# Find greatest value of $P = \left |3 i (z_1 + z_2) + 9 - z_1z_2\right |.$

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 $$\min P =4\sqrt{10}.$$

• Why can you assume $z_1=1$? Commented Jun 3, 2023 at 14:13
• see what? wrong link Commented Jun 3, 2023 at 14:26
• @MathFail I see here math.stackexchange.com/questions/4707363/… Commented Jun 3, 2023 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. Commented Jun 3, 2023 at 14:31
• @MathFail Thank you very much. I am waiting from you a correct solution for this my question. Commented Jun 3, 2023 at 14:34

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,

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

• 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 $P(x_2)=-3\sqrt{2}+10$. Commented Jun 3, 2023 at 18:25
• Yes, I have edited, I miss a factor when expanding it. Thank you for checking this. Commented Jun 3, 2023 at 18:30

$$\left |3 i (z_1 + z_2) + 9 - z_1z_2\right |\leq 3\left |(z_1 + z_2)\right |+ 9 + \left |z_1z_2\right | =10 + 3\sqrt{2}$$

• "the least and the greatest value" Commented Jun 3, 2023 at 17:46
• How does this show that the maximum is attainable? Commented Jun 3, 2023 at 18:27
• @durianice Triangle inequality complex numbers Commented Jun 3, 2023 at 19:02
• @jimbo Yes, but you did not show that $z_1,z_2$ are attainable. Commented Jun 3, 2023 at 19:03
• $z_1$, $z_2$ satisfying the conditions $|z_1| = 1$, $|z_2| = 1$, $\left| z_1 + z_2\right| =\sqrt{2}$ Commented Jun 3, 2023 at 19:04