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z and w are two nonzero complex numbers such that  z  =  w and Arg z + Arg w = π then z equals [2002]
(b) Let  z  =  ω  = r
If  z – 4  <  z – 2 , its solution is given by [2002]
(c) Given  z – 4  <  z – 2 
Let z = x + iy
⇒  (x – 4) + iy)  <  (x – 2) + iy 
⇒ (x – 4)^{2} + y^{2} < (x – 2)^{2} + y^{2}
⇒ x^{2} – 8x + 16 < x^{2} – 4x + 4
⇒ 12 < 4x
⇒ x > 3
⇒ Re(z) > 3
The locus of the centre of a circle which touches the circle  z – z_{1}  = a and  z – z_{2}  = b externally (z , z_{1} & z_{2} are complex numbers) will be [2002]
(b) Let the circle be z – z_{0} = r. Then according to given conditions z_{0} – z_{1} = r + a and z_{0} – z_{2}= r + b.
Eliminating r, we get z_{0 }– z_{1} –z_{0} – z_{2} = a – b.
∴ Locus of centre z_{0} is z – z_{1} –z – z_{2} = a – b,
which represents a hyperbola.
If z and ω are two nonzero complex numbers such that zω = 1 and Arg(z)  Arg(ω) = , then is equal to [2003]
(a)  =  w= z ω=zω= 1
Arg()= arg() + arg(ω)=  arg(z) + argω
= 
∴ =1
Let Z_{1} and Z _{2 }be two roots of the equation Z^{2} + aZ +b= 0 , Z being complex. Further , assume that the origin, Z_{1 }and Z_{2} form an equilateral triangle. Then [2003]
z^{ 2} + az +b= 0 ; z_{1 }+ z_{2 }=  a & z_{1} z_{2}=b 0, z_{1},z_{2 }form an equilateral Δ
∴ 0^{2} + z_{1}^{2} + z_{2}^{2} = 0.z_{1} + z_{1}.z_{2}+z_{2} .0 (for an equilateral triangle,
z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1}z_{2}+ z_{2} z_{3}+ z_{3}z_{1} )
⇒ z_{1}^{2} + z_{2}^{2}= z_{1}z^{2}
⇒ ( z_{1} + z_{2} )^{2}=3 z_{1}z^{2}
∴a^{2}= 3b
⇒
= ⇒ (i)^{x} = 1;
Let z and w be complex numbers such that = 0 and arg zw = π. Then arg z equals [2004]
arg zw = π ⇒ arg z + argw = π...(1)
If z = x iy and = p+ iq, then is equal to [2004]
= p+ iq ⇒ z = p^{3 }+ (iq)^{3} + 3 p (iq)( p+ iq)
⇒ x  iy = p^{3}  3pq^{2} + i(3p^{2}qq^{3})
∴ x = p^{3 } 3pq^{2} ⇒
y = q^{3}  3p^{2}q ⇒
If  z^{2}  1 =z ^{2}+1, then z lies on [2004]
z^{2 } 1=z^{2}+1 ⇒z^{2}  1^{2} =
⇒
⇒
⇒ ⇒
⇒ z is purely imaginary
If the cube roots of unity are 1,ω ,ω^{2 }then the roots of the equation ( x – 1)^{3} + 8 = 0, are [2005]
(x  1)^{3} + 8=0
⇒ (x  1) = (2) (1)^{1 /3 }
⇒ x – 1 = – 2 or 2ω or  2ω^{2}
or x = – 1 or 1 – 2ω or 1 – 2ω^{2} .
If z_{1} and z_{2} are two non zero complex numbers such that  z_{1} +z_{2} =  z1 +  z_{2} , then arg z_{1} – arg z_{2} is equal to [2005]
z_{1 }+z_{2} = z_{1} + z_{2} ⇒ z_{1} and z_{2} are collinear and are to the same side of origin; hence arg z_{1} – arg z_{2} = 0.
If and w = 1, then z lies on [2005]
As given w ⇒ w = = 1
⇒
⇒ distance of z from origin and point is
same hence z lies on bisector of the line joining points (0, 0) and (0, 1/3).
Hence z lies on a straight line.
