Q: All the solutions of z2018+z20181+z2016+z20161=0 lie on the curve

Write (1) as ⇒⇒⇒(z2018+z20161)+(z2016+z20181)=0(z2034+1)+(z2018z2+1)=0z2034=−1 or z2=−1∣z∣=1

Q: Let zk=ak+ibk=0, for k=1,2,…, 2018. Then value of the expression E=(∑k=12018∣zk∣)2∣(∑k=12018(ak−bk))(∑k=12018(ak+bk))∣

For a,b∈R,a+ib=0(a+b)2+(a−b)2=2(a2+b2)⇒∣a+b∣,∣a−b∣≤2a2+b2=2∣a+ib∣ Note that both ∣a+b∣ and ∣a−b∣ cannot be equal to 2a2+b2 Now, ∣∣∑k=12018(ak±bk)∣∣≤∑k=12018∣ak±bk∣≤2∑k=12018∣zk∣ Therefore, E<(∑k=12018∣zk∣)2(2∑k=12018∣zk∣)(2∑k=12018∣zk∣)=2

Q: Let ω be a complex number such that ∣ω∣=2. Let α,β be solution of the equation z2+i(ω+iωˉ)z−2i=0 then

⇒⇒z2+i(ω+iωˉ)z−iωωˉ=0=0[∵ωωˉ=2]z(z+iω)−ωˉ(z+iω)=0(z−ωˉ)(z+iω)=0 ⇒z=ωˉ,−iω Let α=ωˉ,β=−iω, then ∣α∣+∣β∣=∣ωˉ∣+∣−iω∣=∣ω∣+∣ω∣=22

Q: If a,b are distinct complex numbers lying on the unit circle ∣z∣=1, then value of ∣∣1+aba+b∣∣2+∣∣1−aba−b∣∣2

Let a=cosα+isinα b=cosβ+isinβ where −π<α,β≤π,α=β we have 1+aba+b=1+cos(α+β)+isin(α+β)(cosα+cosβ)+i(sinα+sinβ) =2cos(2α+β)[cos2α+β+isin2α+β]2cos(2α−β)[cos2α+β+isin2α+β] =cos((α+β)/2)cos((α−β)/2) ⇒∣∣1+aba+b∣∣≥∣∣cos(2α−β)∣∣[∵∣∣cos(2α+β)∣∣≤1] Similarly, ∣∣1−aba−b∣∣≥∣∣sin(2α−β)∣∣ Thus, ∣∣1+aba+b∣∣2+∣∣1−aba−b∣∣2≥1

Q: Let a1=2,an+1=an+ian−i for n≥1, then ∑k=13nak is equal to

a2⇒a3=2+i2−i=4−i2(2−i)2=53−4i=53−54i=53−54i+i53−54i−i=3+i3−9i=3+i−3i(3+i)=−3i Next, a4=a3+ia3−i=−3i+i−3i−i=2 Thus, a3j+i=ai for 1≤j≤n−1,i=1,2,3 ⇒∑k=13nak=n(a1+a2+a3)=5n(13−19i)

Q: Let a,b,c and d be four distinct complex numbers such that ∣a∣=∣b∣=∣c∣=1. If z=(a−d)(b−c)(a−c)(b−d) is real, then

We have aaˉ=bbˉ=ccˉ=1 As z is real z=zˉ ⇒⇒⇒⇒⇒⇒(a−d)(b−c)(a−c)(b−d)=(aˉ−dˉ)(bˉ−cˉ)(aˉ−cˉ)(bˉ−dˉ)=(a1−dˉ)(b1−c1)(a1−c1)(b1−dˉ)=(1−adˉ)(c−b)(c−a)(1−bdˉ)a−db−d=1−adˉ1−bdˉb−d−abdˉ+addˉ=a−d−abdˉ+bddˉb−a=(b−a)ddˉddˉ=1∣d∣=1

Q: Let p and q be two complex numbers such that q=0. If the roots of the equation x2+px+q2=0 have the same absolute value, then p/q is

