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Question: If \(\sqrt{2}\left( \cos\frac{\pi}{12} - i\sin\frac{\pi}{12} \right)\text{ },\sqrt{2}\left( - \sin\...

If 2(cos⁡π12−isin⁡π12) ,2(−sin⁡π12+icos⁡π12) ,−1−i\sqrt{2}\left( \cos\frac{\pi}{12} - i\sin\frac{\pi}{12} \right)\text{ },\sqrt{2}\left( - \sin\frac{\pi}{12} + i\cos\frac{\pi}{12} \right)\ , - 1 - i then x is equal to.

A

(2−2i)1/3(2 - 2i)^{1/3}

B

Re(z)>0,Im(z)<0{Re}(z) > 0,{Im}(z) < 0

C

Re(z)>0,Im(z)>0{Re}(z) > 0,{Im}(z) > 0

D

Im(z)=0{Im}(z) = 0

Answer

Re(z)>0,Im(z)>0{Re}(z) > 0,{Im}(z) > 0

Explanation

Solution

x+1x=2cos⁡θx + \frac { 1 } { x } = 2 \cos \theta z12+z22+z32=z1z2+z2z3+z3z1z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1}z_{2} + z_{2}z_{3} + z_{3}z_{1}

⇒z1=1,z2=ω,z3=ω2z_{1} = 1,z_{2} = \omega,z_{3} = \omega^{2} ⇒ z12+z22+z32=1+ω2+ω4=0z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 1 + \omega^{2} + \omega^{4} = 0.