Find the maximum of the $| left( w + 2 right) ^3 left( w - 3 right)^2|$ with $|w|=1$
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Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$
Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$
Let $w=cos x+i sin x$. Then we have an ugly form
complex-numbers
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up vote
3
down vote
favorite
Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$
Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$
Let $w=cos x+i sin x$. Then we have an ugly form
complex-numbers
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$
Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$
Let $w=cos x+i sin x$. Then we have an ugly form
complex-numbers
Let $w in mathbbC$, and $left | w right | = 1$. Find the maximum of the function $| left( w + 2 right) ^3 left( w - 3 right)^2|$
Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$
Let $w=cos x+i sin x$. Then we have an ugly form
complex-numbers
edited Aug 14 at 13:49
z100
48848
48848
asked Aug 14 at 8:56
communnites
1,204431
1,204431
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add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Hint:
$$sqrt[5]w+2leqdfracw+25$$
the equality occurs if $|w+2|=|w-3|$
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up vote
0
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Calling
$$
z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
$$
so
$$
|z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
$$
now
$$
fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
$$
for $phi in 0,pmfracpi3$ so the maximum is
$$
108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
$$
for $phi = 0$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint:
$$sqrt[5]w+2leqdfracw+25$$
the equality occurs if $|w+2|=|w-3|$
add a comment |Â
up vote
4
down vote
accepted
Hint:
$$sqrt[5]w+2leqdfracw+25$$
the equality occurs if $|w+2|=|w-3|$
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint:
$$sqrt[5]w+2leqdfracw+25$$
the equality occurs if $|w+2|=|w-3|$
Hint:
$$sqrt[5]w+2leqdfracw+25$$
the equality occurs if $|w+2|=|w-3|$
edited Aug 14 at 9:06
Jack D'Aurizioâ¦
271k31266632
271k31266632
answered Aug 14 at 9:05
lab bhattacharjee
215k14152264
215k14152264
add a comment |Â
add a comment |Â
up vote
0
down vote
Calling
$$
z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
$$
so
$$
|z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
$$
now
$$
fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
$$
for $phi in 0,pmfracpi3$ so the maximum is
$$
108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
$$
for $phi = 0$
add a comment |Â
up vote
0
down vote
Calling
$$
z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
$$
so
$$
|z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
$$
now
$$
fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
$$
for $phi in 0,pmfracpi3$ so the maximum is
$$
108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
$$
for $phi = 0$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Calling
$$
z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
$$
so
$$
|z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
$$
now
$$
fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
$$
for $phi in 0,pmfracpi3$ so the maximum is
$$
108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
$$
for $phi = 0$
Calling
$$
z_1=rho_1 e^iphi_1 = sqrt(cosphi+2)^2+sin^2phi = sqrt5+4cosphi\
z_2=rho_2 e^iphi_2 = sqrt(cosphi-2)^2+sin^2phi = sqrt10-6cosphi
$$
so
$$
|z_1^3z_2^2| = rho_1^3rho_2^2 = left(5+4cosphiright)^frac 32|10-6cosphi|
$$
now
$$
fracddphileft( left(5+4cosphiright)^frac 32(10-6cosphi)right) = 6 sinphi (4 cosphi+5)^3/2-6 sinphi (10-6 cosphi) sqrt4 cosphi+5=0
$$
for $phi in 0,pmfracpi3$ so the maximum is
$$
108=max_w| left( w + 2 right) ^3 left( w - 3 right)^2|;;mboxs. t.;;|w| = 1
$$
for $phi = 0$
edited Aug 14 at 13:23
z100
48848
48848
answered Aug 14 at 13:05
Cesareo
5,8802412
5,8802412
add a comment |Â
add a comment |Â
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