12th Grade > Mathematics
COMPLEX NUMBERS MCQs
Complex Numbers
Total Questions : 60
| Page 3 of 6 pages
Answer: Option A. -> ecosθ [cos(sinθ)]
:
A
eeiθ=ecosθ+isinθ=ecosθeisinθ=ecosθ[cos(sinθ)+isin(sinθ)]∴Realpartofeeiθisecosθ[cos(sinθ)]
:
A
eeiθ=ecosθ+isinθ=ecosθeisinθ=ecosθ[cos(sinθ)+isin(sinθ)]∴Realpartofeeiθisecosθ[cos(sinθ)]
Answer: Option C. -> 2π3
:
C
Here -1+√−3=reiθ ⇒ -1+i√3=reiθ
=rcosθ=-1 And rsinθ=√3
Hence tanθ=-√3 ⇒tanθ=tan2π3.Hence θ=2π3
:
C
Here -1+√−3=reiθ ⇒ -1+i√3=reiθ
=rcosθ=-1 And rsinθ=√3
Hence tanθ=-√3 ⇒tanθ=tan2π3.Hence θ=2π3
Answer: Option B. -> cosπ2-isinπ2
:
B
1−i1+i=(1−i)(1−i)(1+i)(1−i)=1+(i)2−2i1+1=-i
Which can be written as cosπ2-isinπ2
:
B
1−i1+i=(1−i)(1−i)(1+i)(1−i)=1+(i)2−2i1+1=-i
Which can be written as cosπ2-isinπ2
Answer: Option C. -> 6+8i, 6+17i
:
C
We have ∣∣z−12z−8i∣∣=53, ∣∣z−4z−8∣∣=1
Let z=x+iy, then
∣∣z−12z−8i∣∣=53 ⇒ 3|z-12| = 5|z-8i|
⇒ 9(x−12)2+9y2 = 25x2+25(y−8)2 ⋯⋯(i)
∣∣z−4z−8∣∣=1 ⇒ |z-4| = |z-8|
⇒ (x−4)2+y2 = y2+(x−8)2⇒ x=6
Putting x=6 in (i), we get y2-25y+136=0
y=17, 8
Hence z=6+17i or z=6+8i
Trick: Check it with options.
:
C
We have ∣∣z−12z−8i∣∣=53, ∣∣z−4z−8∣∣=1
Let z=x+iy, then
∣∣z−12z−8i∣∣=53 ⇒ 3|z-12| = 5|z-8i|
⇒ 9(x−12)2+9y2 = 25x2+25(y−8)2 ⋯⋯(i)
∣∣z−4z−8∣∣=1 ⇒ |z-4| = |z-8|
⇒ (x−4)2+y2 = y2+(x−8)2⇒ x=6
Putting x=6 in (i), we get y2-25y+136=0
y=17, 8
Hence z=6+17i or z=6+8i
Trick: Check it with options.
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. -> No value of x
:
D
sinx+icos2x and cosx-isin2xare conjugate to each other if sinx=cosxand cos2x=sin2x
Or tan x = 1 ⇒ x =π4,5π4,9π4, .......... ...............(i)
And tan 2x = 1⇒ 2x =π4,5π4,9π4, ..........
or x =π8,5π8,9π8, .......... ...............(ii)
There exists no value of xcommon in (i) and (ii). Therefore there is no value of xfor which the given complex numbers are conjugate.
:
D
sinx+icos2x and cosx-isin2xare conjugate to each other if sinx=cosxand cos2x=sin2x
Or tan x = 1 ⇒ x =π4,5π4,9π4, .......... ...............(i)
And tan 2x = 1⇒ 2x =π4,5π4,9π4, ..........
or x =π8,5π8,9π8, .......... ...............(ii)
There exists no value of xcommon in (i) and (ii). Therefore there is no value of xfor which the given complex numbers are conjugate.
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 A. -> √2 [cos3π4+isin3π4]
:
A
1+7i(2−i)2=(1+7i)(3+4i)(3−4i)(3+4i)=−25+25i25= -1+i
Let z=x+iy = -1+i
∴ rcosθ =-1 And rsinθ =1 ∴ θ=3π4 and r=√2
Thus 1+7i(2−i)2=√2 [cos3π4+isin3π4]
Alter: ∣∣∣1+7i(2−i)2∣∣∣=∣∣1+7i3−4i∣∣=√2
And arg (1+7i(2−i)2)=tan−17−tan−1(−43)
=tan−17+tan−1(43)=3π4
∴ 1+7i(2−i)2=√2 [cos3π4+isin3π4]
:
A
1+7i(2−i)2=(1+7i)(3+4i)(3−4i)(3+4i)=−25+25i25= -1+i
Let z=x+iy = -1+i
∴ rcosθ =-1 And rsinθ =1 ∴ θ=3π4 and r=√2
Thus 1+7i(2−i)2=√2 [cos3π4+isin3π4]
Alter: ∣∣∣1+7i(2−i)2∣∣∣=∣∣1+7i3−4i∣∣=√2
And arg (1+7i(2−i)2)=tan−17−tan−1(−43)
=tan−17+tan−1(43)=3π4
∴ 1+7i(2−i)2=√2 [cos3π4+isin3π4]
Answer: Option A. -> a=1+i
:
A
We have a =√2i=√2(cosπ2+isinπ2)12=√2(eiπ2)12=√2eiπ4=√2(1√2+i√2i)=1+i
Trick:Check with options.
:
A
We have a =√2i=√2(cosπ2+isinπ2)12=√2(eiπ2)12=√2eiπ4=√2(1√2+i√2i)=1+i
Trick:Check with options.
Answer: Option B. -> ±(2-i).
:
B
Let √3−4i=x+iy ⇒ 3-4i= x2-y2+2ixy
⇒ x2-y2=3, 2xy=-4 ..........(i)
⇒ (x2+y2)2=(x2−y2)2+4x2y2=(3)2+(−4)2=25
⇒ (x2+y2)2=5 ............(ii)
From equation (i) and (ii) ⇒ x2=4 ⇒x=±2,
y2=1 ⇒y=±1.Hence the square root of (3-4i)
is ±(2-i).
:
B
Let √3−4i=x+iy ⇒ 3-4i= x2-y2+2ixy
⇒ x2-y2=3, 2xy=-4 ..........(i)
⇒ (x2+y2)2=(x2−y2)2+4x2y2=(3)2+(−4)2=25
⇒ (x2+y2)2=5 ............(ii)
From equation (i) and (ii) ⇒ x2=4 ⇒x=±2,
y2=1 ⇒y=±1.Hence the square root of (3-4i)
is ±(2-i).
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