JEE Maths#4

Algebra Level 4

Let z 1 z_1 , z 2 z_2 and z 3 z_3 be three points on ∣ z ∣ = 1 |z|=1 in the complex plane and z 1 + z 2 + z 3 = 0 z_1 + z_2 + z_3 = 0 . If θ 1 \theta_1 , θ 2 \theta_2 and θ 3 \theta_3 are the arguments of z 1 z_1 , z 2 z_2 and z 3 z_3 respectively, find the value of

cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) \cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos (\theta_3 -\theta_1)

-1.5 2 1 1.5 -2 0

Rahil Sehgal by Rahil Sehgal , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

5 solutions

Chew-Seong Cheong
Mar 29, 2017

Since ∣ z 1 ∣ = ∣ z 2 ∣ = ∣ z 3 ∣ = 1 |z_1| = |z_2| = |z_3| = 1 , ⟹ { z 1 = e i θ 1 = cos ⁡ θ 1 + i sin ⁡ θ 1 z 2 = e i θ 2 = cos ⁡ θ 2 + i sin ⁡ θ 2 z 3 = e i θ 3 = cos ⁡ θ 3 + i sin ⁡ θ 3 \implies \begin{cases} z_1 = e^{i\theta_1} = \cos \theta_1 + i\sin \theta_1 \\ z_2 = e^{i\theta_2} = \cos \theta_2 + i\sin \theta_2 \\ z_3 = e^{i\theta_3} = \cos \theta_3 + i\sin \theta_3 \end{cases}

Therefore, from z 1 + z 2 + z 3 = 0 z_1 + z_2 + z_3 = 0 , ⟹ cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 + i ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) = 0 \implies \cos \theta_1 + \cos \theta_2 + \cos \theta_3 + i(\sin \theta_1 + \sin \theta_2 + \sin \theta_3) = 0 and:

{ cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 = 0 . . . ( 1 ) sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 = 0 . . . ( 2 ) \begin{cases} \cos \theta_1 + \cos \theta_2 + \cos \theta_3 = 0 & ...(1) \\ \sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 0 & ...(2) \end{cases}

Squaring equation (1) and equation (2):

( cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 ) 2 = 0 cos ⁡ 2 θ 1 + cos ⁡ 2 θ 2 + cos ⁡ 2 θ 3 + 2 ( cos ⁡ θ 1 cos ⁡ θ 2 + cos ⁡ θ 2 cos ⁡ θ 3 + cos ⁡ θ 3 cos ⁡ θ 1 ) = 0 . . . ( 3 ) ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) 2 = 0 sin ⁡ 2 θ 1 + sin ⁡ 2 θ 2 + sin ⁡ 2 θ 3 + 2 ( sin ⁡ θ 1 sin ⁡ θ 2 + sin ⁡ θ 2 sin ⁡ θ 3 + sin ⁡ θ 3 sin ⁡ θ 1 ) = 0 . . . ( 4 ) \begin{aligned} (\cos \theta_1 + \cos \theta_2 + \cos \theta_3)^2 & = 0 \\ \cos^2 \theta_1 + \cos^2 \theta_2 + \cos^2 \theta_3 + 2(\cos \theta_1\cos \theta_2 + \cos \theta_2\cos \theta_3 + \cos \theta_3 \cos \theta_1) & = 0 & ...(3) \\ (\sin \theta_1 + \sin \theta_2 + \sin \theta_3)^2 & = 0 \\ \sin^2 \theta_1 + \sin^2 \theta_2 + \sin^2 \theta_3 + 2(\sin \theta_1\sin \theta_2 + \sin \theta_2\sin \theta_3 + \sin \theta_3 \sin \theta_1) & = 0 & ...(4) \end{aligned}

Adding equation (3) and equation (4):

3 + 2 ( cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) ) = 0 ⟹ cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) = − 3 2 = − 1.5 \begin{aligned} 3 + 2(\cos (\theta_1- \theta_2) + \cos (\theta_2- \theta_3) + \cos (\theta_3- \theta_1)) & = 0 \\ \implies \cos (\theta_1- \theta_2) + \cos (\theta_2- \theta_3) + \cos (\theta_3- \theta_1) & = - \frac 32 = \boxed{-1.5} \end{aligned}

15 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir, I did it with the help of geometry...

Since, ∣ z 1 ∣ = ∣ z 2 ∣ = ∣ z 3 ∣ = 1 a n d z 1 + z 2 + z 3 = 0 |z1| = |z2| = |z3| = 1 and z1 + z2 + z3 = 0

Therefore, Triangle formed will be an equilateral triangle.

( θ 1 − θ 2 ) = ( θ 2 − θ 3 ) = ( θ 3 − θ 2 ) = 2 π / 3 ( \theta1- \theta2) = ( \theta2- \theta3) = ( \theta3- \theta2)= 2π/3

c o s ( θ 1 − θ 2 ) + c o s ( θ 2 − θ 3 ) + c o s ( θ 3 − θ 2 ) = − 3 / 2 cos(\theta1- \theta2) + cos( \theta2- \theta3)+ cos( \theta3- \theta2)= -3/2 which is our answer.

