Demo Test
Complex Numbers

Question 1 :


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Let a,b be two real numbers such that ab < 0 . IF the complex number 1+aib+i is of unit modulus and a + ib lies on the circle |z - 1| = |2z| , then a possible value of 1+[a]4b , where [t] is greatest integer function, is :
1) (1+74)
2) 1/2
3) 0
4) -1
3
ans : 3
1+aib+i = 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 = −1+76 and 1-76
⇒ [a] = 0
∴ 1+[a]4b = 0
⇒ similarly when a=−1−76 and b=1+76
⇒ [a] = -1
1+[a]4b = 0 = 0
If the center and radius of the circle |z−2z−3|=2 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
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.
Let z be a complex number such that |z−2iz+i|=2,z≠−i . 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
|z−2iz+i|=2

⇒(z−2i)(z¯+2i)=4(z+i)(z¯−i)

⇒2z¯+2iz−2iz¯+4=4(zz¯−zi+zi―+1)

⇒3zz¯−6iz+6iz¯=0

⇒2z¯−2iz+2iz¯=0

∴ center (−2i) or (0,−2)
Let z1 = 2 + 3 and z 2 = 3 + 4i . The set S={z∈C:|z−z1|2−|z−z2|2=|z1−z2|2} 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
|z−z1|2−|z−z2|2=|z1−z2|2

⇒(x−2)2+(y−3)2−(x−3)2−(y−4)2=1+1

⇒−4x+4+9−6y−9+6x−16+8y=2

⇒2x+2y=14

⇒x+y=7

The value of (1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3 is:
1) −12(1−i3)
2) −12(3−i)
3) 12(1−i3)
4) 12(3+i)
2
solution: ans: 2
z=(1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3

1+sin⁡2π9+icos⁡2π9=1+cos⁡5π18+isin⁡5π18=1+2cos2⁡5π36−1+2isin⁡5π36cos⁡5π36=2cos⁡5π36(cos⁡5π36+isin⁡5π36)=2cos⁡5π36ei5π36z=−32+12i=12(i−3)=−12(3−i)

Let p,q∈R and (1−3i)200=2199(p+iq),i=−1 then p+q+q2 and p−q+q2 are roots of the equation.
1) x2+4x−1=0
2) x2−4x+1=0
3) x2+4x+1=0
4) x2−4x−1=0
2
sol: ans:2 (1−3i)200

=[2(12−32i)]200

=2200(cos⁡π3−isin⁡π3)200

=2200(cos⁡200π3−isin⁡200π3)

=2200(cos⁡(66π+2π3)−isin⁡(66θ+2π3))

=2200(cos⁡2π3−isin⁡2π3)

=2200(−12−32i)

=2199(−1−3i)

=2199(p+iq)

∴ p = -1 and q = −3

p−q+q2=−1+3+3=2+3=α

p+q+q2=−1−3+3=2−3=β

∴ α+β=4

αβ=(2+3)(2−3)=4−3=1

∴ x2−(α+β)x+αβ=0

⇒x2−4x+1=0

If z≠0 be a complex number such that |z−1z|=2 , then the maximum value of |z| is :
1) 2
2) 1
3) 2−1
4) 2+1
4
solution: ans: 4

||z1|−|z2||≤|z1+z2|≤|z1|+|z2|

∴ ||z|−1|z||≤|z−1z|

⇒||z|−1|z||≤2

⇒||z|2−1|z||≤2

⇒−2≤|z|2−1|z|≤2

∴ |z|2−1|z|≤2

⇒|z|2−1≤2|z|

⇒|z|2−2|z|−1≤0

⇒|z|2−2|z|+1−2≤0

⇒(|z|−1)2−2≤0

⇒−2≤|z|−1≤2

⇒1−2≤|z|≤1+2 ------(1)

or

−2≤|z|2−1|z|

⇒|z|2−1≤−2|z|

⇒|z|2+2|z|−1≤0

⇒|z|2+2|z|+1−2≤0

⇒(|z|+1)2−2≤0

⇒−2≤|z|+1≤+2

⇒−2−1≤|z|≤2−1 ------ (2) From (1) and (2) we get, Maximum value of |z|=2+1 and minimum value of |z|=−2−1
Let S={z=x+iy:|z−1+i|≥|z|,|z|<2,|z+i|=|z−1|} . Then the set of all values of x , for which w=2x+iy∈S for some y∈R , is
1) (−2,122]
2) (−12,14]
3) (−2,12]
4) (−12,122]
2
If z=2+3i , then z5+(z¯)5 is equal to :
1) 244
2) 224
3) 245
4) 265
1
ans: 1
z=(2+3i)

