Mixing
Residues of $frac{pi cot(pi z)}{(z+1)(z+2)}$
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I need to find the residue at $z= -1$ and $z=-2$ of the following function:
$$f(z) = frac{pi cot(pi z)}{(z+1)(z+2)}$$
I now that $f(z)$ has a double pole at $z=-1$ because $sin(-pi) = 0 $. Now for a double pole the residue should be:
$$limlimits_{z to -1} frac{d}{dz} frac{(z+1) pi cot(pi z)}{z+2}$$
However evaluating the differentiation of that term can be quite annoying especially if you don't remember the "formula" for $cot(z)$. Is there a more simple to evaluate this?
I know that the answer should be $-1$ (checked with Maple)
complex-analysis residue-calculus
asked Jan 16 at 0:08
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
I need to find the residue at $z= -1$ and $z=-2$ of the following function:
$$f(z) = frac{pi cot(pi z)}{(z+1)(z+2)}$$
I now that $f(z)$ has a double pole at $z=-1$ because $sin(-pi) = 0 $. Now for a double pole the residue should be:
$$limlimits_{z to -1} frac{d}{dz} frac{(z+1) pi cot(pi z)}{z+2}$$
However evaluating the differentiation of that term can be quite annoying especially if you don't remember the "formula" for $cot(z)$. Is there a more simple to evaluate this?
I know that the answer should be $-1$ (checked with Maple)
complex-analysis residue-calculus
asked Jan 16 at 0:08
daljit97daljit97
178111
$endgroup$
$begingroup$
The formula for $frac {d}{dz} cot z = -csc^2 z$ and can be derived in about 30 seconds. You could say $u = z+1, cot pi z = cot (pi u - pi) = cot (pi u) $ then use Taylor series about $u = 0$
$endgroup$
– Doug M
Jan 16 at 0:58
$begingroup$
so do you recommend just differentiate and then check the limit? Because when you differentiate you get: ${frac {cos left( pi,z right) sin left( pi,z right) -pi, left( z+2 right) left( z+1 right) }{ left( sin left( pi,z right) right) ^{2} left( z+2 right) ^{2}}}$ and then to find the limit you would have to differentiate again for the Hopital rule.
$endgroup$
– daljit97
Jan 16 at 1:15
add a comment |
$begingroup$
I need to find the residue at $z= -1$ and $z=-2$ of the following function:
$$f(z) = frac{pi cot(pi z)}{(z+1)(z+2)}$$
I now that $f(z)$ has a double pole at $z=-1$ because $sin(-pi) = 0 $. Now for a double pole the residue should be:
$$limlimits_{z to -1} frac{d}{dz} frac{(z+1) pi cot(pi z)}{z+2}$$
However evaluating the differentiation of that term can be quite annoying especially if you don't remember the "formula" for $cot(z)$. Is there a more simple to evaluate this?
I know that the answer should be $-1$ (checked with Maple)
complex-analysis residue-calculus
asked Jan 16 at 0:08
daljit97daljit97
178111
$endgroup$
I need to find the residue at $z= -1$ and $z=-2$ of the following function:
$$f(z) = frac{pi cot(pi z)}{(z+1)(z+2)}$$
I now that $f(z)$ has a double pole at $z=-1$ because $sin(-pi) = 0 $. Now for a double pole the residue should be:
$$limlimits_{z to -1} frac{d}{dz} frac{(z+1) pi cot(pi z)}{z+2}$$
However evaluating the differentiation of that term can be quite annoying especially if you don't remember the "formula" for $cot(z)$. Is there a more simple to evaluate this?
