Let a,b
be two real numbers such that ab < 0
. IF the complex number
is of unit modulus and a + ib
lies on the circle |z - 1| = |2z|
, then a possible value of
, where [t]
is greatest integer function, is :
1)
2) 1/2
3) 0
4) -1
3
ans : 3
= 1 ⇒ |1 + ai | = |b+ i|
⇒ a 2 + 1 = b 2 + 1
⇒ a = b
⇒ ab < 0
⇒ (a + i b) lies on |z-1| = |2z|
⇒ |a + i b - 1| = 2 |a+ ib|
⇒ 6a2 + 2 a - 1 = 0
⇒ a = and
⇒ [a] = 0
∴ = 0
⇒ similarly when
⇒ [a] = -1
= 0 = 0
= 1 ⇒ |1 + ai | = |b+ i|
⇒ a 2 + 1 = b 2 + 1
⇒ a = b
⇒ ab < 0
⇒ (a + i b) lies on |z-1| = |2z|
⇒ |a + i b - 1| = 2 |a+ ib|
⇒ 6a2 + 2 a - 1 = 0
⇒ a = and
⇒ [a] = 0
∴ = 0
⇒ similarly when
⇒ [a] = -1
= 0 = 0
If the center and radius of the circle
are respectively (P,Q)
and R
, then 3(P+Q+R)
is equal to
1) 12
2) 10
3) 11
4) 9
1
solution:
ans : 1
|z - 2| = 2 |z-3|
⇒ x2 y 2 - 4x +4 = 4x2 4y 2 - 24x +36
⇒ x2 y 2 - 20/3x + 32/3 = 0
⇒ (P,Q) = (10/3 , 0) and using radius formula from the circles, we can find R = 2/3 therefore 3(P+Q+R) = 12
|z - 2| = 2 |z-3|
⇒ x2 y 2 - 4x +4 = 4x2 4y 2 - 24x +36
⇒ x2 y 2 - 20/3x + 32/3 = 0
⇒ (P,Q) = (10/3 , 0) and using radius formula from the circles, we can find R = 2/3 therefore 3(P+Q+R) = 12
For two non-zero complex numbers z1
and z2
, if Re(z1z2)
and
, then which of the following are possible?
A. Im(z1) > 0
and Im(z2) > 0
B. Im(z1) < 0
and Im(z2) > 0
C. Im(z1) > 0
and Im(z2) < 0
D. Im(z1) < 0
and Im(z2 ) < 0
Choose the correct answer from the options given below:
1) A & B
2) A & C
3) B & D
4) B & C
4
sol:
ans : 4
Let, z1 = x1 + i y1
and z2 = x2 + i y2
⇒ z1z2 = x1x2 - y1y2 + i(x1y2 + x1y2)
Given, Re(z1 + z2) = 0
⇒ x1 + x2 = 0 x1x2 - y1y2 = 0 since x1 + x2 = 0 y1y2 = - xx2 So, multiplication of imaginary part's of z1 and z2 is negative. It means sign of y1 and y2 are opposite of each other. ∴ B and C are correct.
⇒ z1z2 = x1x2 - y1y2 + i(x1y2 + x1y2)
Given, Re(z1 + z2) = 0
⇒ x1 + x2 = 0 x1x2 - y1y2 = 0 since x1 + x2 = 0 y1y2 = - xx2 So, multiplication of imaginary part's of z1 and z2 is negative. It means sign of y1 and y2 are opposite of each other. ∴ B and C are correct.
Let z
be a complex number such that
. Then
lies on the circle of radius 2 and centre
1) (0, -2)
2)(0,0)
3) (0,2)
4) (2,0)
1
sol: ans:1
center or
center or
Let z1 = 2 + 3 and z 2 = 3 + 4i
. The set represents a
1) hyperbola with eccentricity = 2
2) straight line with the sum of its intercepts on the coordinates axes is -18
3) hyperbo;a with the length of transverse axis = 7
4) straight line with the sum of its intecepts on the coordinated axes = 14
4
sol:
ans : 4
The value of is:
1)
2)
3)
4)
2
solution:
ans: 2
Let
and
then
and
are roots of the equation.