⇒z5=(2+3i)((2+3i)2)2

=(2+3i)(−5+12i)2

=(2+3i)(−119−120i)

=−238−240i−357i+360

=122−597i

z―5=122+597i

z5+z―5=244

Let S be the set of all (α,β),π<α,β<2π , for which the complex number 1−isin⁡α1+2isin⁡α is purely imaginary and 1+icos⁡β1−2icos⁡β is purely real. Let Zαβ=sin⁡2α+icos⁡2β,(α,β)∈S . Then ∑(α,β)∈S(iZαβ+1iZ¯αβ) is equal to :
1) 3
2) 3i
3) 1
4) 2 - i
3
solution: ans:3
∵ 1−isin⁡α1+2isin⁡α is purely imaginary

1−isin⁡α1+2isin⁡α+1+isin⁡α1−2isin⁡α=0

⇒1−2sin2α=0

∴ α=5π4,7π4

1+icos⁡β1−2icos⁡β

1+icos⁡β1−2icos⁡β−1−icos⁡β1+2icos⁡β=0

⇒cos⁡β=0

∴ β=3π2

∴ S={(5π2,3π2),(7π4,3π2)}

Zαβ=1−i

Zαβ=−1−i

∴ ∑(α,β)∈S(iZαβ+1iZ―αβ)=i(−2i)+1i[11+i+1−1+i]

=2+1i2i−2=1
Let the minimum value v0 of v=|z|2+|z−3|2+|z−6i|2,z∈C is attained at z=z0 . Then |2z02−z¯03+3|2+v02 is equal to
1) 1000
2) 1024
3) 1105
1196
1
solution: ans: 1
z=x+iy

v=x2+y2+(x−3)2+y2+x2+(y−6)2

=(3x2−6x+9)+(3y2−12y+36)

=3(x2+y2−2x−4y+15)

=3[(x−1)2+(y−2)2+10]

vmin

z=1+2i=z0

v0=30

|2(1+2i)2−(1−2i)3+3|2+900

=|2(−3+4i)−(1−8i3−6i(1−2i)+3|2+900

=|−6+8i−(1+8i−6i−12)+3|2+900

=|8+6i|2+900

=1000
If z=x+iy satisfies |z|−2=0 and |z−i|−|z+5i|=0 , then
1) x+2y−4=0
2) x2+y−4=0
3) x+2y+4=0
4) x2−y+3=0
3
sol: ans:3
|z−i|=|z+5i|

So, z lies on ⊥ bisector of (0,1) and (0,−5) i.e., line y=−2

|z|=2

⇒z=−2i

x=0

y=−2

x+2y+4=0
If α,β,γ,δ are the roots of the equation x4+x3+x2+x+1=0 , then α2021+β2021+γ2021+δ2021 is equal to :
1) -4
2) -1
3) 1
4) 4
2
solution: ans= 2
x5=1

x5−1=0

⇒(x−1)(x4+x3+x2+x+1)=0

x4+x3+x2+x+1=0 has roots α , β , γ and 8.
∴ Roots of x5−1=0 are 1, α , β , γ , 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 =1p+αp+βp+γp+8p={0;Ifpisnotmultipleof55;Ifpismultipleof5

p=2021 is not a multiple of 5.

∴ 12021+α2021+β2021+γ2021+82021=0

⇒α2021+β2021+γ2021+82021=−1
For n∈N , let Sn={z∈C:|z−3+2i|=n4} and Tn={z∈C:|z−2+3i|=1n} . Then the number of elements in the set {n∈N:Sn∩Tn=ϕ} is :
1) 0
2) 2
3) 3
4) 4
4
solution: ans: 4
Sn={z∈C:|z−3+2i|=n4} represents a circle with centre C1(3,−2) and radius r1=n4

similarly Tn represents circle with centre C2(2,−3) and radius r2=1n

 As Sn∩Tn=ϕ

C1C2>r1+r2 OR C1C2<|r1−r2|2>n4+1n OR 2<|n4−1n|n=1,2,3,4n may take infinite values 
The real part of the complex number (1+2i)8.(1−2i)2(3+2i).(4−6i)― is equal to :

