12th Grade > Mathematics
COMPLEX NUMBERS MCQs
Complex Numbers
Total Questions : 60
| Page 5 of 6 pages
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 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 B. -> nπ+π6, nϵI
:
B
√3+i=(a+ib)(c+id)
∴ac−bd=√3 and ad+bc=1
Now tan−1(ba)+tan−1(dc)
= tan−1(ab+dc1−ba.dc)= tan−1(bc+adac−bd)=tan−1(1√3)
=nπ+π6, nϵI
:
B
√3+i=(a+ib)(c+id)
∴ac−bd=√3 and ad+bc=1
Now tan−1(ba)+tan−1(dc)
= tan−1(ab+dc1−ba.dc)= tan−1(bc+adac−bd)=tan−1(1√3)
=nπ+π6, nϵI
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).
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 A. -> 2
:
A
(i−1i+1×i−1i−1)n=(−2i−2)n=in
Hence, to make the real number the least positive integar is 2.
:
A
(i−1i+1×i−1i−1)n=(−2i−2)n=in
Hence, to make the real number the least positive integar is 2.
Answer: Option D. -> −¯¯¯ω
:
D
Since |z|=|ω| and arg(z)=π−arg(ω)
Let ω=reiθ, then ¯¯¯ω=re−iθ∴z=rei(π−θ)=reiπ.e−iθ=−re−iθ=−¯¯¯ω
:
D
Since |z|=|ω| and arg(z)=π−arg(ω)
Let ω=reiθ, then ¯¯¯ω=re−iθ∴z=rei(π−θ)=reiπ.e−iθ=−re−iθ=−¯¯¯ω
Answer: Option A. -> π
:
A
arg(z1z2)=arg(z1)−arg(z2)∴arg(−z)−arg(z)=arg(−zz)=arg(−1)=π
:
A
arg(z1z2)=arg(z1)−arg(z2)∴arg(−z)−arg(z)=arg(−zz)=arg(−1)=π
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 C. -> 3+5i
:
C
z = 3+5i ,¯z = 3-5i
⇒z3=(3+5i)3=33+(5i)3+3.3.5i(3+5i)
= -198+10i
Hence, z3+¯z+ 198 = 10i-198+3-5i+198 = 3+5i .
:
C
z = 3+5i ,¯z = 3-5i
⇒z3=(3+5i)3=33+(5i)3+3.3.5i(3+5i)
= -198+10i
Hence, z3+¯z+ 198 = 10i-198+3-5i+198 = 3+5i .
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