I know that the answer should be $-1$ (checked with Maple)
complex-analysis residue-calculus
complex-analysis residue-calculus
asked Jan 16 at 0:08
daljit97daljit97
178111
asked Jan 16 at 0:08
daljit97daljit97
178111
edited Jan 16 at 0:36
daljit97
asked Jan 16 at 0:08
daljit97daljit97
178111
asked Jan 16 at 0:08
daljit97daljit97
178111
asked Jan 16 at 0:08
daljit97daljit97
178111
178111
$begingroup$
The formula for $frac {d}{dz} cot z = -csc^2 z$ and can be derived in about 30 seconds. You could say $u = z+1, cot pi z = cot (pi u - pi) = cot (pi u) $ then use Taylor series about $u = 0$
$endgroup$
– Doug M
Jan 16 at 0:58
$begingroup$
so do you recommend just differentiate and then check the limit? Because when you differentiate you get: ${frac {cos left( pi,z right) sin left( pi,z right) -pi, left( z+2 right) left( z+1 right) }{ left( sin left( pi,z right) right) ^{2} left( z+2 right) ^{2}}}$ and then to find the limit you would have to differentiate again for the Hopital rule.
$endgroup$
– daljit97
Jan 16 at 1:15
add a comment |
$begingroup$
The formula for $frac {d}{dz} cot z = -csc^2 z$ and can be derived in about 30 seconds. You could say $u = z+1, cot pi z = cot (pi u - pi) = cot (pi u) $ then use Taylor series about $u = 0$
$endgroup$
– Doug M
Jan 16 at 0:58
$begingroup$
so do you recommend just differentiate and then check the limit? Because when you differentiate you get: ${frac {cos left( pi,z right) sin left( pi,z right) -pi, left( z+2 right) left( z+1 right) }{ left( sin left( pi,z right) right) ^{2} left( z+2 right) ^{2}}}$ and then to find the limit you would have to differentiate again for the Hopital rule.
$endgroup$
– daljit97
Jan 16 at 1:15
$begingroup$
The formula for $frac {d}{dz} cot z = -csc^2 z$ and can be derived in about 30 seconds. You could say $u = z+1, cot pi z = cot (pi u - pi) = cot (pi u) $ then use Taylor series about $u = 0$
$endgroup$
– Doug M
Jan 16 at 0:58
$begingroup$
The formula for $frac {d}{dz} cot z = -csc^2 z$ and can be derived in about 30 seconds. You could say $u = z+1, cot pi z = cot (pi u - pi) = cot (pi u) $ then use Taylor series about $u = 0$
$endgroup$
– Doug M
Jan 16 at 0:58
$begingroup$
so do you recommend just differentiate and then check the limit? Because when you differentiate you get: ${frac {cos left( pi,z right) sin left( pi,z right) -pi, left( z+2 right) left( z+1 right) }{ left( sin left( pi,z right) right) ^{2} left( z+2 right) ^{2}}}$ and then to find the limit you would have to differentiate again for the Hopital rule.
$endgroup$
– daljit97
Jan 16 at 1:15
$begingroup$
so do you recommend just differentiate and then check the limit? Because when you differentiate you get: ${frac {cos left( pi,z right) sin left( pi,z right) -pi, left( z+2 right) left( z+1 right) }{ left( sin left( pi,z right) right) ^{2} left( z+2 right) ^{2}}}$ and then to find the limit you would have to differentiate again for the Hopital rule.
$endgroup$
– daljit97
Jan 16 at 1:15
add a comment |
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$begingroup$
The formula for $frac {d}{dz} cot z = -csc^2 z$ and can be derived in about 30 seconds. You could say $u = z+1, cot pi z = cot (pi u - pi) = cot (pi u) $ then use Taylor series about $u = 0$
$endgroup$
– Doug M
Jan 16 at 0:58
$begingroup$
so do you recommend just differentiate and then check the limit? Because when you differentiate you get: ${frac {cos left( pi,z right) sin left( pi,z right) -pi, left( z+2 right) left( z+1 right) }{ left( sin left( pi,z right) right) ^{2} left( z+2 right) ^{2}}}$ and then to find the limit you would have to differentiate again for the Hopital rule.
$endgroup$
– daljit97
Jan 16 at 1:15