What is $max(|z|)$ if $Bigl|frac6z - i2 + 3izBigr|leq1$Understanding why $int_0^pi/2 sqrt1+cos^2x geq fracpi4bigl( 1 + sqrt2bigr)$Calculate $Bigl|frac1n^zBigr|$.Show that $Pleft(limsuplimits_ttoinftyleft|B_tright|/t>cright)leq Pleft(suplimits_tin[n,n+1]left|B_tright|/n>c;mboxi.o.right)$Why is my solution incorrect? AM GM inequalityComplex numbers $left(frac1+i1-iright)^k = 1$ what is $k$?What are the points of differentiability of $f(z)=vert z vert^2+iota bar z +1$?Verification involving modulus and complex numbersEquality occurring at two different values for a three way inequalityProof for Floor function exponent algebra IdentityReverse Littlewood-Offord problem: lower bound for the number of choices of signs such that $|pm a_1dotspm a_n| leq max|a_i|.$
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What is $max(|z|)$ if $Bigl|frac6z - i2 + 3izBigr|leq1$
Understanding why $int_0^pi/2 sqrt1+cos^2x geq fracpi4bigl( 1 + sqrt2bigr)$Calculate $Bigl|frac1n^zBigr|$.Show that $Pleft(limsuplimits_ttoinftyleft|B_tright|/t>cright)leq Pleft(suplimits_tin[n,n+1]left|B_tright|/n>c;mboxi.o.right)$Why is my solution incorrect? AM GM inequalityComplex numbers $left(frac1+i1-iright)^k = 1$ what is $k$?What are the points of differentiability of $f(z)=vert z vert^2+iota bar z +1$?Verification involving modulus and complex numbersEquality occurring at two different values for a three way inequalityProof for Floor function exponent algebra IdentityReverse Littlewood-Offord problem: lower bound for the number of choices of signs such that $|pm a_1dotspm a_n| leq max|a_i|.$
$begingroup$
I'm solving a multiple choice question but my answer is not among the choices and I don't know where I am wrong. The question is this:
What is the $max|z|$ if: $$Bigl|frac6z - i2 + 3izBigr|leq1$$ and $z$ is a complex number.
- $frac15$
- $frac14$
- $frac13$
- $frac12$
My solution is this but my answer isn't among them:
$$ biggl|frac6z - i2 + 3izbiggr|leq1Rightarrow |6z-i| leq |2+3iz| Rightarrow |6z-i|^2 leq |2+3iz|^2 $$
$$ Rightarrow (6z-i)overline(6z-i) leq (2+3iz)overline(2+3iz) $$
$$ Rightarrow (6z-i)(6barz+i) leq (2+3iz)(2-3ibarz) $$
$$ Rightarrow 36|z|^2 +6iz - 6ibarz + 1 leq 4 - 6ibarz +6iz + 9|z|^2 $$
$$ Rightarrow 17|z|^2 leq 3 Rightarrow |z|^2 leq frac317 Rightarrow |z| leq sqrtfrac317$$
Where am I wrong? what is the right solution?
Thanks in advance.
inequality complex-numbers
edited Mar 12 at 11:08
Bernard
123k741116
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
$endgroup$
add a comment |
$begingroup$
I'm solving a multiple choice question but my answer is not among the choices and I don't know where I am wrong. The question is this:
What is the $max|z|$ if: $$Bigl|frac6z - i2 + 3izBigr|leq1$$ and $z$ is a complex number.
- $frac15$
- $frac14$
- $frac13$
- $frac12$
My solution is this but my answer isn't among them:
$$ biggl|frac6z - i2 + 3izbiggr|leq1Rightarrow |6z-i| leq |2+3iz| Rightarrow |6z-i|^2 leq |2+3iz|^2 $$
$$ Rightarrow (6z-i)overline(6z-i) leq (2+3iz)overline(2+3iz) $$
$$ Rightarrow (6z-i)(6barz+i) leq (2+3iz)(2-3ibarz) $$
$$ Rightarrow 36|z|^2 +6iz - 6ibarz + 1 leq 4 - 6ibarz +6iz + 9|z|^2 $$
$$ Rightarrow 17|z|^2 leq 3 Rightarrow |z|^2 leq frac317 Rightarrow |z| leq sqrtfrac317$$
Where am I wrong? what is the right solution?
