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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
4 years, 1 month 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 , $$|z|_{max} = \frac{a + \sqrt{a^2 + 4}}{2}$$

- 4 years, 1 month ago

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 $$a$$.

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

Oh, sorry, yes, I fixed it.

- 4 years, 1 month ago

Does $z = a + bi$ or am I misinterpreting something?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

No... $$a$$ is just another real number...

- 4 years, 1 month ago