1) 50013
2) 11013
3) 556
4) 55013
4
solution: ans:4
(1+2i)8.(1−2i)2(3+2i).(4−6i)―

=(1+2i)2(1−2i)2(1+2i)6(3+2i)(4+6i)

=(1−4i2)2(1+2i)612+18i+8i+12i2

=(1+5)2[(1+2i)2]312+26i−12

=25(1+4i2+4i)326i

=25(1−4+4i)326i

=25(−3+4i)326i

=2526i[(−3)3+(4i)3+3.(−3)2.4i+3(−3).(4i)2]

=2526i(−27−64i+108i+144)

=2526i(117+44i)

=25i26i2(117+44i)

=25i−26(117+44i)

=25×117i−26−25×44i226

=25×117i−26+22×2513

=25×117i−26+55013

∴ real part =55013

Let α and β be the roots of the equation x2 + (2i - 1) = 0. Then, the value of |α 8 + β 8| is equal to:
1) 50
2) 250
3) 1250
4) 1500
1
solution: ans: 1
x2+(2i−1)=0

⇒x2=1−2i

as α and β are roots of the equation

α2=1−2i

β2=1−2i

∴ α2=β2=1−2i

∴ |α2|=12+(−2)2=15

|α8+β8|

|α8+α8|

=2|α8|

=2|α2|4

=2(5)4

=2×25

=50
The area of the polygon, whose vertices are the non-real roots of the equation z―=iz2 is :
1) 334
2) 332
3) 32
4) 34
1
sol: ans: 1
z―=iz2

z=x+iy

x−iy=i(x2−y2+2xiy)

x−iy=i(x2−y2)−2xy

∴ x=−2yx

x2−y2=−y

case - 1
x=0

y=−12

x=0

−y2=−y

y=0,1

y=−12
case - 2
⇒x2−14=12⇒x=±32

x={0,i,32−i2,−32−i2}

area of polygon = =12|01132−121−32−121|


=12|−3−32|=334
Let z1 and z2 be two complex numbers such that z―1=iz―2 and arg⁡(z1z―2)=π . Then
1) arg⁡z2=π4
2) arg⁡z2=−3π4
3) arg⁡z1=π4
4) arg⁡z1=−3π4
3
solution: ans:3
z1z2=−i⇒z1=−iz2

⇒arg⁡(z1)=−π2+arg⁡(z2)

arg⁡(z1)−arg⁡(z―2)=π

⇒arg⁡(z1)+arg⁡(z2)=π

By solveing above equations we get, arg⁡(z1)=π4 and arg⁡(z2)=3π4
Let a circle C in complex plane pass through the points z1=3+4i , z2=4+3i and z3=5i . If z(≠z1) 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) tan−1(25)−π
2) tan−1(247)−π
3) tan−1(3)−π
4) tan−1(34)−π
2
solution: ans: 2
z1=3+4i

z2=4+3i

z3=5i

C≡x2+y2=25

Let z(x,y)
z(x,y)

⇒(y−4x−3)(2−4)=−1

⇒y=2x−2≡L

∴ z is intersection of C and L

⇒z≡(−75,−245)

∴ Arg(z)=−π+tan−1(247)
If (3+i)100=299(p+iq) , then p and q are roots of the equation :
1) x2−(3−1)x−3=0
2) x2+(3+1)x+3=0
3) x2+(3−1)x−3=0
4) x2−(3+1)x+3=0
1
solution: ans:1 (2eiπ/6)100=299(p+iq)

2100(cos⁡50π3+isin⁡50π3)=299(p+iq)

p+iq=2(cos⁡2π3+isin⁡2π3)

p = −1

q = 3

x2−(3−1)x−3=0

If z and ω are two complex numbers such that |zω|=1 and arg⁡(z)−arg⁡(ω)=3π2 , then arg⁡(1−2z―ω1+3z―ω) is : (Here arg(z) denotes the principal argument of complex number z)
1) arg⁡(1−2z―ω1+3z―ω)
2) π4
3) −3π4
4) −π4
2
solution: ans: 2
3π4

|zω|=1

⇒

|z|=r

|ω|=1r

let arg⁡(z)=θ

∴ arg⁡(ω)=(θ−3π2)

z=reiθ

⇒z―=reiθ

ω=1rei(θ−3π2)

now consider 1−2z―ω1+3z―ω=1−2ei(−3π2)1+3ei(−3π2)=(1−2i1+3i)

