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Maximum and minimum absolute value of a complex number
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Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$
So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$
Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$
Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
algebra-precalculus complex-numbers maxima-minima
edited 4 hours ago
greedoid
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asked 7 hours ago
ARahman
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up vote
3
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Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$
So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$
Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$
Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
algebra-precalculus complex-numbers maxima-minima
edited 4 hours ago
greedoid
31k94187
asked 7 hours ago
ARahman
1085
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$
So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$
Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$
Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
algebra-precalculus complex-numbers maxima-minima
edited 4 hours ago
greedoid
31k94187
asked 7 hours ago
ARahman
1085
Let, $z in mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$left| z - frac1z right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-dfrac1z=a-dfraca4+ileft(b+dfracb4right)$
So
$$
beginsplit
left|z-frac1zright|
&=sqrtleft(a-dfraca4right)^2+left(b+dfracb4right)^2\
&=sqrt4+dfrac14-dfraca^22+dfracb^22\
&=sqrt4+dfrac14+dfrac12(b^2-a^2)
endsplit
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $sqrt2+dfrac14=dfrac32$
Now, comes the maximum value.
We can write $$sqrt4+dfrac14+dfrac12(b^2-a^2)$$
$$=sqrt4+dfrac14+dfrac12(4-2a^2)$$
$$=sqrt4+dfrac14+2-a^2$$
$$=sqrt6+dfrac14-a^2$$
Setting $a=0$ gives the maximum value $sqrt6+dfrac14=dfrac52$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
algebra-precalculus complex-numbers maxima-minima
algebra-precalculus complex-numbers maxima-minima
edited 4 hours ago
greedoid
31k94187
asked 7 hours ago
ARahman
1085
edited 4 hours ago
greedoid
31k94187
asked 7 hours ago
ARahman
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edited 4 hours ago
greedoid
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edited 4 hours ago
greedoid
31k94187
edited 4 hours ago
greedoid
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31k94187
asked 7 hours ago
ARahman
1085
asked 7 hours ago
ARahman
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ARahman
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A (very) faster way:
We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+frac12=frac52$$
Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.
Can you deal with the minimum now?
answered 6 hours ago
ajotatxe
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$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$
answered 6 hours ago
Henry Lee
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You could try like this (with help of triangle inequality):
$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$
clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.
answered 6 hours ago
greedoid
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Your idea is good, but you lose yourself in some computations (the maximum is correct, though).
Consider the square of the modulus:
$$
f(z)=left|z-frac1zright|^2=fracz^2-1^2
$$
Since $|z|=2$ by assumption, we can as well consider
$$
g(z)=|z^2-1|^2=(z^2-1)(barz^2-1)=z^2barz^2-z^2-barz^2+1=5-z^2-barz^2
$$
If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $barz^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then
$$
g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2
$$
The maximum is for $a=0$, the minimum for $a=pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.
We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.
answered 6 hours ago
egreg
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$ z - frac1z = fracz^2-1z $
So that, we get $left| z - frac1z right| $ = $fracz^2-12 $
If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:
$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $
Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $
We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $
$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $
So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.
Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$
if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$
answered 6 hours ago
Dominik Kutek
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Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]
Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.
So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.
Intuitively[2] I'd say that as $frac 2516 > frac 916$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.
As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.
So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.
====
[1]. There probably no need for this, but we can create an identity:
$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.
[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.
Actually formally was easier than my intuition.
answered 4 hours ago
fleablood
1
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
A (very) faster way:
We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+frac12=frac52$$
Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.
Can you deal with the minimum now?
answered 6 hours ago
ajotatxe
50.5k13185
add a comment |Â
up vote
3
down vote
A (very) faster way:
We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+frac12=frac52$$
Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.
Can you deal with the minimum now?
answered 6 hours ago
ajotatxe
50.5k13185
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A (very) faster way:
We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+frac12=frac52$$
Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.
Can you deal with the minimum now?
answered 6 hours ago
ajotatxe
50.5k13185
A (very) faster way:
We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+frac12=frac52$$
Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.