1)
2)
3)
4)
2
sol: ans:2
p = -1 and q =
p = -1 and q =
If
be a complex number such that
, then the maximum value of
is :
1)
2) 1
3)
4)
4
solution:
ans: 4
------(1)
or
------ (2) From (1) and (2) we get, Maximum value of and minimum value of
------(1)
or
------ (2) From (1) and (2) we get, Maximum value of and minimum value of
Let
. Then the set of all values of
, for which
for some
, is
1)
2)
3)
4)
2
If
, then
is equal to :
1) 244
2) 224
3) 245
4) 265
1
ans: 1
Let S be the set of all
, for which the complex number
is purely imaginary and
is purely real. Let
. Then
is equal to :
1) 3
2) 3i
3) 1
4) 2 - i
3
solution: ans:3
is purely imaginary
is purely imaginary
Let the minimum value
of
is attained at
. Then
is equal to
1) 1000
2) 1024
3) 1105
1196
1
solution:
ans: 1
If
satisfies
and
, then
1)
2)
3)
4)
3
sol: ans:3
So, lies on bisector of and i.e., line
So, lies on bisector of and i.e., line
If
are the roots of the equation
, then
is equal to :
1) -4
2) -1
3) 1
4) 4
2
solution:
ans= 2
has roots , , and 8.
Roots of are 1, , , , 8.We know, Sum of pth power of nth roots of unity = 0. (If p is not multiple of n) or n (If p is multiple of n)
Here, Sum of pth power of nth roots of unity
is not a multiple of 5.
has roots , , and 8.
Roots of are 1, , , , 8.We know, Sum of pth power of nth roots of unity = 0. (If p is not multiple of n) or n (If p is multiple of n)
Here, Sum of pth power of nth roots of unity
is not a multiple of 5.
For
, let
and
. Then the number of elements in the set
is :
1) 0
2) 2
3) 3
4) 4
4
solution:
ans: 4
represents a circle with centre and radius
similarly represents circle with centre and radius
represents a circle with centre and radius
similarly represents circle with centre and radius
The real part of the complex number
is equal to :
1)
2)
3)
4)
4
solution:
ans:4
real part
real part
Let
and
be the roots of the equation x2 + (2i - 1) = 0. Then, the value of | 8 +
8| is equal to:
1) 50
2) 250
3) 1250
4) 1500
1
solution:
ans: 1
as and are roots of the equation
as and are roots of the equation
The area of the polygon, whose vertices are the non-real roots of the equation
is :
1)
2)
3)
4)
1
sol:
ans: 1
case - 1
case - 2
area of polygon =
case - 1
case - 2
area of polygon =
Let z1 and z2 be two complex numbers such that
and
. Then
1)
2)
3)
4)
3
solution: ans:3
By solveing above equations we get, and
By solveing above equations we get, and
Let a circle C in complex plane pass through the points
,
and
. If
is a point on C such that the line through z and z1 is perpendicular to the line through z2 and z3, then arg(z)
is equal to:
1)
2)
3)
4)
2
solution: ans: 2
Let z(x,y)
z is intersection of C and L
Let z(x,y)
z is intersection of C and L
If
, then p and q are roots of the equation :
1)
2)
3)
4)
1
solution:
ans:1
p =
q =
p =
q =
If z and
are two complex numbers such that
and
, then
is :
(Here arg(z) denotes the principal argument of complex number z)
1)
2)
3)
4)
2
solution:
ans: 2
let
now consider
let
now consider
The least value of |z| where z is complex number which satisfies the inequality
, is equal to :
1) 8
2) 3
3) 2
4)
2
solution: ans: 2
Let |z| = t and t 0
t+1 0
Let |z| = t and t 0
t+1 0
Let a complex number z, |z|
1,
satisfy
. Then, the largest value of |z| is equal to ____________.
