JEE maths #8

Algebra Level 5

Find the number of complex numbers z z satisfying the conditions ∣ z z ˉ + z ˉ z ∣ = 1 \left| \dfrac z{\bar z} + \dfrac {\bar z}z \right| = 1 and ∣ z ∣ = 1 |z| = 1 .

1 2 8 4 5

Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Chew-Seong Cheong
Mar 30, 2017

From ∣ z ∣ = 1 |z|=1 , ⟹ z = e θ i \implies z = e^{\theta i} . Then

∣ z z ˉ + z ˉ z ∣ = 1 ∣ e θ i e − θ i + e − θ i e θ i ∣ = 1 ∣ e 2 θ i + e − 2 θ i ∣ = 1 ∣ 2 cos ⁡ 2 θ ∣ = 1 cos ⁡ 2 θ = ± 1 2 2 θ = ( n ± 1 3 ) π where n = 0 , 1 , 2... θ = 3 n ± 1 6 π = π 6 , π 3 , 2 π 3 , 5 π 6 , 7 π 6 , 4 π 3 , 5 π 3 , 11 π 6 \begin{aligned} \left| \frac z{\bar z} + \frac {\bar z}z \right| & = 1 \\ \left| \frac {e^{\theta i}}{e^{-\theta i}} + \frac {e^{-\theta i}}{e^{\theta i}} \right| & = 1 \\ \left| e^{2\theta i} + e^{-2\theta i} \right| & = 1 \\ \left| 2\cos 2\theta \right| & = 1 \\ \cos 2\theta & = \pm \frac 12 \\ 2\theta & = \left({\color{#3D99F6}n} \pm \frac 13\right) \pi \quad \quad \small \color{#3D99F6} \text{where }n=0,1,2... \\ \theta & = \frac {3n \pm 1}6 \pi \\ & = \frac \pi 6, \frac \pi 3, \frac {2\pi}3, \frac {5\pi}6, \frac {7\pi}6, \frac {4\pi}3, \frac {5\pi}3, \frac {11\pi}6 \end{aligned}

Therefore, there are 8 \boxed{8} solutions.

7 Helpful 0 Interesting 0 Brilliant 0 Confused

∣ z z ˉ + z ˉ z ∣ = 1. ⟹ a + i b a − i b + a − i b a + i b = ± 1 ∴ ( a + i b ) 2 + ( a − i b ) 2 a 2 + b 2 = + 1..... ( 1 ) a n d ( a + i b ) 2 + ( a − i b ) 2 a 2 + b 2 = − 1........ ( 2 ) ( 1 ) 2 ( a 2 − b 2 ) = a 2 + b 2 , ⟹ a 2 − 3 b 2 = 0. ∴ a 2 = 3 b 2 ⟹ ± a a n d ± b . . . 4 n u m b e r s . ( 2 ) 2 ( a 2 − b 2 ) = − a 2 − b 2 , ⟹ 3 a 2 − b 2 = 0. ∴ 3 a 2 = b 2 ⟹ ± a a n d ± b . . . 4 n u m b e r s . T o t a l 8 n u m b e r s . \left| \dfrac z{\bar z} + \dfrac {\bar z}z \right| = 1.\\ ~~~~\\ \implies~\dfrac{a+ib}{a-ib} +\dfrac{a-ib}{a+ib} =\pm 1 \\ ~~~~\\ \therefore~\dfrac{(a+ib)^2+(a-ib)^2}{a^2+b^2}=+ 1.....(1)\\ ~~~~\\ and~\dfrac{(a+ib)^2+(a-ib)^2}{a^2+b^2}=-1........(2)\\ ~~~~\\ (1)~~2(a^2-b^2)=a^2+b^2,~~\implies~a^2-3b^2=0.\\ ~~~~\\ \therefore~a^2=3b^2~~\implies~\pm~ a~ and~\pm~b...4~numbers.\\ ~~~~\\ (2)~~2(a^2-b^2)=-a^2-b^2,~~\implies~3a^2-b^2=0.\\ ~~~~\\ \therefore~3a^2=b^2~~\implies~\pm~ a~ and~\pm~b...4~numbers.\\ ~~~~\\ Total~~8~numbers.

Niranjan Khanderia - 3 years, 2 months ago
Rahil Sehgal
Mar 30, 2017

4 Helpful 0 Interesting 0 Brilliant 0 Confused

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