Help Please!

Q1) Since $\alpha$, $\beta$, $\gamma$ are the cube root so $p$, it implies that they are the roots of the equation $x^3 = p$. This implies that $x= \sqrt[3]{p}$ $\implies$ $\{ \alpha, \beta, \gamma \} = \{ k , k\omega, k \omega ^2 \}$, where $k$ is the real cube root of $p$ and $\omega$ is the primitive cube root of unity. So, WLOG let $\alpha = k$, $\beta = k \omega$ and $\gamma = k \omega^2$.

Now consider the equation $x \beta + y \gamma + z \alpha = kx \omega + ky \omega^2 + kz$. But since $\omega ^3 = 1$.

$\begin{aligned} x \beta + y \gamma + z \alpha &= kx \omega + ky \omega^2 + kz \\ &= kx \omega + ky \omega^2 + kz \omega^3 \\ &= \omega (kx + ky \omega + kz \omega^2) \\ &= \omega (x \alpha + y \beta + z \gamma ) \end{aligned}$

So, we get that

$\dfrac{x \alpha + y \beta + z \gamma}{x \beta + y \gamma + z \alpha} = \dfrac{1}{\omega} = \omega^2$

For Q3 use rotation of complex number. Let O(0,0) be the centre of circle then by rotating AO (A=$(1,\sqrt{3})$ by 120 degrees clockwise and anticlockwise obtain $z_{2}$ and $z_{3}$

Q4:

Let $Z = r_1 cis(\theta_1)$ and $W = r_2 cis(\theta_2)$.

Let's consider: $|Z - W|^2 = (r_1 cos(\theta_1) - r_2 cos(\theta_2))^2 + (r_1sin(\theta_1) - r_2sin(\theta_2))^2$ $|Z - W|^2 = r_1^2 cos^2(\theta_1) + r_2^2 cos^2(\theta_2) - 2r_1 r_2 cos(\theta_1) cos(\theta_2) + r_1^2 sin^2(\theta_1) + r_2^2 sin(\theta_2) - 2r_1 r_2 sin(\theta_1) sin(\theta_2)$

Doing a bit of algebra, and knowing some trig. identities, we obtain

$|Z - W|^2 = r_1^2 + r_2^2 - 2r_1 r_2 cos(\theta_1 - \theta_2)$

Subtracting and adding $2r_1 r_2$ allows us to factorise as follows:

$|Z - W|^2 = (r_1 - r_2)^2 + 2r_1 r_2 - 2r_1 r_2 cos(\theta_1 - \theta_2)$

$|Z - W|^2 = (r_1 - r_2)^2 + 2r_1 r_2(1 - cos(\theta_1 - \theta_2))$.

Using the double-angle identity for cosine,

$|Z - W|^2 = (r_1 - r_2)^2 + 4r_1 r_2(sin^2(\frac{\theta_1 - \theta_2}{2}))$

Now, $r_1, r_2 \leq 1$ and we allegedly know that $|\theta| \geq |sin(\theta)|$ which can be easily proven.

Thus, we can say, $|Z - W|^2 \leq (r_1 - r_2)^2 + 4(\frac{\theta_1 - \theta_2}{2})^2$

$|Z - W|^2 \leq (r_1 - r_2)^2 + (\theta_1 - \theta_2)^2$, by how we defined $Z$ and $W$, $\theta_1$ and $\theta_2$ are their principal arguments and $r_1$ and $r_2$ being their modulus respectively.

Therefore, $|Z - W|^2 \leq (|Z| - |W|)^2 + (Arg(Z) - Arg(W))^2$.

As required. 🌝🌚

Please kindly point out any errors if there are any.

Is it clear now brother?

I'll make it as concise as possible. :P

Q2: $|z_1 - z_2| = |z_2 - z_3| = |z_3 - z_1|$ because it's an equilateral triangle.

$|(a - 1) + (1-b)i| = |1 + bi|= |-a - i|$

$a^2 - 2a + 1 + 1 - 2b + b^2 = 1 + b^2 = a^2 + 1$

Which gives us two systems of equations, namely, $a^2 - 2a + 2 - 2b + b^2 = 1 + b^2$ and $1 + b^2 = a^2 + 1$.

Simplifying them further yields, $a^2 - 2a + 1 - 2b = 0$ and $a^2 = b^2$.

Since $0 < a,b < 1 \Rightarrow a = b$ from the second equation.

Noting the constraint and solving the system gives us the following solution: $a = b = 2 - \sqrt{3}$

Q3: Sorry man, I forgot how to include a diagram. Hopefully it was articulate enough to be of any use.

$z_1 = 1 + i \sqrt{3}$ can be written as $2e^{i \pi / 3}$.

Note, that the circle is about the origin, so we can get the principal argument using some circle properties (i.e., angle subtended at the circumference is half the angle subtended at the centre) as $z_2$ and $z_3$ as $\pi$ and $-\pi/3$.

Also, they all lie on the circumference of the circle $\Rightarrow |z_2| = |z_3| = 2$

Hence, $z_2 = 2cis(\pi)$ and $z_3 = 2cis(-\pi/3)$

In Cartesian form, $z_2 = -2$ and $z_3 = 1 - i \sqrt{3}$

@Vishnuram Leonardodavinci @Surya Prakash @Tanishq Varshney A Great Thanks for your time and help :)

Also I got Q5 ;)

@Surya Prakash @Chew-Seong Cheong @Akhil Bansal @Brian Charlesworth @Sudeep Salgia