Mixing
What precisely does the notation $frac {partial h} {partial z}$ mean in this context?
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I am given the following problem.
Let $Ωsubset mathbb{C} $ and let $h in C^1(Ω)$ be such that $frac {partial h} {partial z}=0$. Show that $h(z)=overline {f(z)}$ for some $f$ analytic in $Ω$.
The $frac {partial h} {partial z}$ notation is mysterious to me. But, after looking at this, I have a guess as to what it means.
Write $h(x+iy)=u(x,y)+iv(x,y).$ Then $frac {partial h} {partial z}=frac {[u_x+iv_x]-i[u_y+iv_y]} 2=frac {u_x+v_y+iv_x-iu_y} 2$. Is this interpretation correct?
complex-analysis derivatives notation partial-derivative
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
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add a comment |
$begingroup$
I am given the following problem.
Let $Ωsubset mathbb{C} $ and let $h in C^1(Ω)$ be such that $frac {partial h} {partial z}=0$. Show that $h(z)=overline {f(z)}$ for some $f$ analytic in $Ω$.
The $frac {partial h} {partial z}$ notation is mysterious to me. But, after looking at this, I have a guess as to what it means.
Write $h(x+iy)=u(x,y)+iv(x,y).$ Then $frac {partial h} {partial z}=frac {[u_x+iv_x]-i[u_y+iv_y]} 2=frac {u_x+v_y+iv_x-iu_y} 2$. Is this interpretation correct?
complex-analysis derivatives notation partial-derivative
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
$endgroup$
$begingroup$
It's just the ordinary derivative $frac{partial}{partial z} = frac{1}{2}left(frac{partial}{partial x} - frac{partial}{partial y}right)$.
$endgroup$
– anomaly
Jan 24 at 23:16
add a comment |
$begingroup$
I am given the following problem.
Let $Ωsubset mathbb{C} $ and let $h in C^1(Ω)$ be such that $frac {partial h} {partial z}=0$. Show that $h(z)=overline {f(z)}$ for some $f$ analytic in $Ω$.
The $frac {partial h} {partial z}$ notation is mysterious to me. But, after looking at this, I have a guess as to what it means.
Write $h(x+iy)=u(x,y)+iv(x,y).$ Then $frac {partial h} {partial z}=frac {[u_x+iv_x]-i[u_y+iv_y]} 2=frac {u_x+v_y+iv_x-iu_y} 2$. Is this interpretation correct?
complex-analysis derivatives notation partial-derivative
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
$endgroup$
I am given the following problem.
Let $Ωsubset mathbb{C} $ and let $h in C^1(Ω)$ be such that $frac {partial h} {partial z}=0$. Show that $h(z)=overline {f(z)}$ for some $f$ analytic in $Ω$.
The $frac {partial h} {partial z}$ notation is mysterious to me. But, after looking at this, I have a guess as to what it means.
Write $h(x+iy)=u(x,y)+iv(x,y).$ Then $frac {partial h} {partial z}=frac {[u_x+iv_x]-i[u_y+iv_y]} 2=frac {u_x+v_y+iv_x-iu_y} 2$. Is this interpretation correct?
complex-analysis derivatives notation partial-derivative
complex-analysis derivatives notation partial-derivative
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
asked Jan 24 at 23:01
Pascal's WagerPascal's Wager
371315
371315
$begingroup$
It's just the ordinary derivative $frac{partial}{partial z} = frac{1}{2}left(frac{partial}{partial x} - frac{partial}{partial y}right)$.
$endgroup$
– anomaly
Jan 24 at 23:16
add a comment |
$begingroup$
It's just the ordinary derivative $frac{partial}{partial z} = frac{1}{2}left(frac{partial}{partial x} - frac{partial}{partial y}right)$.
$endgroup$
– anomaly
Jan 24 at 23:16
$begingroup$
It's just the ordinary derivative $frac{partial}{partial z} = frac{1}{2}left(frac{partial}{partial x} - frac{partial}{partial y}right)$.
$endgroup$
– anomaly
Jan 24 at 23:16
$begingroup$
It's just the ordinary derivative $frac{partial}{partial z} = frac{1}{2}left(frac{partial}{partial x} - frac{partial}{partial y}right)$.
$endgroup$
– anomaly
Jan 24 at 23:16
add a comment |
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$begingroup$
It's just the ordinary derivative $frac{partial}{partial z} = frac{1}{2}left(frac{partial}{partial x} - frac{partial}{partial y}right)$.
$endgroup$
– anomaly
Jan 24 at 23:16