1) 5
2) 8
3) 6
4) 7
4
solution:
ans: 4
If
,
R are such that 1 -
2i (here i2 = -
1) is a root of z2 +
z +
= 0, then ( -
) is equal to :
1) -7
2) 7
3) 3
4) -3
1
solution:
ans: 1
1 - 2i is the root of the equation. So other root is 1 + 2i Sum of roots = 1 - 2i + 1 + 2i = 2 = - Product of roots = (1 - 2i)(1 + 2i) = 1 - 4i2 = 5 = - = -7
1 - 2i is the root of the equation. So other root is 1 + 2i Sum of roots = 1 - 2i + 1 + 2i = 2 = - Product of roots = (1 - 2i)(1 + 2i) = 1 - 4i2 = 5 = - = -7
Let the lines (2 -
i)z = (2 + i)
and (2 +
i)z + (i -
2)
- 4i = 0, (here i2 = -
1) be normal to a circle C. If the line iz +
+ 1 + i = 0 is tangent to this circle C, then its radius is :
1)
2)
3)
4)
1
solution:
ans:1
Solve L1 and L2: x = 1,
centre
Radius = distance from to
Solve L1 and L2: x = 1,
centre
Radius = distance from to
The value of is :
- 215i
- 215
215i
65
1
solution:
ans:1
=
=
=
=
= - 215i
=
=
=
=
= - 215i
If a and b are real numbers such that
where
then a + b is equal to:
1) 33
2) 9
3) 24
4) 57
2
solution:
ans: 2
given
given
Let
, z = x + iy and k > 0. If the curve represented
by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is :
2
4
1/2
3/2
1
solution:
ans:1
given z =x + iy
Given,
This curve intersect the y-axis at point P and Q, so at point P and Q x = 0 Putting x = 0 at the above equation,
Let roots of this quadratic equation y1 and y2 Point P (0, y1) and Q (0, y2)
Given PQ = 5
give k is greater than 0 so, k= 2
Given,
This curve intersect the y-axis at point P and Q, so at point P and Q x = 0 Putting x = 0 at the above equation,
Let roots of this quadratic equation y1 and y2 Point P (0, y1) and Q (0, y2)
Given PQ = 5
give k is greater than 0 so, k= 2
If z1 , z2 are complex numbers such that
Re(z1) = |z1 – 1|, Re(z2) = |z2 – 1| , and
arg(z1 - z2) =
, then Im(z1 + z2 ) is equal to :
1)
2)
3)
4)
4
solution:
ans: 4
Given Re(z1) = |z1 – 1| x1 = |(x1 - 1) + iy1|
x1
Also given Re(z2) = |z2 – 1| x2 = |(x2 - 1) + iy2|
Performing equation (1) - (2),
given,
Given Re(z1) = |z1 – 1| x1 = |(x1 - 1) + iy1|
x1
Also given Re(z2) = |z2 – 1| x2 = |(x2 - 1) + iy2|
Performing equation (1) - (2),
given,
The imaginary part of
can be :
1) -2
2) 6
3)
4) -
1
solution:
ans:1
Similarly,
Possible imaginary parts are
Similarly,
Possible imaginary parts are
The value of
is :
1)
2) -
3)
4)
2
solution:
ans: 2
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
Let z be complex number such that
and |z| =
.
Then the value of |z + 3i| is :
1)
2)
3)
4)
4
solution:
ans: 4
|z – i| = |z + 2i|
(let z = x + iy) x2 + (y – 1)2 = x2 + (y + 2)2
y =
Also given |z| =
x2 + y2 =
x2 = 6
z =
|z + 3i| = =
|z – i| = |z + 2i|
(let z = x + iy) x2 + (y – 1)2 = x2 + (y + 2)2
y =
Also given |z| =
x2 + y2 =
x2 = 6
z =
|z + 3i| = =
If
, where z = x + iy, then the point (x, y) lies on a:
1) straight line whose slope is
2) straight line whose slope is
3) circle whose diameter is
4) circle whose centre is at
4
solution:
ans: 3
Put z = x + iy
Real part of this equation is = 1
2x2 + 2y2 +2x + 3y + 1 = 0
This is an equation of circle. Locus is a circle whose center is and radius
Diameter = 2
=
Put z = x + iy
Real part of this equation is = 1
2x2 + 2y2 +2x + 3y + 1 = 0
This is an equation of circle. Locus is a circle whose center is and radius
Diameter = 2
=
Let z
C with Im(z) = 10 and it satisfies
= 2i - 1 for some natural number n. Then :
1) n = 20 and Re(z) = -10
2) n = 40 and Re(z) = 10
3) n = 40 and Re(z) = -10
