Find the minimum of $|z+frac{1}{z}|$ for $|z|ge2$
$begingroup$
Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$
Attempt
By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$
$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$
$|z|-|frac{1}{z}|ge2-frac{1}{2}$
$|z|-|frac{1}{z}|gefrac{3}{2}$
so $|z+frac{1}{z}|ge frac{3}{2}$
But the answer given in my book is $-frac{1}{8}$
complex-numbers
$endgroup$
add a comment |
$begingroup$
Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$
Attempt
By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$
$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$
$|z|-|frac{1}{z}|ge2-frac{1}{2}$
$|z|-|frac{1}{z}|gefrac{3}{2}$
so $|z+frac{1}{z}|ge frac{3}{2}$
But the answer given in my book is $-frac{1}{8}$
complex-numbers
$endgroup$
1
$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01
$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01
add a comment |
$begingroup$
Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$
Attempt
By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$
$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$
$|z|-|frac{1}{z}|ge2-frac{1}{2}$
$|z|-|frac{1}{z}|gefrac{3}{2}$
so $|z+frac{1}{z}|ge frac{3}{2}$
But the answer given in my book is $-frac{1}{8}$
complex-numbers
$endgroup$
Find the least value assumed by the function $w=|z+frac{1}{z}|$ where $,|z|ge2$
Attempt
By triangular inequality,$|z+frac{1}{z}|ge ||z|-|frac{1}{z}||$
$|z|ge2......................(1)$
$|frac{1}{z}|lefrac{1}{2}$
$-|frac{1}{z}|gefrac{-1}{2}.........(2)$
$|z|-|frac{1}{z}|ge2-frac{1}{2}$
$|z|-|frac{1}{z}|gefrac{3}{2}$
so $|z+frac{1}{z}|ge frac{3}{2}$
But the answer given in my book is $-frac{1}{8}$
complex-numbers
complex-numbers
edited Dec 18 '18 at 14:58
Did
249k23226466
249k23226466
asked Dec 18 '18 at 13:58
user984325user984325
246112
246112
1
$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01
$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01
add a comment |
1
$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01
$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01
1
1
$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01
$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01
$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01
$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $r = |z| ge 2$ then
$w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.
Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get
$w ge r |1 -frac{1}{r^2}| = r - 1/r$.
Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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votes
$begingroup$
Let $r = |z| ge 2$ then
$w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.
Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get
$w ge r |1 -frac{1}{r^2}| = r - 1/r$.
Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.
$endgroup$
add a comment |
$begingroup$
Let $r = |z| ge 2$ then
$w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.
Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get
$w ge r |1 -frac{1}{r^2}| = r - 1/r$.
Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.
$endgroup$
add a comment |
$begingroup$
Let $r = |z| ge 2$ then
$w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.
Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get
$w ge r |1 -frac{1}{r^2}| = r - 1/r$.
Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.
$endgroup$
Let $r = |z| ge 2$ then
$w=|z+frac{1}{z}| = |r e^{i phi}+frac{1}{r} e^{-i phi}|= r |1 +frac{1}{r^2} e^{-2 i phi}|$.
Now the minimum w.r.t. $phi$ is obtained for $phi = pi/2$ and we get
$w ge r |1 -frac{1}{r^2}| = r - 1/r$.
Now for $r ge 2$, the last expression takes its lowest value at the boundary, i.e. at $r=2$, which gives $3/2$. This is tight, since $z = 2 i $ gives exactly that result.
edited Dec 19 '18 at 7:36
answered Dec 18 '18 at 14:44
AndreasAndreas
8,3961137
8,3961137
add a comment |
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1
$begingroup$
$-frac 18$ or $+frac 18$? Absolute value cannot be negative.
$endgroup$
– user376343
Dec 18 '18 at 14:01
$begingroup$
A negative answer must be wrong, and I find your calculations persuasive: $z=pm2i$ will give $frac32$
$endgroup$
– Henry
Dec 18 '18 at 14:01