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Let be two complexs numbers $z_1$, $z_2$ satisfying the conditions $|z_1| = 1$, $|z_2| = 1$, $\left| z_1 + z_2\right| =\sqrt{2}$. Find the least and the greatest value of the value $$P = \left |3 i (z_1 + z_2) + 9 - z_1z_2\right |.$$ Solution. Suppose $z_1 = 1$, $z_2 = x + i y. $ we find $z_2$ by the system of equations $$\begin{cases} |z_2| = 1,\\ \left| z_1 + z_2\right| =\sqrt{2} \end{cases}$$ that is mean $$\begin{cases} x^2 + y^2 = 1,\\ (x+1)^2 + y^2 =2. \end{cases}$$
Solve this system, we have $(x,y)=(0,1) $ or $(x,y)=(0,-1) $, and then $z_2 = i$ or $z_2 = -i$

With $z_1 = 1$, $z_2 = i$, we have $$P = \left |3 i (1 + i) + 9 - i\right | = 2\sqrt{10}.$$

With $z_1 = 1$, $z_2 = i$, we have $$P = \left |3 i (1 - i) + 9 + i\right | = 4\sqrt{10}.$$

Thus $\max P =4\sqrt{10} $ and $\min P =4\sqrt{10}. $

MathFail
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John Paul Peter
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  • Why can you assume $z_1=1$? – MathFail Jun 03 '23 at 14:13
  • see what? wrong link – MathFail Jun 03 '23 at 14:26
  • @MathFail I see here https://math.stackexchange.com/questions/4707363/complex-numbers-z-w-u-with-z-1-w-2-u-3-find-the-maximum-v/4707432#4707432 – John Paul Peter Jun 03 '23 at 14:28
  • In your OP, it contains some fixed constant, i.e, $9$, but in the link you referenced, there is no such a constant, all complex numbers are in relative positions. Have you seen my asnwer in that post? I gave a general solution without assuming any fixed points. To be safe, never assuming some fixed points unless you are 100% sure it won't change the essence of the problem. – MathFail Jun 03 '23 at 14:31
  • @MathFail Thank you very much. I am waiting from you a correct solution for this my question. – John Paul Peter Jun 03 '23 at 14:34

2 Answers2

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Let $z_1=e^{ix}, z_2=e^{iy}$, take the principle branch $\arg z\in (-\pi, \pi]$. Since $z_1, z_2$ are symmetric, without loss of generality, let $y\ge x$

$$|z_1+z_2|=\sqrt2\Longrightarrow\cos(x-y)=0\Longrightarrow y=x+\frac\pi2$$

Next, plug in and simplify, we get

$$P=2\sqrt{(5-3\sin x)(5-3\sin y)}=2\sqrt{(5-3\sin x)(5-3\cos x)}=2\sqrt Q$$

take derivative for $Q'=0$ to find extrema points

$$(\cos x-\sin x)(-5+3\cos x+3 \sin x)=0\tag{1}$$

and note that $\cos x+\sin x\le\sqrt2<\frac53$, hence, (1) implies

$$\sin x=\cos x\Longrightarrow x_{1}=-\frac{3\pi}4,~~~x_{2}=\frac\pi4,$$

therefore,

$$\max P=P(x_1)=10+3\sqrt{2},~~~~\min P=P(x_2)=10-3\sqrt2$$

MathFail
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  • Should be $\sin(x_3)+\cos(x_3)=5/3$, which has no solution. $\max P$ should be $3\sqrt{2}+10$ and $\min P$ should be $P(x_2)=-3\sqrt{2}+10$. – durianice Jun 03 '23 at 18:25
  • Yes, I have edited, I miss a factor when expanding it. Thank you for checking this. – MathFail Jun 03 '23 at 18:30
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$$\left |3 i (z_1 + z_2) + 9 - z_1z_2\right |\leq 3\left |(z_1 + z_2)\right |+ 9 + \left |z_1z_2\right | =10 + 3\sqrt{2} $$

jimbo
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