4) n= 20 and Re(z) = 10
3
solution:
ans: 3
Let Re (z) = x, then
& 20 = -20 + 2n x = -10 & n = 20
Let Re (z) = x, then
& 20 = -20 + 2n x = -10 & n = 20
The equation |z – i| = |z – 1|, i =
, represents :
1) a circle of radius of 1
2) the line through the origin with slope -1
3) a circle of radius
4) the line through the origine with slope 1
4
solution: ans: 4
Let the complex number z = x + iy Now given | (x + iy) - 1 | = | (x+iy) - i | (x - 1)2 + y2 = x2 + (y - 1)2 x = y
Let the complex number z = x + iy Now given | (x + iy) - 1 | = | (x+iy) - i | (x - 1)2 + y2 = x2 + (y - 1)2 x = y
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) =
, then:
1)
2)
3)
4)
4
solution:
ans: 4
let
then z =
let
then z =
If a > 0 and z =
, has magnitude
, then
is equal to :
1)
2)
3)
4)
2
solution:
ans: 2
-------(1)
given
so from equation(1)
a = 3
a is greater than 3 a = 3
-------(1)
given
so from equation(1)
a = 3
a is greater than 3 a = 3
Let
be complex numbers satisfying
. Then the least value of
, such that
, is equal to __________.
6
solution:
ans: 6
----(1)
---(2)
then -----(3)
Now again from equation (1), equation (2), equation (3) we get :
and
and
or
then the minimum value of n is 6
----(1)
---(2)
then -----(3)
Now again from equation (1), equation (2), equation (3) we get :
and
and
or
then the minimum value of n is 6
Let
. Then
is equal to ______________.
0
solution: ans = 0
Let
and
if then
and if
then
Let
and
if then
and if
then
If
,
, then
is equal to _________.
2
solution: answer = 2
or 2
or 2
Let
,
. Then the value of
is ______________.
13
solution: answer = 13
If the real part of the complex number
is zero, then the value of sin 2 3
+ cos 2
is equal to _______________.
1
solution: ans = 1
since Re(z) = 0, calculate the real part of the z and equate it to zero
on solving this we get = so sin 2 3 =1
since Re(z) = 0, calculate the real part of the z and equate it to zero
on solving this we get = so sin 2 3 =1
Let z1, z2 be the roots of the equation z2 + az + 12 = 0 and z1, z2 form an equilateral triangle with origin. Then, the value of |a| is
6
solution: ans=6
For equilateral triangle with vertices 1, z2 and z2,
Here one vertex z3 is 0
For equilateral triangle with vertices 1, z2 and z2,
Here one vertex z3 is 0
Let z and
be two complex numbers such that
and Re(
) has minimum value. Then, the minimum value of n
N for which n
n is real, is equal to ______________.
4
solution: ans = 4
Let z = x + iy | z + i | = | z 3i |
y = 1
= x2 + y2 - 2x - 2iy + 2
= x2 + 12 - 2x - 2i + 2
Re() = x 2 - 2x + 3
Re() = (x - 1) 2 + 2
Re()min at x = 1 z = 1 + i
= 2(1 - i) =
n =
for n to be real least value of n in N must be 4
Let z = x + iy | z + i | = | z 3i |
y = 1
= x2 + y2 - 2x - 2iy + 2
= x2 + 12 - 2x - 2i + 2
Re() = x 2 - 2x + 3
Re() = (x - 1) 2 + 2
Re()min at x = 1 z = 1 + i
= 2(1 - i) =
n =
for n to be real least value of n in N must be 4
If the least and the largest real values of a, for which the
equation z + α
|z – 1| + 2i = 0 (z ∈
C and i 2= - 1
) has a solution, are p and q respectively; then 4(p2 + q2) is equal to ______.
10
solution:
ans = 10
y + 2 =0
y = 2 and
y + 2 =0
y = 2 and
If
, (m, n
N) then the greatest common divisor of the least values of m and n is _______ .
4
m = 8 k1 and n = 12 k2
Least value of m = 8 and n = 12.
GCD (8, 12) = 4
Let
. If
, and
be the greatest integral part of | k |. Then
is equal to _________.
310
Solution: ans = 310
we know,
and
= 310
we know,
and
= 310