Can you deal with the minimum now?
answered 6 hours ago
ajotatxe
50.5k13185
answered 6 hours ago
ajotatxe
50.5k13185
answered 6 hours ago
ajotatxe
50.5k13185
answered 6 hours ago
ajotatxe
50.5k13185
50.5k13185
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up vote
2
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$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$
answered 6 hours ago
Henry Lee
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up vote
2
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$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$
answered 6 hours ago
Henry Lee
1,23816
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up vote
2
down vote
up vote
2
down vote
$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$
answered 6 hours ago
Henry Lee
1,23816
$$f(z)=left|z-frac1zright|=fracz^2-1right2$$
since we know $|z|=2$
answered 6 hours ago
Henry Lee
1,23816
answered 6 hours ago
Henry Lee
1,23816
answered 6 hours ago
Henry Lee
1,23816
answered 6 hours ago
Henry Lee
1,23816
1,23816
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add a comment |Â
up vote
2
down vote
You could try like this (with help of triangle inequality):
$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$
clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.
answered 6 hours ago
greedoid
31k94187
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up vote
2
down vote
You could try like this (with help of triangle inequality):
$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$
clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.
answered 6 hours ago
greedoid
31k94187
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up vote
2
down vote
up vote
2
down vote
You could try like this (with help of triangle inequality):
$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$
clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.
answered 6 hours ago
greedoid
31k94187
You could try like this (with help of triangle inequality):
$$|z-1over z|= |z^2-1over z| = over 2 geq z^2 =3over 2$$
clearly this can be achieved at $z = 2$ and $$ over 2 leq z^2 = 5over 2$$ which can be achieved at $z=-2$.
answered 6 hours ago
greedoid
31k94187
edited 6 hours ago
answered 6 hours ago
greedoid
31k94187
answered 6 hours ago
greedoid
31k94187
answered 6 hours ago
greedoid
31k94187
31k94187
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add a comment |Â
up vote
1
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Your idea is good, but you lose yourself in some computations (the maximum is correct, though).
Consider the square of the modulus:
$$
f(z)=left|z-frac1zright|^2=fracz^2-1^2
$$
Since $|z|=2$ by assumption, we can as well consider
$$
g(z)=|z^2-1|^2=(z^2-1)(barz^2-1)=z^2barz^2-z^2-barz^2+1=5-z^2-barz^2
$$
If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $barz^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then
$$
g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2
$$
The maximum is for $a=0$, the minimum for $a=pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.
We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.
answered 6 hours ago
egreg
169k1283191
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up vote
1
down vote
Your idea is good, but you lose yourself in some computations (the maximum is correct, though).
Consider the square of the modulus:
$$
f(z)=left|z-frac1zright|^2=fracz^2-1^2
$$
Since $|z|=2$ by assumption, we can as well consider
$$
g(z)=|z^2-1|^2=(z^2-1)(barz^2-1)=z^2barz^2-z^2-barz^2+1=5-z^2-barz^2
$$
If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $barz^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then
$$
g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2
$$
The maximum is for $a=0$, the minimum for $a=pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.
We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.
answered 6 hours ago
egreg
169k1283191
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your idea is good, but you lose yourself in some computations (the maximum is correct, though).
Consider the square of the modulus:
$$
f(z)=left|z-frac1zright|^2=fracz^2-1^2
$$
Since $|z|=2$ by assumption, we can as well consider
$$
g(z)=|z^2-1|^2=(z^2-1)(barz^2-1)=z^2barz^2-z^2-barz^2+1=5-z^2-barz^2
$$
If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $barz^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then
$$
g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2
$$
The maximum is for $a=0$, the minimum for $a=pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.
We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.
answered 6 hours ago
egreg
169k1283191
Your idea is good, but you lose yourself in some computations (the maximum is correct, though).
Consider the square of the modulus:
$$
f(z)=left|z-frac1zright|^2=fracz^2-1^2
$$
Since $|z|=2$ by assumption, we can as well consider
$$
g(z)=|z^2-1|^2=(z^2-1)(barz^2-1)=z^2barz^2-z^2-barz^2+1=5-z^2-barz^2
$$
If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $barz^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then
$$
g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2
$$
The maximum is for $a=0$, the minimum for $a=pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.
