The least value of |z| where z is a complex n | vedclass

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Question

(A) Given the inequality: $\exp \left(\frac{(|z|+3)(|z|-1)}{|z|+1} \ln 2\right) \geq \log _{\sqrt{2}}|5 \sqrt{7}+9 i |$First,simplify the right side:

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given the inequality: $\exp \left(\frac{(|z|+3)(|z|-1)}{|z|+1} \ln 2\right) \geq \log _{\sqrt{2}}|5 \sqrt{7}+9 i |$First,simplify the right side: $|5 \sqrt{7}+9 i| = \sqrt{(5 \sqrt{7})^2 + 9^2} = \sqrt{175 + 81} = \sqrt{256} = 16$.So,$\log _{\sqrt{2}}(16) = \log _{2^{1/2}}(2^4) = 8 \log _{2}(2) = 8$.Now,the inequality becomes: $2^{\frac{(|z|+3)(|z|-1)}{|z|+1}} \geq 8$.Since $8 = 2^3$,we have: $\frac{(|z|+3)(|z|-1)}{|z|+1} \geq 3$.Let $x = |z|$,where $x \geq 0$. Then $\frac{(x+3)(x-1)}{x+1} \geq 3$.$(x^2 + 2x - 3) \geq 3(x+1)$.$x^2 + 2x - 3 \geq 3x + 3$.$x^2 - x - 6 \geq 0$.$(x-3)(x+2) \geq 0$.Since $x = |z| \geq 0$,$x+2$ is always positive,so $x-3 \geq 0$,which means $|z| \geq 3$.Thus,the least value of $|z|$ is $3$.

The least value of $|z|$ where $z$ is a complex number satisfying the inequality $\exp \left(\frac{(|z|+3)(|z|-1)}{|z|+1} \log _{ e } 2\right) \geq \log _{\sqrt{2}}|5 \sqrt{7}+9 i |$,where $i=\sqrt{-1}$,is equal to:

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