12th Grade > Mathematics
COMPLEX NUMBERS MCQs
Complex Numbers
Total Questions : 60
| Page 1 of 6 pages
Answer: Option A. -> 1
:
A
Let z = x + iy, ¯z = x -iy and z−1=1x+iy
⇒¯¯¯¯¯¯¯¯z−1=x+iyx2+y2 ; ∴¯¯¯¯¯¯¯¯z−1¯z=x+iyx2+y2(x−iy) = 1
:
A
Let z = x + iy, ¯z = x -iy and z−1=1x+iy
⇒¯¯¯¯¯¯¯¯z−1=x+iyx2+y2 ; ∴¯¯¯¯¯¯¯¯z−1¯z=x+iyx2+y2(x−iy) = 1
Answer: Option D. -> 4
:
D
Let z = x+iy, so that ¯z = x - iy, therefore
z2+¯z=0⇔(x2−y2+x)+i(2xy−y) = 0
Equating real and imaginary parts , we get
x2−y2+x = 0 .......(i)
And 2xy - y = 0⇒ y = 0 or x = 12
if y = 0 , then (i) gives x2 + x = 0⇒ x = 0 or
x = -1
If x = 12,
Then x2−y2+x=0⇒y2=14+12=34⇒y=±√32
Hence, there are four solutions in all.
:
D
Let z = x+iy, so that ¯z = x - iy, therefore
z2+¯z=0⇔(x2−y2+x)+i(2xy−y) = 0
Equating real and imaginary parts , we get
x2−y2+x = 0 .......(i)
And 2xy - y = 0⇒ y = 0 or x = 12
if y = 0 , then (i) gives x2 + x = 0⇒ x = 0 or
x = -1
If x = 12,
Then x2−y2+x=0⇒y2=14+12=34⇒y=±√32
Hence, there are four solutions in all.
Answer: Option B. -> a2+b2
:
B
we have
(1+i)(1+2i)(1+3i)......(1+ni) = a+ib .......(i)
⇒ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)
Multiplying (i) and (ii), we get 2.5.10.....(1+n2) = a2+b2
:
B
we have
(1+i)(1+2i)(1+3i)......(1+ni) = a+ib .......(i)
⇒ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)
Multiplying (i) and (ii), we get 2.5.10.....(1+n2) = a2+b2
Answer: Option A. -> z=0
:
A
Let z = x+iy,¯z = x-iy
∴ z¯z = 0⇒ (x+iy)(x-iy) = 0 ⇒ x2+y2 = 0
It is possible onle when x and y oth simultaneously zero
i.e, z = 0 +0i = 0
:
A
Let z = x+iy,¯z = x-iy
∴ z¯z = 0⇒ (x+iy)(x-iy) = 0 ⇒ x2+y2 = 0
It is possible onle when x and y oth simultaneously zero
i.e, z = 0 +0i = 0
Answer: Option B. -> e−rsinθ
:
B
if z=reiθ=r(cosθ + isinθ)
⇒ iz=ir(cosθ + isinθ)=-rcosθ + irsinθ
or eiz = e−rcosθ+irsinθ=e−sinθericosθ
or |eiz|=|e−rsinθ||eircosθ|
=e−rsinθ[cos2(rcosθ)+sin2(rcosθ)]2=e−rsinθ
:
B
if z=reiθ=r(cosθ + isinθ)
⇒ iz=ir(cosθ + isinθ)=-rcosθ + irsinθ
or eiz = e−rcosθ+irsinθ=e−sinθericosθ
or |eiz|=|e−rsinθ||eircosθ|
=e−rsinθ[cos2(rcosθ)+sin2(rcosθ)]2=e−rsinθ
Answer: Option C. -> 12(2√(a2+4)+a)
:
C
let z=r (cosθ+isinθ)
Then ∣∣z+1z∣∣=a ⇒ ∣∣z+1z∣∣2=a2
⇒ r2+1r2+2cosθ = a2 ⋯ ⋯(i)
Differentiating w.r.t θ we get
2rdrdθ-2r3drdθ-4sin2θ
Putting drdθ=0, we get θ=0,π2
r is maximum for θ = 0, π2, therefore from (i)
r2+1r2−2=a2⇒r−1r=a⇒r=a+2√a2+42
:
C
let z=r (cosθ+isinθ)
Then ∣∣z+1z∣∣=a ⇒ ∣∣z+1z∣∣2=a2
⇒ r2+1r2+2cosθ = a2 ⋯ ⋯(i)
Differentiating w.r.t θ we get
2rdrdθ-2r3drdθ-4sin2θ
Putting drdθ=0, we get θ=0,π2
r is maximum for θ = 0, π2, therefore from (i)
r2+1r2−2=a2⇒r−1r=a⇒r=a+2√a2+42
Answer: Option A. -> 1
:
A
c+ic−i = a+ib .........(i)
∴ c−ic+i = a-ib .........(ii)
Multiplying (i) and (ii), we get
c2+1c2+1=a2+b2⇒a2+b2=1.
:
A
c+ic−i = a+ib .........(i)
∴ c−ic+i = a-ib .........(ii)
Multiplying (i) and (ii), we get
c2+1c2+1=a2+b2⇒a2+b2=1.
Answer: Option A. -> 1
:
A
c+ic−i = a+ib .........(i)
∴ c−ic+i = a-ib .........(ii)
Multiplying (i) and (ii), we get
c2+1c2+1=a2+b2⇒a2+b2=1.
:
A
c+ic−i = a+ib .........(i)
∴ c−ic+i = a-ib .........(ii)
Multiplying (i) and (ii), we get
c2+1c2+1=a2+b2⇒a2+b2=1.
Answer: Option A. -> zero
:
A
Since1-i = √2 [cosπ4−isinπ4], |1-i|
∴ |1−i|x=2x ⇒ (√2)x=2x ⇒ 2x/2 =2x
⇒ x2=x then x=0.
Therefore, the number of non-zero integral solutions is nil or Zero.
:
A
Since1-i = √2 [cosπ4−isinπ4], |1-i|
∴ |1−i|x=2x ⇒ (√2)x=2x ⇒ 2x/2 =2x
⇒ x2=x then x=0.
Therefore, the number of non-zero integral solutions is nil or Zero.
Answer: Option C. -> θ
:
C
Given, |z1|=1, arg z=θ∴z=e−iθ
But ¯¯¯z=1z∴arg(1+z1+1z)=arg(z)=θ
:
C
Given, |z1|=1, arg z=θ∴z=e−iθ
But ¯¯¯z=1z∴arg(1+z1+1z)=arg(z)=θ
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