Let x1,x2 be the roots of x2+px+q2=0 and ∣x1∣=∣x2∣=r. Note that x1x2=q2=0. Now, q2p2=x1x2(x1+x2)2=x2x1+x1x2+2=∣x2∣2x1xˉ2+∣x1∣2xˉ1x2+2=r21(x1xˉ2+xˉ1x2)+2=r22Re(x1xˉ2)+2 ⇒q2p2 is a real number. Next, as −Re(x1xˉ2)≤∣x1xˉ2∣=∣x1x2∣=∣x1∣∣x2∣=r2, we get ⇒q2p2≥−r22r2+2=−2+2=0 Therefore, p/q is a real number.

Q: If ∣∣z3+z31∣∣≤2, then ∣∣z+z1∣∣ cannot exceed

Let ∣∣z+z1∣∣=a. Now, the identity, z3+z31=(z+z1)3−3(z+z1) gives us ⇒⇒⇒∣∣z+z1∣∣3≤∣∣z3+z31∣∣+3∣∣z+z1∣∣a3≤2+3a⇒a3−3a−2≤0(a−2)(a+1)2≤0⇒a−2≤0a≤2

Q: If ω=cosnπ+isinnπ, then value of 1+ω+ω2+…+ωn−1 is

We have S=1+ω+ω2+⋯+ωn−1=1−ω1−ωn But ωn=cos(nnπ)+isin(nnπ)=−1 Thus, S=1−ω2 But 1−ω=1−cos(nπ)−isin(nπ)=2sin2(2nπ)−2isin(2nπ)cos(2nπ)=−2isin(2nπ)[cos(2nπ)+isin(2nπ)] Thus, S=2isin(π/2n)−2[cos(2nπ)+isin(2nπ)]−1 =1+icot(2nπ) Alternatively, we write S=e−πi/2n−eπi/2n2e−πi/2n=isin(2nπ)−1[cos(2nπ)−isin(2nπ)]=1+icot(2nπ)

Q: For any complex number z, the minimum value of ∣z∣+∣z−2i∣ is

We have, for z∈C ⇒2∣2i∣≤∣z∣+∣z−2i∣=∣z+(2i−z)∣≤∣z∣+∣2i−z∣ Thus, minimum value of ∣z∣+∣z−2i∣ is 2 and it is attained for any z lying on the segment joining z=0 and z=2i.

Question 1

All the solutions of z2018+z20181​+z2016+z20161​=0 lie on the curve

Accepted Answer

Write (1) as

⇒⇒⇒​(z2018+z20161​)+(z2016+z20181​)=0(z2034+1)+(z2018z2+1​)=0z2034=−1 or z2=−1∣z∣=1​

Question 2

Let zk​=ak​+ibk​=0, for k=1,2,…, 2018. Then value of the expression E=(∑k=12018​∣zk​∣)2∣(∑k=12018​(ak​−bk​))(∑k=12018​(ak​+bk​))∣​

Accepted Answer

For a,b∈R,a+ib=0(a+b)2+(a−b)2=2(a2+b2)⇒∣a+b∣,∣a−b∣≤2​a2+b2​=2​∣a+ib∣​

Note that both ∣a+b∣ and ∣a−b∣ cannot be equal to

2​a2+b2​

Now,

∣∣​∑k=12018​(ak​±bk​)∣∣​≤∑k=12018​∣ak​±bk​∣≤2​∑k=12018​∣zk​∣
Therefore,

E<(∑k=12018​∣zk​∣)2(2​∑k=12018​∣zk​∣)(2​∑k=12018​∣zk​∣)​=2

Question 3

Let ω be a complex number such that ∣ω∣=2​. Let α,β be solution of the equation z2+i(ω+iωˉ)z−2i=0 then

Accepted Answer

⇒⇒​z2+i(ω+iωˉ)z−iωωˉ=0=0[∵ωωˉ=2]z(z+iω)−ωˉ(z+iω)=0(z−ωˉ)(z+iω)=0​
⇒z=ωˉ,−iω
Let α=ωˉ,β=−iω, then
∣α∣+∣β∣=∣ωˉ∣+∣−iω∣=∣ω∣+∣ω∣=22​