=(1−2i)(1−3i)(1+3i)(1−3i)=−12(1+i)

∴ prinarg⁡(1−2z―ω1+3z―ω)

=prinarg⁡(1−2z―ω1+3z―ω)

=(−12(1+i))

=−(π−π4)=−3π4

The least value of |z| where z is complex number which satisfies the inequality exp⁡((|z|+3)(|z|−1)||z|+1|loge2)≥log2|57+9i|,i=−1 , is equal to :
1) 8
2) 3
3) 2
4) 5
2
solution: ans: 2
Let |z| = t and t ≥ 0

e(t+3)(t−1)t+1loge2≥log216=8 ∵t+1 > 0

2(t+3)(t−1)t+1≥23

(t+3)(t−1)t+1≥3

t2+2t−3≥3t+3

t2−t−6≥0

t∈(−∞,−2)∪[3,∞)

≥

∴ t∈[3,∞)
Let a complex number z, |z| ≠ 1, satisfy log12(|z|+11(|z|−1)2)≤2 . Then, the largest value of |z| is equal to ____________.
1) 5
2) 8
3) 6
4) 7
4
solution: ans: 4 log12(|z|+11(|z|−1)2)≤2

|z|+11(|z|−1)2≥12

2|z|+22≥(|z|−1)2

2|z|+22≥|z|2−2|z|+1

|z|2−4|z|−21≤0

(|z|−7)(|z|+3)≤0

⇒|z|≤7

∴

|z|max=7
If α , β∈ R are such that 1 - 2i (here i2 = - 1) is a root of z2 + α z + β = 0, then (α - β ) 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 = - α Product of roots = (1 - 2i)(1 + 2i) = 1 - 4i2 = 5 = β ∴ α - β = -7
Let the lines (2 - i)z = (2 + i)z― and (2 + i)z + (i - 2)z― - 4i = 0, (here i2 = - 1) be normal to a circle C. If the line iz + z― + 1 + i = 0 is tangent to this circle C, then its radius is :
1) 322
2) 32
3) 122
4) 32
1
solution: ans:1
(2−i)z=(2+i)z―

⇒(2−i)(x+iy)=(2+i)(x−iy)

⇒2x−ix+2iy+y=2x+ix−2−iy+y

⇒2ix−4iy=0

L1:x−2y=0

⇒(2+i)z+(i−2)z―−4i=0

⇒(2+i)(x+iy)+(i−2)(x−iy)−4i=0

⇒2x+ix+2iy−y+ix−2x+y+2iy−4i=0

⇒2ix+4iy−4i=0

L2:x+2y−2=0

Solve L1 and L2: x = 1, 4y=2,y=12

∴ x=1

centre (1,12)

L3:iz+z―+1+i=0

⇒i(x+iy)+x−iy+1+i=0

⇒ix−y+x−iy+1+i=0

⇒(x−y+1)+i(x−y+1)=0

Radius = distance from (1,12) to x−y+1=0

⇒

r=1−12+12

⇒

r=322
The value of (−1+i31−i)30 is :
- 215i
- 215
215i
65
1
solution: ans:1


(−1+i31−i)30

=(2ω1−i)30

=230.ω30((1−i)2)15

=230.1(1+i2−2i)15

=230.1−215.i15

= - 215i
If a and b are real numbers such that (2+α)4=a+bα where α=−1+i32 then a + b is equal to:
1) 33
2) 9
3) 24
4) 57
2
solution: ans: 2
α=ω

given α=−1+i32

⇒(2+ω)4=a+bω(ω3=1)

⇒24+4.23ω+6.22ω3+4.2.ω3+ω4=a+bω

⇒16+32ω+24ω2+8+ω=a+bω

⇒24+24ω2+33ω=a+bω

⇒−24ω+33ω=a+bω

⇒a=0,b=9
Let u=2z+iz−ki , 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

u=2z+iz−ki

=2(x+iy)+i(x+iy)−ki

=2x+i(2y+1)x+i(y−k)×x−i(y−k)x−i(y−k)

