I have few doubts regarding complex numbers. Help is as always , greatly appreciated. Nice solutions or hints are welcome.

Q1) If $\alpha , \beta , \gamma$ are cube roots of $p$ such that $p<0$ , then for any $x,y,z$ , the expression $\dfrac{x\alpha+y\beta+z\gamma}{x\beta+y\gamma+z\alpha} = ?$

Q2) If $0 such that the points $z_1=a+i$ , $z_2=1+bi$ and $z_3=0$ form an equilateral triangle , find the value of $a$ and $b$.

Q3) Suppose $z_1,z_2,z_3$ are vertices of an equilateral triangle inscribed in a circle given by $|z|=2$. If $z_1=1+i\sqrt{3}$ , then find $z_2$ and $z_3$.

Q4) If $|Z|\leq 1 \ , \ |W| \leq 1$ then show that $|Z-W|^2 \leq (|Z|-|W|)^2+(\text{arg}(Z) - \text{arg}(W))^2$

Q5) Let $\overline{b}z+b\overline{z}=c$ , $b\neq 0$ , be a line in complex plane , where $b$ is complex conjugate of $b$. If a point $z_1$ is reflection of point $z_2$ in this line , show that $c=\overline{z_1}b+z_2\overline{b}$.

Note by Nihar Mahajan
4 years, 2 months ago

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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

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

- 4 years, 2 months ago

Nice!!!

- 4 years, 2 months ago

Great one! But if you care just for the answer ignoring the approach, then just set $y=z=0$ and $x=1$. With this you will just be left with $\dfrac{\alpha}{\beta}=\dfrac{p^{1/3}}{p^{1/3}\omega}=\omega^2$.

Clarification: $\omega$ can take any value of the two possible values: $\dfrac{-1+i\sqrt 3}{2}$ and $\dfrac{-1-i\sqrt 3}{2}$.

- 4 years, 2 months ago

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 $z_{3}$

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

Wow, I am impressed by this approach _

- 4 years, 2 months ago

- 4 years, 2 months ago

Is it clear now brother?

- 4 years, 2 months ago

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: $a = b = 2 - \sqrt{3}$

- 4 years, 2 months ago

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 $z_3 = 1 - i \sqrt{3}$

- 4 years, 2 months ago

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

Also I got Q5 ;)

- 4 years, 2 months ago

Anytime ;)

- 4 years, 2 months ago

Ur welcome $\Huge \ddot{\smile}$

- 4 years, 2 months ago

No worries. :D Glad to be of help.

- 4 years, 2 months ago