Free Complex Numbers 02 Practice Test - 11th Grade - Commerce
Question 1
The value of i1+3+5+......+(2n+1) is
i if n is even, - i if n is odd
1 if n is even, - 1 if n is odd
1 if n is odd, i if n is even
i if n is even, - 1 if n is odd
SOLUTION
Solution : C
Let z = i[1+3+5+......+(2n+1)]
Clearly series is A.P with common difference = 2
∵ Tn = 2n-1 And Tn+1 = 2n+1
So, number of terms in A.P. = n+1
Now, sn+1=n+12 [2.1+(n+1-1)2]
⇒ Sn+1=n+12[2+2n]=(n+1)2 i.e., i(n+1)2
Now put n = 1,2,3,4,5,........
n = 1, z = i4 = 1, n = 2, z = i5 = -1 ,
n = 3, z = i8 = 1, n = 4, z = i10 = -1 ,
n = 5, z = i12 = 1, .........
Question 2
If z is a complex number ¯¯¯¯¯¯¯¯z−1(¯z) = , then
1
-1
0
None of these
SOLUTION
Solution : 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
Question 3
The complex numbers sinx+icos2x and cosx-isin2x are conjugate to each other for
x = nπ
x = (n+12)π
x=0
No value of x
SOLUTION
Solution : D
sinx+icos2x and cosx-isin2xare conjugate to each other if sinx=cosx and 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 x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate.
Question 4
The maximum distance from the origin of coordinates to the point z satisfying the equation ∣∣z+1z∣∣=a is
12(2√(a2+1)+a)
12(2√(a2+2)+a)
12(2√(a2+4)+a)
None of these
SOLUTION
Solution : 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
Question 5
√3+i=(a+ib)(c+id),thentan−1ba+tan−1dc
has the value
2nπ+π3, nϵI
nπ+π6, nϵI
nπ-π3, nϵI
2nπ-π6, nϵI
SOLUTION
Solution : 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
Question 6
The number of non-zero integral solutions of the equation |1−i|x=2x is
zero
1
2
None of these
SOLUTION
Solution : 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.
Question 7
If c+ic−i = a+ib, where a,b,c are real, then a2+b2 =
1
-1
c2
−c2
SOLUTION
Solution : 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.
Question 8
The square root of 3 - 4i is
±(2+i).
±(2-i).
±(1-2i).
±(1+2i).
SOLUTION
Solution : 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).
Question 9
The conjugate of (2+i)23+i , in the form of a+ib, is
132+i(152)
1310+i(−152)
1310+i(−910)
1310+i(910)
SOLUTION
Solution : C
z = (2+i)23+i=3+4i3+i×3−i3−i=1310+i910
Conjugate = 1310−i910.
Question 10
If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+n2) is equal to
a2-b2
a2+b2
√a2+b2
√a2−b2
SOLUTION
Solution : 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