Let z_1, z_2 and z_3 be three complex numbers on the circle |z|=1 with (z_1)=(-)/(4), (z_2…

Question

Let $z_1, z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $\arg \left(z_1\right)=\frac{-\pi}{4}, \arg \left(z_2\right)=0$ and $\arg \left(z_3\right)=\frac{\pi}{4}$. If $\left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2=\alpha+\beta \sqrt{2}, \alpha, \beta \in Z$, then the value of $\alpha^2+\beta^2$ is :

Accepted Answer

To solve the problem, we start with the given information about the complex numbers $ z_1, z_2, $ and $ z_3 $, which lie on the unit circle $ |z| = 1 $. Their arguments are as follows:

$ \arg(z_1) = -\frac{\pi}{4} $

$ \arg(z_2) = 0 $

$ \arg(z_3) = \frac{\pi}{4} $

Thus, the complex numbers can be represented as:

$ z_1 = e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_2 = e^{i\cdot 0} = 1 $

$ z_3 = e^{i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

Next, calculate the conjugates needed:

$ \bar{z}_2 = 1 $

$ \bar{z}_3 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ \bar{z}_1 = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} $

We need to evaluate:

$ z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 $

Calculate each term separately:

$ z_1 \bar{z}_2 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) \cdot 1 = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_2 \bar{z}_3 = 1 \cdot \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} $

$ z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) \cdot \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \frac{(1 + i)^2}{2} = \frac{1 + 2i - 1}{2} = i $

Sum the evaluated terms:

$ z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1 = \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) + i $

Simplify:

$ = \sqrt{2} + i - \sqrt{2}i = \sqrt{2} + i(1-\sqrt{2}) $

Calculate the modulus squared:

$ \left| \sqrt{2} + i (1 - \sqrt{2}) \right|^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 $

$ = 2 + (1 - 2\sqrt{2} + 2) = 5 - 2\sqrt{2} $

Thus, the expression $ |z_1 \bar{z}_2 + z_2 \bar{z}_3 + z_3 \bar{z}_1|^2 $ simplifies as follows:

$ \alpha = 5 $

$ \beta = -2 $

Finally, compute $ \alpha^2 + \beta^2 $:

$ \alpha^2 + \beta^2 = 5^2 + (-2)^2 = 25 + 4 = 29 $

Therefore, the value of $ \alpha^2 + \beta^2 $ is 29.

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