Problem in the complex plane

Find the maximum distance from the origin to the point zz satisfying

∣z+1z∣=a\displaystyle |z+\frac{1}{z}|=a

#NumberTheory #MathProblem #Math

Note by Krishna Jha
7 years, 9 months ago

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8 votes

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Comments

I'm getting a+a2+42\frac{a+\sqrt{a^2+4}}{2}, actually. Solution below.

Define z=reiθz = r e^{i \theta}. Thus, ∣z+1z∣=∣reiθ+e−iθr∣=r2+1r2+2cos⁡(2θ)=a⇒r=a2+a4−4a2cos⁡(2θ)+2cos⁡(4θ)−2−2cos⁡(2θ)2\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 θ=π2⇒r=12(a2+a2(a2+4)+2)=14(a2+2aa2+4+a2+4)=a+a2+42\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 zz satisfying this maximum is z=a+a2+42iz = \frac{a+\sqrt{a^2+4}}{2} i. We can verify this answer by evaluating 12(a2+4+a)−112(a2+4+a)\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 aa.

Michael Lee - 7 years, 9 months ago

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Nice approach.. But i think you have assumed z=reiθz=re^{i\theta}. Correct me if I am wrong.

BTW Thanks for the solution.. I was stuck on this one really.

Krishna Jha - 7 years, 9 months ago

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Oh, sorry, yes, I fixed it.

Michael Lee - 7 years, 9 months ago

We know that ∣∣z1∣−∣z2∣∣≤∣z1+z2∣≤∣z1∣+∣z2∣||z_{1}| - |z_{2}|| \leq |z_{1} + z_{2}| \leq |z_{1}| + |z_{2}|

Hence we get that ∣∣z∣−∣1z∣∣≤a ||z| - |\frac{1}{z}|| \leq a

Consider,∣z∣≥1⇒∣z∣≥∣1z∣ |z| \geq 1 \Rightarrow |z| \geq |\frac{1}{z}|

⇒∣z∣2−a∣z∣−1≤0\Rightarrow |z|^2 - a|z| - 1 \leq 0

⇒∣z∣∈[a−a2+42,a+a2+42]\Rightarrow |z| \in [ \frac{ a - \sqrt{a^2 + 4}}{2} , \frac{a + \sqrt{a^2 + 4}}{2} ] or ∣z∣∈[1,a+a2+42]|z| \in [1 , \frac{a + \sqrt{a^2 + 4}}{2}] ( as ∣z∣≥1)|z| \geq 1) .....(i) .....(i)

Similarly , consider 0≤∣z∣≤1 0 \leq |z| \leq 1,

∣z∣2+a∣z∣−1≥0 |z|^2 + a|z| - 1 \geq 0

⇒∣z∣∈[−a+a2+42,1]\Rightarrow |z| \in [\frac{ -a + \sqrt{a^2 + 4}}{2} , 1] ....(ii)....(ii)

From (i)(i) and (ii)(ii) , ∣z∣∈[−a+a2+42,a+a2+42] |z| \in [\frac{ -a + \sqrt{a^2 + 4}}{2} , \frac{a + \sqrt{a^2 + 4}}{2}] .....(iii) ..... (iii)

Now, ∣z∣+∣1z∣≥∣z+1z∣=a |z| + |\frac{1}{z}| \geq |z + \frac{1}{z}| = a

⇒∣z∣2−a∣z∣+1≥0\Rightarrow |z|^2 - a|z| + 1 \geq 0

⇒∣z∣∈(−∞,a−a2−42]∪[a+a2−42,∞).....(iv)\Rightarrow |z| \in (-\infty , \frac{a - \sqrt{a^2 - 4}}{2}] \cup [\frac{a + \sqrt{a^2 - 4}}{2} , \infty) .....(iv)

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

We get , ∣z∣∈[−a+a2+42,a−a2−42]∪[a+a2−42,a+a2+42] |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=a+a2+42|z|_{max} = \frac{a + \sqrt{a^2 + 4}}{2}

jatin yadav - 7 years, 9 months ago

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

Jess Smith - 7 years, 9 months ago

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

Bob Krueger - 7 years, 9 months ago

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No... aa is just another real number...

Krishna Jha - 7 years, 9 months ago
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