=2x2+(2y+1)(y−k)+i(2xy+x−2xy+2kx)x2+(y−k)2

Given, Re(u)+Im(u)=1

⇒2x2+(2y+1)(y−k)x2+(y−k)2+x+2kxx2+(y−k)2=1

⇒2x2+(2y+1)(y−k)+x+2kx=x2+(y−k)2

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, ∴ (2y+1)(y−k)=(y−k)2

⇒2y2+y−2yk−k=y2+k2−2ky

⇒y2+y−(k+k2)=0

Let roots of this quadratic equation y1 and y2 ∴ Point P (0, y1) and Q (0, y2) y1+y2=1

y1y2=−k−k2

∴ (y1−y2)2=(y1+y2)2−4y1y2

=1+4k+4k2

⇒|y1−y2|=1+4k+4k2

Given PQ = 5 ⇒|y1−y2|=5

⇒1+4k+4k2=5

⇒k2+k−6=0

⇒k=−3,2

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) = π6 , then Im(z1 + z2 ) is equal to :
1) 32
2) 13
3) 23
4) 23
4
solution: ans: 4
z1=x1+iy1,z2=x2+iy2

Given Re(z1) = |z1 – 1| ∴x1 = |(x1 - 1) + iy1|

⇒ x1 (x1−1)2+y12

⇒

x12=(x1−1)2+y12

⇒y12−2x1+1=0

Also given Re(z2) = |z2 – 1| ∴ x2 = |(x2 - 1) + iy2| ⇒

Performing equation (1) - (2), x22=(x2−1)2+y22

⇒

y22−2x2−1=0

(y12−y22)+2(x2−x1)=0

⇒

(y1+y2)(y1−y2)=2(x1−x2)

⇒

y1+y2=2(x1−x2y1−y2)

given, arg⁡(z1−z2)=π6

⇒

tan−1(y1−y2x1−x2)=π6

⇒y1−y2x1−x2=13

∴ y1+y2=23
The imaginary part of (3+2−54)12−(3−2−54)12 can be :
1) -26
2) 6
3) 6
4) - 6
1
solution: ans:1

3+2−54

=9−6+2−54

=9+(6i)2+2.3.6i

=32+(6i)2+2.3.6i

=(3+6i)2

Similarly, (3−2−54)=(3−6i)2

∴(3+2−54)12−(3−2−54)12

=±(3+6i)−[±(3−6i)]

=6,−6,26i,−26i

∴ Possible imaginary parts are 26i,−26i
The value of (1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3 is :
1) 12(3−i)
2) -12(3−i)
3) −12(1−i3)
4) 12(1−i3)
2
solution: ans: 2
(1+sin⁡2π9+icos⁡2π91+sin⁡2π9−icos⁡2π9)3

=(1+sin⁡(π2−5π18)+icos⁡(π2−5π18)1+sin⁡(π2−5π18)−icos⁡(π2−5π18))3

=(1+cos⁡(5π18)+isin⁡(5π18)1+cos⁡(5π18)−isin⁡(5π18))3

=(2cos2(5π36)+2isin⁡(5π36)cos⁡(5π36)2cos2(5π36)−2isin⁡(5π36)cos⁡(5π36))3

=(cos⁡(5π36)+isin⁡(5π36)cos⁡(5π36)−isin⁡(5π36))3

=(ei5π36e−i5π36)3

=(ei5π18)3

=ei5π18×3

=ei5π6

=cos⁡5π6+isin⁡5π6

=−32+i2
Let z be complex number such that |z−iz+2i|=1 and |z| = 52 . Then the value of |z + 3i| is :
1) 23
2) 10
3) 154
4) 72
4
solution: ans: 4
|z−iz+2i|=1

⇒|z – i| = |z + 2i|

(let z = x + iy) ⇒x2 + (y – 1)2 = x2 + (y + 2)2

y = −12

Also given |z| = 52

⇒x2 + y2 = 254

⇒ x2 = 6

∴ z = ±6 −12i

|z + 3i| = 6+254 = 72
If Re(z−12z+i)=1 , where z = x + iy, then the point (x, y) lies on a:
1) straight line whose slope is 32
2) straight line whose slope is −23
3) circle whose diameter is 52
4) circle whose centre is at (−12,−32)
4
solution: ans: 3
Re(z−12z+i)=1