Rahil Sehgal - 4 years, 2 months ago

Log in to reply

You should put a backslash "\" in front of all functions in LaTex. See if I enter "cos x" in LaTex I get " c o s x cos x ". cos is in italic and there is no space between cos and x. Now if I enter "\cos x", I get " cos ⁡ x \cos x ". Note that cos is not in italic, which is for variable and not function, and there is a space between cos and x.

Chew-Seong Cheong - 4 years, 2 months ago

Log in to reply

Thank you sir... I will take care of it the next time...

Rahil Sehgal - 4 years, 2 months ago

The vectors are three radius at 120 degrees with each other in a unit circle. 3 * Cos120=-1.5. up voted.

Niranjan Khanderia - 3 years, 2 months ago

Log in to reply

Thanks. Glad that you like the solution.

Chew-Seong Cheong - 3 years, 2 months ago
Abhiram Bondada
Apr 2, 2017

Imagine they are vectors of equal magnitude(as it is given so) and as their sum is 0 (also given in question z1 +z2 +z3 =0) so by seeing these things we can say that the vectors are long the medians of an equilateral triangle

The angle between medians is 120 so take three vectors a,b,c at angles 0, 120 ,240 (for getting answer easily) Now substitute values in the given equation of cos(z1-z2) blah blah and you get -1.5

Ans -1.5

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution (+1).... I did the same way :)

Rahil Sehgal - 4 years, 2 months ago

choose 3 points as 60 -60 180.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@shubham dhull :- did the same....

Quasar An - 4 years, 2 months ago
Aman Bhandare
Mar 29, 2017

Answer is -1.5 I just plugged in cube roots of unity for z1, z2 and z3. It also gave the correct answer but the basic method is always preferable.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

S i n c e ∣ z 1 ∣ = ∣ z 2 ∣ = ∣ z 3 ∣ = 1. Since~~ |z_1| = |z_2| = |z_3| = 1. { z 1 = cos ⁡ θ 1 + i sin ⁡ θ 1 z 2 = cos ⁡ θ 2 + i sin ⁡ θ 2 z 3 = cos ⁡ θ 3 + i sin ⁡ θ 3 F r o m z 1 + z 2 + z 3 = 0. ⟹ cos ⁡ θ 1 + cos ⁡ θ 2 + cos ⁡ θ 3 + i ( sin ⁡ θ 1 + sin ⁡ θ 2 + sin ⁡ θ 3 ) = 0 ∴ cos ⁡ θ a + cos ⁡ θ b = − cos ⁡ θ c . . . . . . . . . . ( P ) & sin ⁡ θ a + sin ⁡ θ b = − sin ⁡ θ c . . . . . . . . . . . ( Q ) ∴ ( P ) 2 + ( Q ) 2 : 2 + 2 ∗ ( cos ⁡ θ a ∗ cos ⁡ θ b + sin ⁡ θ a ∗ sin ⁡ θ b ) = 1. ⟹ cos ⁡ ( θ a − θ b ) = − 0.5. ⟹ cos ⁡ ( θ 1 − θ 2 ) = cos ⁡ ( θ 2 − θ 3 ) = cos ⁡ ( θ 3 − θ 1 ) = − 0.5. ∴ cos ⁡ ( θ 1 − θ 2 ) + cos ⁡ ( θ 2 − θ 3 ) + cos ⁡ ( θ 3 − θ 1 ) = − 1.5 . I n c i d e n t a l l y i f z 1 + z 2 + z 3 = 0 , t h e n t h e v e c t o r s a r e r a d i i e a c h a t 12 0 o w i t h e a c h o t h e r . \begin{cases}z_1 = \cos \theta_1 + i\sin \theta_1 \\ z_2 = \cos \theta_2 + i\sin \theta_2 \\ z_3 = \cos \theta_3 + i\sin \theta_3 \end{cases}\\ ~~~\\ ~~~ \\ From~~z_1 + z_2 + z_3 = 0. \\ \implies \cos \theta_1 + \cos \theta_2 + \cos \theta_3 + i(\sin \theta_1 + \sin \theta_2 + \sin \theta_3) = 0\\ ~~~~\\ ~~~\\ \therefore~~ \cos \theta_a + \cos \theta_b = - \cos \theta_c..........(P)\\ \&~~~\sin \theta_a + \sin \theta_b = - \sin \theta_c. ..........(Q)\\ ~~~~\\ ~~~~\\ \therefore~~(P)^2 + (Q)^2 :~~~~~~~ 2+2*(\cos \theta_a * \cos \theta_b + \sin \theta_a * \sin \theta_b)=1.\\ \implies~\cos( \theta_a- \theta_b)= - 0.5.\\ ~~~~\\ ~~~~\\ \implies~\cos(\theta_1-\theta_2) = \cos(\theta_2-\theta_3) = \cos (\theta_3 -\theta_1)= - 0.5.\\ \therefore~~\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos (\theta_3 -\theta_1)=\huge~~\color{#D61F06}{- 1.5}.\\ ~~~~~\\ ~~~\\ Incidentally~ if~~z_1 + z_2 + z_3 = 0,~then~the~vectors~are~radii~each~at~120^o~with~ each~ other.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...