Question :What is the maximum value of $\lvert z\rvert$ when $z$ satisfies the condition $\lvert z+\frac{2}{z}\rvert=2$
My try: By removing modulus, and simplifying as $-2<z+\frac{2}{z}<2$, I am not getting the required answer. Someone please help.
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3$z$ is a complex number, your inequality makes no sense at all. – Zain Patel Apr 20 '17 at 15:51
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-2<z+2/z doesnt make sense for complex z – axioman Apr 20 '17 at 15:53
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Have edited the question, Z is not a complex number. – Tanuj Apr 20 '17 at 15:53
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Please let me know how can I proceed from there? – Tanuj Apr 20 '17 at 15:55
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There is no such real $z$. – Zain Patel Apr 20 '17 at 15:57
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How can you say that? My book shows the answer as $3^{0.5}+1$. – Tanuj Apr 20 '17 at 15:59
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Wolfram Alpha offers $z=\pm\sqrt{\frac32}\pm i\sqrt{\frac52}$ with $|z|=2$ – Henry Apr 20 '17 at 15:59
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I don't see things getting any clearer I'm sorry, someone with a concrete approach is welcomed. – Tanuj Apr 20 '17 at 16:00
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@user38227 how does that satisfy the condition on $z$ if $z$ is a real number? (that answer makes perfect sense if $z$ is a complex number though) – Zain Patel Apr 20 '17 at 16:02
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@user38227 You can confirm then with a calculator that your textbook is incorrect. $3^{0.5} + 1 + \frac{2}{3^{0.5} +1} \approx 3.46$ – orlp Apr 20 '17 at 16:04
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Anyway: here's the answer to the correct question https://math.stackexchange.com/questions/853992/max-and-min-value-of-z-in-leftz-frac2z-right-2 – Zain Patel Apr 20 '17 at 16:07
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http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumberInequalities.shtml – lab bhattacharjee Apr 20 '17 at 16:19