The value of is [2006]
= i × 0 – i [∵ e^{2πi }= 1] = – i
If z^{2} +z + 1=0 , where z is complex number, then the value of
[2006]
z^{2} +z + 1=0 ⇒ z = ω or ω^{2}
So,
and
∴ The given sum = 1+1 + 4 + 1 + 1 + 4 = 12
If  z + 4  ≤ 3, then the maximum value of  z + 1  is[2007]
z lies on or inside the circle with centre (–4, 0) and radius 3 units.
From the Argand diagram maximum value of  z + 1 is 6
The conjugate of a complex n umber then that complex number is [2008]
Let R be the real line. Consider the following subsets of the plane R × R:
S ={(x, y): y = x + 1 and 0 < x < 2}
T ={(x, y): x – y is an integer},
Which one of the following is true? [2008]
Given S = {(x , y) : y = x + 1 and 0 < x < 2}
∵ x ≠ x + 1 for any x ∈(0, 2) ⇒ (x, x) ∉ S
∴ S is not reflexive.
Hence S in not an equivalence relation.
Also T ={x, y): x – y is an integer}
∵ x – x = 0 is an integer " x ∈ R
∴ T is reflexive.
If x – y is an integer then y – x is also an integer
∴T is symmetric If x – y is an integer and y – z is an integer then (x – y) + (y– z) = x – z is also an integer.
∴ T is transitive Hence T is an equivalence relation.
The number of complex numbers z such that z – 1 = z + 1 = z – i equals [2010]
Let z = x+ iy
z 1 = z +1 (x 1)^{2} + y^{2} = (x +1)^{2}+y^{2}
⇒ Re z =0 ⇒ x = 0
z 1 = zi (x1)^{2}+ y^{2}= x^{2} +(y 1)^{2 }
⇒ x =y
z+1 = zi (x+1)^{2} + y^{2 }= x^{2} +(y1)^{2 }
Only (0, 0) will satisfy all conditions.
⇒ Number of complex number z = 1
Let α, β be real and z be a complex number. If z^{2} + αz + β = 0 has two distinct roots on the line Re z =1, then it is necessary that :[2011]
∵ Real part of roots is 1 Let roots are 1 + pi, 1 + q
∴ sum of roots = 1 + pi + 1 + qi = – a which is real
⇒ q = – p or root are 1 + pi and 1 – pi product of roots = 1 + p^{2} = β∈ (1,)
p ≠ 0 as roots are distinct.
If ω(≠1) is a cube root of unity, and (1 + ω)^{7} = A +Bω.Then (A, B) equals [2011]
(1 + ω)^{7} = A + Bω;
(–ω^{2})^{7} = A + Bω – ω^{2 }= A + Bω;
1 + ω = A + Bω
⇒ A = 1, B = 1.
If z ≠ 1 and is real, then the point represented by the complex number z lies : [2012]
Now is real ⇒ Im = 0
⇒
⇒ Im [(x^{2} – y^{2} + 2ixy) (x – 1) – iy)] = 0
⇒ 2xy (x – 1) – y (x^{2} – y^{2}) = 0
⇒ y(x^{2} + y^{2} – 2x) = 0 ⇒ y = 0; x^{2} + y^{2} – 2x = 0
∴ z lies either on real axis or on a circle through origin.
If z is a complex number of unit modulus and argument θ, then arg equals: [JEE M 2013]
Given  z  = 1, arg z = θ
As we know,
∴ = arg (z) = θ
If z is a complex number such that z ≥ 2, then the minimum value of
We know minimum value of
Thus minimum value of
Since,  Z ≥2 therefore
⇒
A complex number z is said to be unimodular if z = 1.
Suppose z_{1} and z_{2 }are complex numbers such that is unimodular and z_{2} is not unimodular. Then the point z_{1} lies on a: [JEE M 2015]
⇒
⇒
⇒
⇒
⇒
⇒
⇒ Point z_{1} lies on circle of radius 2.
A value of θ for which is purely imaginary, is: [JEE M 2016]
Rationalizing the given expression
For the given expression to be purely imaginary, real part of the above expression should be equal to zero.
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