Question 4

If a,b are distinct complex numbers lying on the unit circle ∣z∣=1, then value of ∣∣​1+aba+b​∣∣​2+∣∣​1−aba−b​∣∣​2

Accepted Answer

Let a=cosα+isinα
b=cosβ+isinβ
where −π<α,β≤π,α=β
we have
1+aba+b​=1+cos(α+β)+isin(α+β)(cosα+cosβ)+i(sinα+sinβ)​
=2cos(2α+β​)[cos2α+β​+isin2α+β​]2cos(2α−β​)[cos2α+β​+isin2α+β​]​
=cos((α+β)/2)cos((α−β)/2)​
⇒∣∣​1+aba+b​∣∣​≥∣∣​cos(2α−β​)∣∣​[∵∣∣​cos(2α+β​)∣∣​≤1]
Similarly, ∣∣​1−aba−b​∣∣​≥∣∣​sin(2α−β​)∣∣​
Thus, ∣∣​1+aba+b​∣∣​2+∣∣​1−aba−b​∣∣​2≥1

Question 5

Let a1​=2,an+1​=an​+ian​−i​ for n≥1, then ∑k=13n​ak​ is equal to

Accepted Answer

a2​⇒a3​​=2+i2−i​=4−i2(2−i)2​=53−4i​=53​−54​i=53​−54​i+i53​−54​i−i​=3+i3−9i​=3+i−3i(3+i)​=−3i​

Next,

a4​=a3​+ia3​−i​=−3i+i−3i−i​=2

Thus, a3j+i​=ai​ for 1≤j≤n−1,i=1,2,3

⇒∑k=13n​ak​=n(a1​+a2​+a3​)=5n​(13−19i)

Question 6

Let a,b,c and d be four distinct complex numbers such that ∣a∣=∣b∣=∣c∣=1. If z=(a−d)(b−c)(a−c)(b−d)​ is real, then

Accepted Answer

We have aaˉ=bbˉ=ccˉ=1
As z is real z=zˉ

⇒⇒⇒⇒⇒⇒​(a−d)(b−c)(a−c)(b−d)​=(aˉ−dˉ)(bˉ−cˉ)(aˉ−cˉ)(bˉ−dˉ)​=(a1​−dˉ)(b1​−c1​)(a1​−c1​)(b1​−dˉ)​=(1−adˉ)(c−b)(c−a)(1−bdˉ)​a−db−d​=1−adˉ1−bdˉ​b−d−abdˉ+addˉ=a−d−abdˉ+bddˉb−a=(b−a)ddˉddˉ=1∣d∣=1​

Question 7

Let p and q be two complex numbers such that q=0. If the roots of the equation x2+px+q2=0 have the same absolute value, then p/q is

Accepted Answer

Let x1​,x2​ be the roots of x2+px+q2=0 and ∣x1​∣=∣x2​∣=r. Note that x1​x2​=q2=0. Now,

q2p2​​=x1​x2​(x1​+x2​)2​=x2​x1​​+x1​x2​​+2=∣x2​∣2x1​xˉ2​​+∣x1​∣2xˉ1​x2​​+2=r21​(x1​xˉ2​+xˉ1​x2​)+2=r22​Re(x1​xˉ2​)+2​

⇒q2p2​ is a real number.
Next, as −Re(x1​xˉ2​)≤∣x1​xˉ2​∣=∣x1​x2​∣=∣x1​∣∣x2​∣=r2, we get

⇒q2p2​≥−r22r2​+2=−2+2=0
Therefore, p/q is a real number.

