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 \(a\).

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWe 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 , \(|z|_{max} = \frac{a + \sqrt{a^2 + 4}}{2} \)

Log in to reply

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 \(a\).

Log in to reply

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.

Log in to reply

Oh, sorry, yes, I fixed it.

Log in to reply

Does \[ z = a + bi \] or am I misinterpreting something?

Log in to reply

Important distinction. Is the \(a\) in the problem referring to the real part of \(z\), or is it just some real number?

Log in to reply

No... \(a\) is just another real number...

Log in to reply