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If z(2-2sqrt(3i))^2=i(sqrt(3)+i)^4, then...

If `z(2-2sqrt(3i))^2=i(sqrt(3)+i)^4,` then `a r g(z)=`

A

`(5pi)/6`

B

`-pi/6`

C

`pi/6`

D

`(7pi)/6`

Text Solution

Verified by Experts

The correct Answer is:
B
We have,
`z(2-2sqrt(3)i)^(2)=i(sqrt(3)+i)^(4)`
`rArr 4z(1-isqrt(3))^(2)=i^(5)(1-isqrt(3))^(4)`
`rArr 4z=i(1-isqrt(3))^(2)`
`rArr 4z=i(-2omega)^(2)=4iomega^(2) rArr z=iomega^(2)=sqrt(30/2-1/2i`
Clearly, arg(z) `=-pi/6`
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OBJECTIVE RD SHARMA-COMPLEX NUMBERS -Chapter Test
  1. If z(2-2sqrt(3i))^2=i(sqrt(3)+i)^4, then a r g(z)=

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  2. The locus of the center of a circle which touches the circles |z-z1|=a...

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  3. If n1, n2 are positive integers, then (1 + i)^(n1) + ( 1 + i^3)^(n1) +...

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  4. The modulus of sqrt(2i)-sqrt(-2i) is

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  5. Prove that the triangle formed by the points 1,(1+i)/(sqrt(2)),a n di ...

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  6. The value of (1+isqrt(3))/(1-isqrt(3))^(6)+(1-isqrt(3))/(1+isqrt(3))^(...

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  7. If alpha+ibeta=tan^(-1) (z), z=x+iy and alpha is constant, the locus o...

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  8. If cosA+cosB+cosC=0,sinA + sinB + sinC=0and A+B+C=180^0, then the valu...

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  9. The value of the expression 1.(2-omega).(2-omega^2)+2.(3-omega)(3-omeg...

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  10. The value of the expression (1+1/omega)(1+1/omega^(2))+(2+1/omega)(2+...

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  11. The condition that x^(n+1)-x^(n)+1 shall be divisible by x^(2)-x+1 is ...

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  12. The expression (1+i)^(n1)+(1+i^(3))^(n(2)) is real iff

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  13. |{:("6i " "-3i " "1" ),("4 " " 3i" " -1"),("20 " "3 " " i"):}|=x+iy th...

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  14. The centre of a square ABCD is at z0dot If A is z1 , then the centroid...

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  15. If cosalpha+2cosbeta+3cosgamma=sinalpha+2sinbeta+3singamma=0 and alpha...

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  16. If cosalpha+2cosbeta+3cosgamma=sinalpha+2sinbeta+3singamma=0 and alpha...

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  17. Sum of the series sum(r=0)^n (-1)^r ^nCr[i^(5r)+i^(6r)+i^(7r)+i^(8r)] ...

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  18. If az(1)+bz(2)+cz(3)=0 for complex numbers z(1),z(2),z(3) and real num...

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  19. If 2z1-3z2 + z3=0, then z1, z2 and z3 are represented by

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  20. Re((z+4)/(2z-1)) = 1/2, then z is represented by a point lying on

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  21. The vertices of a square are z1,z2,z3 and z4 taken in the anticlockwis...

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