Question 8

If ∣∣​z3+z31​∣∣​≤2, then ∣∣​z+z1​∣∣​ cannot exceed

Accepted Answer

Let ∣∣​z+z1​∣∣​=a.
Now, the identity,

z3+z31​=(z+z1​)3−3(z+z1​)
gives us

⇒⇒⇒​∣∣​z+z1​∣∣​3≤∣∣​z3+z31​∣∣​+3∣∣​z+z1​∣∣​a3≤2+3a⇒a3−3a−2≤0(a−2)(a+1)2≤0⇒a−2≤0a≤2​

Question 9

If ω=cosnπ​+isinnπ​, then value of 1+ω+ω2+…+ωn−1 is

Accepted Answer

We have

S=1+ω+ω2+⋯+ωn−1=1−ω1−ωn​
But ωn=cos(nnπ​)+isin(nnπ​)=−1
Thus, S=1−ω2​
But

1−ω​=1−cos(nπ​)−isin(nπ​)=2sin2(2nπ​)−2isin(2nπ​)cos(2nπ​)=−2isin(2nπ​)[cos(2nπ​)+isin(2nπ​)]​

Thus, S=2isin(π/2n)−2​[cos(2nπ​)+isin(2nπ​)]−1

=1+icot(2nπ​)
Alternatively, we write

S​=e−πi/2n−eπi/2n2e−πi/2n​=isin(2nπ​)−1​[cos(2nπ​)−isin(2nπ​)]=1+icot(2nπ​)​

Question 10

For any complex number z, the minimum value of ∣z∣+∣z−2i∣ is

Accepted Answer

We have, for z∈C

⇒​2​∣2i∣≤∣z∣+∣z−2i∣​=∣z+(2i−z)∣≤∣z∣+∣2i−z∣
Thus, minimum value of ∣z∣+∣z−2i∣ is 2 and it is attained for any z lying on the segment joining z=0 and z=2i.

Question 11

If z1​,z2​,z3​ are complex numbers such that ∣z1​∣=∣z2​∣=∣z3​∣=∣∣​z1​1​+z2​1​+z3​1​∣∣​=1, then ∣z1​+z2​+z3​∣ is

Accepted Answer

Since ∣z1​∣=∣z2​∣=∣z3​∣=1, we get

z1​zˉ1​=z2​zˉ2​=z3​zˉ3​=1
Now,

1⇒1​=∣∣​z1​1​+z2​1​+z3​1​∣∣​=∣zˉ1​+zˉ2​+zˉ3​∣=∣z1​+z2​+z3​​∣=∣z1​+z2​+z3​∣​

Question 12

If ω is an imaginary cube root of unity, then value of the expression 2(1+ω)(1+ω2)+3(2+ω)(2+ω2) +…+(n+1)(n+ω)(n+ω2) is

Accepted Answer

r th term of the given expression is (r+1)(r+ω)(r+ω2)=r3+1
∴ value of the given expression is
∑r=1n​(r3+1)=41​n2(n+1)2+n

Question 13

The greatest and the least value of ∣z1​+z2​∣ if z1​=24+7i and ∣z2​∣=6 are respectively

Accepted Answer

Note that ∣z∣=6 represents a circle. As ∣z2​∣=6 ∣z1​+z2​∣=∣z2​−(−24−7i)∣ represent distance between a point on the circle ∣z∣=6 and the point C(−24−7i).

∣z1​+z2​∣ will be greatest and least at points B and A which are the end points of the diameter of the circle through C. As OC=25,CA=OC−OA=25−6=19 and CB= OC+OB=25+6=31.

∣z2​∣=∴∣z1​+z2​∣2=−====​6⇒z2​=6eiθ where θεR.∣24+7i+6(cosθ+isinθ)∣2(24+6cosθ)2+(7+6sinθ)2576+36cos2θ+288cosθ+49+36sin2θ+84sinθ625+36+12(24cosθ+7sinθ)661+12(25)sin(θ+α)[ put 7=rcosα and 24=rsinα]661+300sin(θ+α)​

Thus, greatest possible value of ∣z1​+z2​∣2 is 661+300= 961 and the least possible value of ∣z1​+z2​∣2 is 361 .
∴ greatest and least possible values of ∣z1​+z2​∣ are 31 and 19.