Put z = x + iy ∴ Re((x+iy)−12(x+iy)+i)=1

⇒

Re(((x−1)+iy2x+i(2y+1))(2x−i(2y+1)2x−i(2y+1)))=1

⇒

Re({(x−1)+iy}{2x−i(2y+1)}4x2+(2y+1)2)=1

Real part of this equation is = 1 ∴ 2x(x−1)+y(2y+1)4x2+(2y+1)2

⇒2x2 + 2y2 +2x + 3y + 1 = 0

This is an equation of circle. ∴Locus is a circle whose center is (−12,−34) and radius 54

∴ Diameter = 2 ×

54

= 52
Let z ∈ 2z−n2z+n 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 2(x+10i)−n2(x+10i)+n=2i−1

⇒(2x−n)+20i=−(2x+n)−40−20i+2ni

⇒2x−n=2x−n−40

& 20 = -20 + 2n ⇒x = -10 & n = 20
The equation |z – i| = |z – 1|, i = −1 , represents :
1) a circle of radius of 1
2) the line through the origin with slope -1
3) a circle of radius 12
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
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = π2 , then:
1) zw―=1−i2
2) z―w=i
3) zw―=−1+i2
4) z―w=−i
4
solution: ans: 4 |zw|=1

let w=1reiθ

then z = rei(θ+π2)

z―w=e−i(θ+π2).eiθ=e−i(π/2)=−i

zw―=ei(θ+π2).e−iθ=eiπ/2=i
If a > 0 and z = (1+i)2a−i , has magnitude 25 , then z― is equal to :
1) −15+35i
2) −15−35i
3) 15−35i
4) −35−15i
2
solution: ans: 2
z=(1+i)2a−i×a+ia+i

⇒z=(1−1+2i)(a+i)a2+1=2ai−2a2+1

⇒|z|=(−2a2+1)2+(2aa2+1)2=4+4a2(a2+1)2

⇒4(1+a2)(a2+1)2=2a2+1 -------(1)

given |z|=25

so 25=21+a2 from equation(1)

⇒25=41+a2

⇒1+a2=10

a = ± 3

∵ a is greater than 3 a = 3
z―=(1−i)23+i

(1+i2−2i)3+i

(1+i2−2i)(3−i)(3+i)(3−i)

(−2i)(3−i)(9−i2)

(−2i)(3−i)10

−6i+2i210

−6i−210

−15−35i

Let z=a+ib,b≠0 be complex numbers satisfying z2=z¯⋅21−z . Then the least value of n∈N , such that zn=(z+1)n , is equal to __________.
6
solution: ans: 6

∵ z2=z―.21−|z|

⇒|z|2=|z―|.21−|z| ----(1)

⇒|z|=21−|z|

∵ b≠0⇒|z|≠0

∴ |z|=1 ---(2)

∵ z=a+ib then a2+b2=1 -----(3)

Now again from equation (1), equation (2), equation (3) we get : a2−b2+i2ab=(a−ib)20

∴ a2−b2=a and 2ab=−b

∴ a=−12 and b=±32

z=−12+32i or z=−12−32i

zn=(z+1)n⇒(z+1z)n=1

(1+1z)n=1

(1+3i2)=1

then the minimum value of n is 6
Let S={z∈C:z2+z¯=0} . Then ∑z∈S(Re(z)+Im(z)) is equal to ______________.
0
solution: ans = 0

∵ z2+z―=0

Let z=x+iy

∴ x2−y2+2ixy+x−iy=0

(x2−y2+x)+i(2xy−y)=0

∴ x2+y2=0 and (2x−1)y=0

if x=+12 then y=±32

and if y=0

then x=0,−1

∴ z=0+0i,−1+0i,12+32i,12−32i

∴ ∑(Re(z)+m(z))=0
If z2+z+1=0 , z∈C , then|∑n=115(zn+(−1)n1zn)2| is equal to _________.
2
solution: answer = 2

∵ z2+z+1=0

⇒ ω or ω2

∵ |∑n=115(zn+(−1)n1zn)2|

=|∑n=115z2n+∑n=115z−2n+2.∑n=115(−1)n|

=|0+0−2|

=2
Let z=1−i32 , i=−1 . Then the value of 21+(z+1z)3+(z2+1z2)3+(z3+1z3)3+....+(z21+1z21)3 is ______________.
13
solution: answer = 13

z=1−3i2=e−iπ3

zr+1zr=2cos⁡(−π3)r=2cos⁡rπ3

⇒21+∑r=121(zr+1zr)3=8(cos3rπ3)=2(cos⁡rπ+3cos⁡rπ3)

