter van de Berg
198
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Consider the function f(z) = $({z^2} + 1)^{1/2}$, that is, the principal value of the function f(z) = $({z^2} + 1)^{1/2}$.
Determine the branch cuts of f(z); check the discontinuity of the branch along the branch cuts.
How do I determine those branch cuts?
complex-analysis
asked Nov 22 at 12:49
Peter van de Berg
198
Consider the function f(z) = $({z^2} + 1)^{1/2}$, that is, the principal value of the function f(z) = $({z^2} + 1)^{1/2}$.
Determine the branch cuts of f(z); check the discontinuity of the branch along the branch cuts.
How do I determine those branch cuts?
complex-analysis
complex-analysis
asked Nov 22 at 12:49
Peter van de Berg
198
asked Nov 22 at 12:49
Peter van de Berg
198
asked Nov 22 at 12:49
Peter van de Berg
198
asked Nov 22 at 12:49
Peter van de Berg
198
asked Nov 22 at 12:49
Peter van de Berg
198
198
add a comment |
add a comment |
1 Answer
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They are where $(z^2+1)$ is negative real, so the square-root has a discontinuity.
answered Nov 22 at 13:13
Empy2
33.4k12261
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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up vote
1
down vote
They are where $(z^2+1)$ is negative real, so the square-root has a discontinuity.
answered Nov 22 at 13:13
Empy2
33.4k12261
add a comment |
up vote
1
down vote
They are where $(z^2+1)$ is negative real, so the square-root has a discontinuity.
answered Nov 22 at 13:13
Empy2
33.4k12261
add a comment |
up vote
1
down vote
up vote
1
down vote
They are where $(z^2+1)$ is negative real, so the square-root has a discontinuity.
answered Nov 22 at 13:13
Empy2
33.4k12261
They are where $(z^2+1)$ is negative real, so the square-root has a discontinuity.
answered Nov 22 at 13:13
Empy2
33.4k12261
answered Nov 22 at 13:13
Empy2
33.4k12261
answered Nov 22 at 13:13
Empy2
33.4k12261
answered Nov 22 at 13:13
Empy2
33.4k12261
33.4k12261
add a comment |
add a comment |
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