Question 14

For complex numbers z1​,z2​ and z3​ satisfying z2​−z3​z1​−z3​​=21−i3​​ are the vertices of a triangle which is

Accepted Answer

We have
∣∣​z2​−z3​z1​−z3​​∣∣​=∣∣​21−i3​​∣∣​=41​(1+3)​=1
⇒∣z1​−z3​∣=∣z2​−z3​∣
 Also, z2​−z3​z1​−z3​​−1=21−i3​​−1⇒z2​−z3​z1​−z2​​=2−1−i3​​
⇒∣z2​−z3​∣∣z1​−z2​∣​=41​(1+3)​=1
⇒∣z1​−z2​∣=∣z2​−z3​∣
 Thus, ∣z1​−z3​∣=∣z2​−z3​∣=∣z2​−z1​∣
Hence, z1​,z2​ and z3​ are the vertices of an equilateral triangle.

Question 15

If a>0, and the equation ∣∣​z−a2∣∣​+ ∣z−2a∣=3 represents an ellipse, then a lies in

Accepted Answer

The equation ∣∣​z−a2∣∣​+∣z−2a∣=3 will represent an ellipse if
∣∣​a2−2a∣∣​<3
[See Theory]
⇔−3<a2−2a<3
⇔−2<(a−1)2<4
⇔(a−1)2<4⇔−2<a−1<2
⇔−1<a<3
⇒a∈(0,3)[∵a>0]

Question 16

If θ is real and z1​,z2​ are connected by z12​+z22​+2z1​z2​cosθ=0, then triangle with vertices 0,z1​ and z2​ is

Accepted Answer

z12​+z22​+2z1​z2​cosθ=0
⇒(z2​z1​​)2+2(z2​z1​​)cosθ+1=0
⇒(z2​z1​​+cosθ)2=−(1−cos2θ)=−sin2θ
⇒z2​z1​​=−cosθ±isinθ
⇒∣∣​z2​z1​​∣∣​=(−cosθ)2+sin2θ​=1
⇒∣z1​∣=∣z2​∣⇒∣z1​−0∣=∣z2​−0∣
Thus, triangle with vertices 0, z1 and z2 are vertices of an isosceles triangle.

Question 17

If ∣∣​z−z4​∣∣​=2, then the greatest value of ∣z∣ is

Accepted Answer

We have ∣z∣=∣∣​z−z4​+z4​∣∣​≤∣∣​z−z4​∣∣​+∣z∣4​⇒∣z∣≤2+∣z∣4​⇒∣z∣2≤2∣z∣+4⇒⇒∣z∣−1)2≤5⇒∣z∣−1≤5​⇒∣z∣≤5​+1 Also, for z=5​+1,∣∣​z−z4​∣∣​=2​
Therefore, the greatest value of ∣z∣ is 5​+1 which is attained when z=5​+1

Question 18

Reflection of the line aˉz+azˉ=0 in the real axis is

Accepted Answer

Let a=α+iβ and z=x+iy, then aˉz+azˉ=0 becomes αx+βy=0 or y=(β−α​)x. Its reflection in the x-axis is
y=βα​x or αx−βy=0
or (2a+aˉ​)(2z+zˉ​)−(2ia−aˉ​)(2iz−zˉ​)=0 or az+aˉzˉ=0

Question 19

If A(z1​),B(z2​) and C(z3​) are the vertices of a triangle such that ∣z1​∣=∣z2​∣=∣z3​∣>0, then orthocentre of △ABC is

Accepted Answer

As ∣z1​∣=∣z2​∣=∣z3​∣, circumcentre of △ABC is zS​=0. Also, centroid of △ABC is zG​=31​(z1​+z2​+z3​)
If zH​ denotes the orthocentre of △ABC, then

zG​=31​(zH​+2zS​)⇒31​(z1​+z2​+z3​)=31​(zH​+2(0))⇒zH​=z1​+z2​+z3​​

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Solutions for IIT JEE Advanced Comprehensive Mathematics by Ravi Prakash

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