Help Please!

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

Q1) If α,β,γ\alpha , \beta , \gamma are cube roots of pp such that p<0p<0 , then for any x,y,zx,y,z , the expression xα+yβ+zγxβ+yγ+zα=?\dfrac{x\alpha+y\beta+z\gamma}{x\beta+y\gamma+z\alpha} = ?

Q2) If 0<a,b<10<a,b<1 such that the points z1=a+iz_1=a+i , z2=1+biz_2=1+bi and z3=0z_3=0 form an equilateral triangle , find the value of aa and bb.

Q3) Suppose z1,z2,z3z_1,z_2,z_3 are vertices of an equilateral triangle inscribed in a circle given by z=2|z|=2. If z1=1+i3z_1=1+i\sqrt{3} , then find z2z_2 and z3z_3.

Q4) If Z1 , W1|Z|\leq 1 \ , \ |W| \leq 1 then show that ZW2(ZW)2+(arg(Z)arg(W))2|Z-W|^2 \leq (|Z|-|W|)^2+(\text{arg}(Z) - \text{arg}(W))^2

Q5) Let bz+bz=c\overline{b}z+b\overline{z}=c , b0b\neq 0 , be a line in complex plane , where bb is complex conjugate of bb. If a point z1z_1 is reflection of point z2z_2 in this line , show that c=z1b+z2bc=\overline{z_1}b+z_2\overline{b}.

Note by Nihar Mahajan
3 years, 11 months ago

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Q1) Since α\alpha, β\beta, γ\gamma are the cube root so pp, it implies that they are the roots of the equation x3=px^3 = p. This implies that x=p3x= \sqrt[3]{p}     \implies {α,β,γ}={k,kω,kω2}\{ \alpha, \beta, \gamma \} = \{ k , k\omega, k \omega ^2 \}, where kk is the real cube root of pp and ω\omega is the primitive cube root of unity. So, WLOG let α=k\alpha = k, β=kω\beta = k \omega and γ=kω2\gamma = k \omega^2.

Now consider the equation xβ+yγ+zα=kxω+kyω2+kzx \beta + y \gamma + z \alpha = kx \omega + ky \omega^2 + kz. But since ω3=1\omega ^3 = 1.

xβ+yγ+zα=kxω+kyω2+kz=kxω+kyω2+kzω3=ω(kx+kyω+kzω2)=ω(xα+yβ+zγ)\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

xα+yβ+zγxβ+yγ+zα=1ω=ω2\dfrac{x \alpha + y \beta + z \gamma}{x \beta + y \gamma + z \alpha} = \dfrac{1}{\omega} = \omega^2

Surya Prakash - 3 years, 11 months ago

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Nice!!!

Abdur Rehman Zahid - 3 years, 11 months ago

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Great one! But if you care just for the answer ignoring the approach, then just set y=z=0y=z=0 and x=1x=1. With this you will just be left with αβ=p1/3p1/3ω=ω2\dfrac{\alpha}{\beta}=\dfrac{p^{1/3}}{p^{1/3}\omega}=\omega^2.

Clarification: ω\omega can take any value of the two possible values: 1+i32\dfrac{-1+i\sqrt 3}{2} and 1i32\dfrac{-1-i\sqrt 3}{2}.

Sandeep Bhardwaj - 3 years, 11 months ago

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For Q3 use rotation of complex number. Let O(0,0) be the centre of circle then by rotating AO (A=(1,3)(1,\sqrt{3}) by 120 degrees clockwise and anticlockwise obtain z2z_{2} and z3z_{3}

Tanishq Varshney - 3 years, 11 months ago

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Q4:

Let Z=r1cis(θ1)Z = r_1 cis(\theta_1) and W=r2cis(θ2)W = r_2 cis(\theta_2).

