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Problem in the complex plane

Find the maximum distance from the origin to the point \(z\) satisfying

\(\displaystyle |z+\frac{1}{z}|=a\)

Note by Krishna Jha
3 years, 11 months ago

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We know that \(||z_{1}| - |z_{2}|| \leq |z_{1} + z_{2}| \leq |z_{1}| + |z_{2}|\)

Hence we get that \( ||z| - |\frac{1}{z}|| \leq a \)

Consider,\( |z| \geq 1 \Rightarrow |z| \geq |\frac{1}{z}|\)

\(\Rightarrow |z|^2 - a|z| - 1 \leq 0\)

\(\Rightarrow |z| \in [ \frac{ a - \sqrt{a^2 + 4}}{2} , \frac{a + \sqrt{a^2 + 4}}{2} ]\) or \(|z| \in [1 , \frac{a + \sqrt{a^2 + 4}}{2}]\) ( as \(|z| \geq 1) \) \( .....(i) \)

Similarly , consider \( 0 \leq |z| \leq 1\),

\( |z|^2 + a|z| - 1 \geq 0 \)

\(\Rightarrow |z| \in [\frac{ -a + \sqrt{a^2 + 4}}{2} , 1] \) \(....(ii)\)

From \((i)\) and \((ii)\) , \( |z| \in [\frac{ -a + \sqrt{a^2 + 4}}{2} , \frac{a + \sqrt{a^2 + 4}}{2}] \) \( ..... (iii)\)

Now, \( |z| + |\frac{1}{z}| \geq |z + \frac{1}{z}| = a\)

\(\Rightarrow |z|^2 - a|z| + 1 \geq 0\)

\(\Rightarrow |z| \in (-\infty , \frac{a - \sqrt{a^2 - 4}}{2}] \cup [\frac{a + \sqrt{a^2 - 4}}{2} , \infty) .....(iv) \)

Combining \((iii)\) and \((iv)\) ,

We get , \( |z| \in [\frac{ -a + \sqrt{a^2 + 4}}{2} , \frac{a - \sqrt{a^2 - 4}}{2}] \cup [\frac{a + \sqrt{a^2 - 4}}{2} , \frac{a + \sqrt{a^2 + 4}}{2}] \)

Hence we conclude that , \(|z|_{max} = \frac{a + \sqrt{a^2 + 4}}{2} \) Jatin Yadav · 3 years, 11 months ago

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I'm getting \(\frac{a+\sqrt{a^2+4}}{2}\), actually. Solution below.

Define \(z = r e^{i \theta}\). Thus, \(\left|z + \frac{1}{z}\right| = \left|r e^{i \theta}+\frac{e^{-i \theta}}{r}\right| = \sqrt{r^2 + \frac{1}{r^2} + 2 \cos(2 \theta)} = a \Rightarrow r = \sqrt{\frac{a^2+\sqrt{a^4-4 a^2 \cos (2 \theta)+2 \cos (4 \theta)-2}-2 \cos (2 \theta)}{2}}\). By inspection of the preceding equation, this has a maximum at \(\theta = \frac{\pi}{2} \Rightarrow r = \sqrt{\frac{1}{2} \left(a^2+\sqrt{a^2 \left(a^2+4\right)}+2\right)} = \sqrt{\frac{1}{4} \left(a^2 + 2a \sqrt{a^2+4} + a^2 + 4\right)} = \boxed{\frac{a+\sqrt{a^2+4}}{2}}\). This signifies that the complex number \(z\) satisfying this maximum is \(z = \frac{a+\sqrt{a^2+4}}{2} i\). We can verify this answer by evaluating \(\frac{1}{2} \left(\sqrt{a^2+4}+a\right) - \frac{1}{\frac{1}{2} \left(\sqrt{a^2+4}+a\right)}\). With some algebraic simplification, we can show this to be equal to \(a\). Michael Lee · 3 years, 11 months ago

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@Michael Lee Nice approach.. But i think you have assumed \(z=re^{i\theta}\). Correct me if I am wrong.

BTW Thanks for the solution.. I was stuck on this one really. Krishna Jha · 3 years, 11 months ago

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@Krishna Jha Oh, sorry, yes, I fixed it. Michael Lee · 3 years, 11 months ago

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Does \[ z = a + bi \] or am I misinterpreting something? Jess Smith · 3 years, 11 months ago

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@Jess Smith Important distinction. Is the \(a\) in the problem referring to the real part of \(z\), or is it just some real number? Bob Krueger · 3 years, 11 months ago

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@Bob Krueger No... \(a\) is just another real number... Krishna Jha · 3 years, 11 months ago

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