We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.
answered 6 hours ago
egreg
169k1283191
answered 6 hours ago
egreg
169k1283191
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egreg
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egreg
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169k1283191
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$ z - frac1z = fracz^2-1z $
So that, we get $left| z - frac1z right| $ = $fracz^2-12 $
If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:
$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $
Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $
We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $
$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $
So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.
Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$
if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$
answered 6 hours ago
Dominik Kutek
1013
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Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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$ z - frac1z = fracz^2-1z $
So that, we get $left| z - frac1z right| $ = $fracz^2-12 $
If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:
$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $
Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $
We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $
$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $
So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.
Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$
if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$
answered 6 hours ago
Dominik Kutek
1013
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Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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$ z - frac1z = fracz^2-1z $
So that, we get $left| z - frac1z right| $ = $fracz^2-12 $
If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:
$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $
Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $
We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $
$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $
So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.
Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$
if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$
answered 6 hours ago
Dominik Kutek
1013
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$ z - frac1z = fracz^2-1z $
So that, we get $left| z - frac1z right| $ = $fracz^2-12 $
If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:
$ frac2 = fracsqrt(x^2-y^2-1)^2 + (2xy)^22 $
Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $
We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x in [-2,2] $
$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $
So, all we have to do is looking at $25-4x^2, for x^2 in[0,4] $ and say what's the min and max.
Clearly if $x=0$ then we get $25$ and the max value is $fracsqrt252 = frac52$
if $x^2 = 4$ we get $9$ and the min value is $fracsqrt92 = frac32$
answered 6 hours ago
Dominik Kutek
1013
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 6 hours ago
Dominik Kutek
1013
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 6 hours ago
Dominik Kutek
1013
answered 6 hours ago
Dominik Kutek
1013
1013
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dominik Kutek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]
Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.
So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.
Intuitively[2] I'd say that as $frac 2516 > frac 916$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.
As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.
So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.
====
[1]. There probably no need for this, but we can create an identity:
$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.
[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.
Actually formally was easier than my intuition.
answered 4 hours ago
fleablood
1
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up vote
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down vote
Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]
Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.
So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.
Intuitively[2] I'd say that as $frac 2516 > frac 916$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.
As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.
So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.
====
[1]. There probably no need for this, but we can create an identity:
$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.
[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.
Actually formally was easier than my intuition.
answered 4 hours ago
fleablood
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]
Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.
So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.
Intuitively[2] I'd say that as $frac 2516 > frac 916$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.
As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.
So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.
====
[1]. There probably no need for this, but we can create an identity:
$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.
[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.
Actually formally was easier than my intuition.
answered 4 hours ago
fleablood
1
Hint: $z - frac 1z = z - frac overline zzoverline z = z - frac overline z^2 = z - frac overline z4 = frac 34 Re(z) -ifrac 54 Im(z)$[1]
Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.
So you are being asked to find the maximum and minimum possible values of $sqrt frac 916a^2 + frac 2516b^2$ given that $a^2 + b^2 = 4$.
Intuitively[2] I'd say that as $frac 2516 > frac 916$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.
As $a^2, b^2 ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.
So max is $sqrt frac 2516*4 = frac 52$ and min is $sqrtfrac 916*4 =frac 32$.
====
[1]. There probably no need for this, but we can create an identity:
$z pm frac 1z = (1pm frac 1^2)Re(z) + i(1 mp frac 1^2)Im(z)$ for any $zne 0$.
[2] And formally I'd say $sqrt frac 916a^2 + frac 2516b^2 =sqrt frac 2516b^2 + frac 916(4-b^2)=sqrtb^2 + frac 94$ which is max/min when $b$ is max/min.
Actually formally was easier than my intuition.
answered 4 hours ago
fleablood
1
edited 4 hours ago
answered 4 hours ago
fleablood
1
answered 4 hours ago
fleablood
1
answered 4 hours ago
fleablood
1
1
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