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If z be any complex number (z!=0) then ...

If z be any complex number `(z!=0)` then `arg((z-i)/(z+i))=pi/2`represents the curve

A

`|z|=1`

B

`|z|=1, "Re"(z) gt 0`

C

`|z|=1, "Re"(z) lt 0`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
We have,
`"arg"(z-i)/(z+i)=pi/2`
`rArr` z lies on the lift side of the circle having (0,1) and (0,-2) as the end point of a diameter.
`rArr |z|=1` and Re(z) `lt 0`.
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OBJECTIVE RD SHARMA-COMPLEX NUMBERS -Chapter Test
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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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  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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  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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