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

Question

Let ${z}_{1},{z}_{2}$ and ${z}_{3}$ be three complex numbers on the circle $|z|=1$ with $arg({z}_{1})=\frac{-\pi }{4},arg({z}_{2})=0$ and $arg({z}_{3})=\frac{\pi }{4}$. If ${|{z}_{1}{\vec{z}}_{2}+{z}_{2}{\vec{z}}_{3}+{z}_{3}{\vec{z}}_{1}|}^{2}=\alpha +\beta \sqrt{2},\alpha ,\beta \in Z$, 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:

${\vec{z}}_{2}=1$

${\vec{z}}_{3}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$

${\vec{z}}_{1}=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$

We need to evaluate:

${z}_{1}{\vec{z}}_{2}+{z}_{2}{\vec{z}}_{3}+{z}_{3}{\vec{z}}_{1}$

Calculate each term separately:

${z}_{1}{\vec{z}}_{2}=(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})\cdot 1=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$

${z}_{2}{\vec{z}}_{3}=1\cdot (\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$

${z}_{3}{\vec{z}}_{1}=(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})\cdot (\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})=\frac{(1+i{)}^{2}}{2}=\frac{1+2i-1}{2}=i$

Sum the evaluated terms:

${z}_{1}{\vec{z}}_{2}+{z}_{2}{\vec{z}}_{3}+{z}_{3}{\vec{z}}_{1}=(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})+(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})+i$

Simplify:

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

Calculate the modulus squared:

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

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

Thus, the expression $|{z}_{1}{\vec{z}}_{2}+{z}_{2}{\vec{z}}_{3}+{z}_{3}{\vec{z}}_{1}{|}^{2}$ 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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