Let's consider: ZW2=(r1cos(θ1)r2cos(θ2))2+(r1sin(θ1)r2sin(θ2))2|Z - W|^2 = (r_1 cos(\theta_1) - r_2 cos(\theta_2))^2 + (r_1sin(\theta_1) - r_2sin(\theta_2))^2 ZW2=r12cos2(θ1)+r22cos2(θ2)2r1r2cos(θ1)cos(θ2)+r12sin2(θ1)+r22sin(θ2)2r1r2sin(θ1)sin(θ2)|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

ZW2=r12+r222r1r2cos(θ1θ2)|Z - W|^2 = r_1^2 + r_2^2 - 2r_1 r_2 cos(\theta_1 - \theta_2)

Subtracting and adding 2r1r22r_1 r_2 allows us to factorise as follows:

ZW2=(r1r2)2+2r1r22r1r2cos(θ1θ2)|Z - W|^2 = (r_1 - r_2)^2 + 2r_1 r_2 - 2r_1 r_2 cos(\theta_1 - \theta_2)

ZW2=(r1r2)2+2r1r2(1cos(θ1θ2))|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,

ZW2=(r1r2)2+4r1r2(sin2(θ1θ22))|Z - W|^2 = (r_1 - r_2)^2 + 4r_1 r_2(sin^2(\frac{\theta_1 - \theta_2}{2}))

Now, r1,r21r_1, r_2 \leq 1 and we allegedly know that θsin(θ)|\theta| \geq |sin(\theta)| which can be easily proven.

Thus, we can say, ZW2(r1r2)2+4(θ1θ22)2|Z - W|^2 \leq (r_1 - r_2)^2 + 4(\frac{\theta_1 - \theta_2}{2})^2

ZW2(r1r2)2+(θ1θ2)2|Z - W|^2 \leq (r_1 - r_2)^2 + (\theta_1 - \theta_2)^2, by how we defined ZZ and WW, θ1\theta_1 and θ2\theta_2 are their principal arguments and r1r_1 and r2r_2 being their modulus respectively.

Therefore, ZW2(ZW)2+(Arg(Z)Arg(W))2|Z - W|^2 \leq (|Z| - |W|)^2 + (Arg(Z) - Arg(W))^2.

As required. 🌝🌚

Please kindly point out any errors if there are any.

Vishnuram Leonardodavinci - 3 years, 11 months ago

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Wow, I am impressed by this approach _

Nihar Mahajan - 3 years, 11 months ago

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^^ Glad you liked it.

Vishnuram Leonardodavinci - 3 years, 11 months ago

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Is it clear now brother?

Mohaiminul Adil - 3 years, 11 months ago

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

Q2: z1z2=z2z3=z3z1|z_1 - z_2| = |z_2 - z_3| = |z_3 - z_1| because it's an equilateral triangle.

(a1)+(1b)i=1+bi=ai|(a - 1) + (1-b)i| = |1 + bi|= |-a - i|

a22a+1+12b+b2=1+b2=a2+1a^2 - 2a + 1 + 1 - 2b + b^2 = 1 + b^2 = a^2 + 1

Which gives us two systems of equations, namely, a22a+22b+b2=1+b2a^2 - 2a + 2 - 2b + b^2 = 1 + b^2 and 1+b2=a2+11 + b^2 = a^2 + 1.

Simplifying them further yields, a22a+12b=0a^2 - 2a + 1 - 2b = 0 and a2=b2a^2 = b^2.

Since 0<a,b<1a=b0 < a,b < 1 \Rightarrow a = b from the second equation.

Noting the constraint and solving the system gives us the following solution: a=b=23a = b = 2 - \sqrt{3}

Vishnuram Leonardodavinci - 3 years, 11 months ago

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

z1=1+i3z_1 = 1 + i \sqrt{3} can be written as 2eiπ/32e^{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 z2z_2 and z3z_3 as π\pi and π/3-\pi/3.

Also, they all lie on the circumference of the circle z2=z3=2\Rightarrow |z_2| = |z_3| = 2

Hence, z2=2cis(π)z_2 = 2cis(\pi) and z3=2cis(π/3)z_3 = 2cis(-\pi/3)

In Cartesian form, z2=2z_2 = -2 and z3=1i3z_3 = 1 - i \sqrt{3}

Vishnuram Leonardodavinci - 3 years, 11 months ago

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

Also I got Q5 ;)

Nihar Mahajan - 3 years, 11 months ago

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Anytime ;)

Tanishq Varshney - 3 years, 11 months ago

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Ur welcome ¨\Huge \ddot{\smile}

Surya Prakash - 3 years, 11 months ago

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No worries. :D Glad to be of help.

Vishnuram Leonardodavinci - 3 years, 11 months ago

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