The continuity of $f(z)=left{begin{smallmatrix}frac1{|z|^2}(operatorname{Re}z)^2(operatorname{Im}z);&zne...
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It is an advance complex question… Need you suggestion kindly help me to describe the continuity of this $f(z)$ at all point of $mathbb C$.
complex-analysis continuity
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It is an advance complex question… Need you suggestion kindly help me to describe the continuity of this $f(z)$ at all point of $mathbb C$.
complex-analysis continuity
I've edited your title to look more readable (although not perfect). You can edit your question (click "edit" just below it to do so) to actaully write out the function in the question itself, and describe your own work so far on the problem, so that we know how much you know, and can be more helpful.
– John Hughes
Nov 26 at 1:59
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up vote
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up vote
-1
down vote
favorite
It is an advance complex question… Need you suggestion kindly help me to describe the continuity of this $f(z)$ at all point of $mathbb C$.
complex-analysis continuity
It is an advance complex question… Need you suggestion kindly help me to describe the continuity of this $f(z)$ at all point of $mathbb C$.
complex-analysis continuity
complex-analysis continuity
edited Nov 26 at 2:19
user302797
19.4k92251
19.4k92251
asked Nov 26 at 1:55
Fiza Irshad
62
62
I've edited your title to look more readable (although not perfect). You can edit your question (click "edit" just below it to do so) to actaully write out the function in the question itself, and describe your own work so far on the problem, so that we know how much you know, and can be more helpful.
– John Hughes
Nov 26 at 1:59
add a comment |
I've edited your title to look more readable (although not perfect). You can edit your question (click "edit" just below it to do so) to actaully write out the function in the question itself, and describe your own work so far on the problem, so that we know how much you know, and can be more helpful.
– John Hughes
Nov 26 at 1:59
I've edited your title to look more readable (although not perfect). You can edit your question (click "edit" just below it to do so) to actaully write out the function in the question itself, and describe your own work so far on the problem, so that we know how much you know, and can be more helpful.
– John Hughes
Nov 26 at 1:59
I've edited your title to look more readable (although not perfect). You can edit your question (click "edit" just below it to do so) to actaully write out the function in the question itself, and describe your own work so far on the problem, so that we know how much you know, and can be more helpful.
– John Hughes
Nov 26 at 1:59
add a comment |
1 Answer
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Write $z=x+iy$ and so the top part of the equation becomes $f(x+iy)=frac{x^2y}{x^2+y^2}$. Can you use this to determine the continuity of $f$ at $z=0$? i.e. when $(x,y)=(0,0)$?
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Write $z=x+iy$ and so the top part of the equation becomes $f(x+iy)=frac{x^2y}{x^2+y^2}$. Can you use this to determine the continuity of $f$ at $z=0$? i.e. when $(x,y)=(0,0)$?
add a comment |
up vote
0
down vote
accepted
Write $z=x+iy$ and so the top part of the equation becomes $f(x+iy)=frac{x^2y}{x^2+y^2}$. Can you use this to determine the continuity of $f$ at $z=0$? i.e. when $(x,y)=(0,0)$?
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Write $z=x+iy$ and so the top part of the equation becomes $f(x+iy)=frac{x^2y}{x^2+y^2}$. Can you use this to determine the continuity of $f$ at $z=0$? i.e. when $(x,y)=(0,0)$?
Write $z=x+iy$ and so the top part of the equation becomes $f(x+iy)=frac{x^2y}{x^2+y^2}$. Can you use this to determine the continuity of $f$ at $z=0$? i.e. when $(x,y)=(0,0)$?
answered Nov 26 at 2:45
thedilated
1,0491615
1,0491615
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I've edited your title to look more readable (although not perfect). You can edit your question (click "edit" just below it to do so) to actaully write out the function in the question itself, and describe your own work so far on the problem, so that we know how much you know, and can be more helpful.
– John Hughes
Nov 26 at 1:59