⇒21+(z+12)3+(z2+1z2)3+....(z21+1z21)3

=21+∑r=121(zr+1zr)3

=21+∑r=121(2cos⁡rπ+6cos⁡rπ3)

=21−2−6

=13
If the real part of the complex number z=3+2icos⁡θ1−3icos⁡θ,θ∈(0,π2) is zero, then the value of sin 2 3 θ + cos 2 θ is equal to _______________.
1
solution: ans = 1
since Re(z) = 0, calculate the real part of the z and equate it to zero

z=3+2icos⁡θ1−3icos⁡θ . 1+3icos⁡θ1+3icos⁡θ

on solving this we get θ = π4 so sin 2 3 θ =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, z12+z22+z33=z1z2+z2z3+z3z1

Here one vertex z3 is 0

∴ z12+z22=z1z2+0+0

z2+az+12=0

∴ z1+z2=−a

z1z2=12

∴

z12+z22+2z1z2=z1z2+2z1z2

⇒(z1+z2)2=3z1z2

⇒(−a)2=3×12

⇒a2=36

⇒a=±6

⇒|a|=6
Let z and ω be two complex numbers such that ω=zz―−2z+2,|z+iz−3i|=1 and Re(ω ) has minimum value. Then, the minimum value of n ∈ N for which ωn n is real, is equal to ______________.
4
solution: ans = 4

Let z = x + iy | z + i | = | z − 3i |

⇒ y = 1

ω = x2 + y2 - 2x - 2iy + 2

ω = x2 + 12 - 2x - 2i + 2

Re(ω) = x 2 - 2x + 3

Re(ω) = (x - 1) 2 + 2

Re(ω)min at x = 1 ⇒ z = 1 + i



⇒ ω = 2(1 - i) = 2ei(−π4)

ωn = 2ei(−nπ4)

for ωn 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


x+iy+α(x−1)2+y2+2i=0

∴ y + 2 =0 x+α(x−1)2+y2=0

y = − 2 and

x2=α2(x2−2x+1+4)

α2=x2x2−2x+5⇒x2(α2−1)−2xα2+5α2=0

x∈R⇒D≥0

4α4−4(α2−1)5α2≥0

α2[4α2−2α2+20]≥0

α2[−16α2+20]≥0

α2[α2−54]≤0

0≤α2≤54

∴ α2∈[0,54]

∴ α∈[−52,52]

4[(q)2+(p)2]=4[54+54]=10

If (1+i1−i)m2=(1+i1−i)n3=1 , (m, n ∈ N) then the greatest common divisor of the least values of m and n is _______ .
4
(1+i1−i)m/2=(1+i1−i)n/3=1

⇒((1+i)22)m/2=((1+i)2−2)n/3=1

⇒(i)m/2=(−i)n/3=1

⇒m2=4k1andn3=4k2

⇒m = 8 k1 and n = 12 k2

Least value of m = 8 and n = 12.

∴

GCD (8, 12) = 4
Let i=−1 . If (−1+i3)21(1−i)24+(1+i3)21(1+i)24=k , and n=[|k|] be the greatest integral part of | k |. Then ∑j=0n+5(j+5)2−∑j=0n+5(j+5) is equal to _________.
310
Solution: ans = 310

(1+i)2=1+i2+2i=1−1+2i=2i

(1−i)2=1+i2−2i=1−1−2i=−2i

we know, −12+i32=ω

⇒−1+i3=2ω

and −12−i32=ω2

⇒−1−i3=2ω2

⇒1+i3=−2ω2

K=(−1+i3)21(1−i)24+(1+i3)21(1+i)24

=(2ω)21((1−i)2)12+(−2ω)21((1+i)2)12

=221.ω21(−2i)12+(−2)21(ω2)21(2i)12

ω3=1

i4=1

=221212−221212=0

∴ n=[|K|]=[|0|]=0

∑j=05(j+5)2−∑j=05(j+5)

∑j=05(j2+25+10j−j−5)

∑j=05(j2+9j+20)

∑j=05j2+9∑j=05j+20∑j=051

5×6×116+9(5×62)+20×6

= 310