Thanks in advance.
inequality complex-numbers
edited Mar 12 at 11:08
Bernard
123k741116
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
$endgroup$
add a comment |
$begingroup$
I'm solving a multiple choice question but my answer is not among the choices and I don't know where I am wrong. The question is this:
What is the $max|z|$ if: $$Bigl|frac6z - i2 + 3izBigr|leq1$$ and $z$ is a complex number.
- $frac15$
- $frac14$
- $frac13$
- $frac12$
My solution is this but my answer isn't among them:
$$ biggl|frac6z - i2 + 3izbiggr|leq1Rightarrow |6z-i| leq |2+3iz| Rightarrow |6z-i|^2 leq |2+3iz|^2 $$
$$ Rightarrow (6z-i)overline(6z-i) leq (2+3iz)overline(2+3iz) $$
$$ Rightarrow (6z-i)(6barz+i) leq (2+3iz)(2-3ibarz) $$
$$ Rightarrow 36|z|^2 +6iz - 6ibarz + 1 leq 4 - 6ibarz +6iz + 9|z|^2 $$
$$ Rightarrow 17|z|^2 leq 3 Rightarrow |z|^2 leq frac317 Rightarrow |z| leq sqrtfrac317$$
Where am I wrong? what is the right solution?
Thanks in advance.
inequality complex-numbers
edited Mar 12 at 11:08
Bernard
123k741116
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
$endgroup$
I'm solving a multiple choice question but my answer is not among the choices and I don't know where I am wrong. The question is this:
What is the $max|z|$ if: $$Bigl|frac6z - i2 + 3izBigr|leq1$$ and $z$ is a complex number.
- $frac15$
- $frac14$
- $frac13$
- $frac12$
My solution is this but my answer isn't among them:
$$ biggl|frac6z - i2 + 3izbiggr|leq1Rightarrow |6z-i| leq |2+3iz| Rightarrow |6z-i|^2 leq |2+3iz|^2 $$
$$ Rightarrow (6z-i)overline(6z-i) leq (2+3iz)overline(2+3iz) $$
$$ Rightarrow (6z-i)(6barz+i) leq (2+3iz)(2-3ibarz) $$
$$ Rightarrow 36|z|^2 +6iz - 6ibarz + 1 leq 4 - 6ibarz +6iz + 9|z|^2 $$
$$ Rightarrow 17|z|^2 leq 3 Rightarrow |z|^2 leq frac317 Rightarrow |z| leq sqrtfrac317$$
Where am I wrong? what is the right solution?
Thanks in advance.
inequality complex-numbers
inequality complex-numbers
edited Mar 12 at 11:08
Bernard
123k741116
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
edited Mar 12 at 11:08
Bernard
123k741116
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
edited Mar 12 at 11:08
Bernard
123k741116
edited Mar 12 at 11:08
Bernard
123k741116
edited Mar 12 at 11:08
Bernard
123k741116
123k741116
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
asked Mar 12 at 9:38
Peyman mohseni kiasariPeyman mohseni kiasari
12811
12811
add a comment |
add a comment |
1 Answer
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$begingroup$
$36-9=27$. You have taken it as $17$. The correct answer is 3).
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
add a comment |
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$begingroup$
$36-9=27$. You have taken it as $17$. The correct answer is 3).
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
add a comment |
$begingroup$
$36-9=27$. You have taken it as $17$. The correct answer is 3).
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
add a comment |
$begingroup$
$36-9=27$. You have taken it as $17$. The correct answer is 3).
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
$endgroup$
$36-9=27$. You have taken it as $17$. The correct answer is 3).
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
answered Mar 12 at 9:42
Kavi Rama MurthyKavi Rama Murthy
67.6k53067
67.6k53067
add a comment |
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