# Free Objective Test 02 Practice Test - 11th and 12th

1+7i(2i)2=

A.

2 [cos3π4+isin3π4]

B.

2 [cosπ4+isinπ4]

C.

[cos3π4+isin3π4]

D.

None of these

#### SOLUTION

Solution : A

1+7i(2i)2=(1+7i)(3+4i)(34i)(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(2i)2=2 [cos3π4+isin3π4]

Alter: 1+7i(2i)2=1+7i34i=2

And arg (1+7i(2i)2)=tan17tan1(43)

=tan17+tan1(43)=3π4

1+7i(2i)2=2 [cos3π4+isin3π4]

If z = 3+5i, then z3¯z + 198 =

A.

-3-5i

B.

-3+5i

C.

3+5i

D.

3-5i

#### SOLUTION

Solution : 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 .

The least positive integer n which will reduce (i1i+1)nto a real number , is

A.

2

B.

3

C.

4

D.

5

#### SOLUTION

Solution : A

(i1i+1×i1i1)n=(2i2)n=in

Hence, to make the real number the least positive integar is 2.

If a=22i then which of the following is correct?

A.

a=1+i

B.

a=2×(1+i)

C.

1

D.

None of these

#### SOLUTION

Solution : A

We have a =2i=2(cosπ2+isinπ2)12=2(eiπ2)12=2eiπ4=2(12+i2i)=1+i
Trick:Check with options.

The value of i1+3+5+......+(2n+1) is

A.

i if n is even, - i if n is odd

B.

1 if n is even, - 1 if n is odd

C.

1 if n is odd, i if n is even

D.

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, .........

If z is a complex number ¯¯¯¯¯¯¯¯z1(¯z) = , then

A.

1

B.

-1

C.

0

D.

None of these

#### SOLUTION

Solution : A

Let z = x + iy, ¯z = x -iy  and  z1=1x+iy

⇒ ¯¯¯¯¯¯¯¯z1=x+iyx2+y2 ; ∴ ¯¯¯¯¯¯¯¯z1¯z=x+iyx2+y2(xiy)  = 1

The complex numbers sinx+icos2x and cosx-isin2x are conjugate to each other for

A.

x = nπ

B.

x = (n+12)π

C.

x=0

D.

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.

If a=cosα+isinα, b=cosβ+isinβ,

c=cosγ+isinγ and bc+ca+ab=1,Then

cos(β-γ)+cos(γ-α)+cos(α-β) is equal to

A. 32
B. -32
C. 0
D. 1

#### SOLUTION

Solution : D

bc=cosβ+isinβcosγ+isinγ×cosγisinγcosγisinγ

bc=cos(βγ)+isin(βγ).....(i)

Similarly, ca=cos(γα)+isin(γα).......(ii)

And ab=cos(αβ)+isin(αβ).......(ii)

Equating real and imaginary parts,

cos(βγ)+cos(γα)+cos(αβ)=1

If cosα+cosβ+cosγ=0=sinα+sinβ+sinγ then

cos2α+cos2β+cos2γ equals

A.

2cos(α+β+γ)

B.

cos2(α+β+γ)

C.

0

D.

1

#### SOLUTION

Solution : C

cosα+cosβ+cosγ=0 ......(i)

sinα+sinβ+sinγ=0 .......(ii)

Let a=cosα+isinα, b=cosβ+isinβ

c=cosγ+isinγ

a+b+c=0 .......(iii)

1a+1b+1c

=[cosα+isinα]1+[cosβ+isinβ]1+[cosγ+isinγ]1

=cosα-isinα+cosβ-isinβ+cosγ-isinγ

ab+bc+ca=0 .....(iv) [by (i) and (ii)]

Squaring both sides of equations (iii),

We get a2+b2+c2+2ab+2bc+2ca=0

or a2+b2+c2 [by (iv)]

[cosα+isinα]2+[cosβ+isinβ]2+[cosγ+isinγ]2=0

(cos2α+cos2β+cos2γ) + i(sin2α+sin2β+sin2γ)

Equating the real part to zero, we have

cos2α+cos2β+cos2γ=0

If  sinα+sinβ+sinγ=0= cosα+cosβ+cosγ,

value of  sin2α+sin2β+sin2γ

A.

23

B.

32

C.

12

D.

1

#### SOLUTION

Solution : B

cosα+cosβ+cosγ = 0............(i)

sinα+sinβ+sinγ  =0...........(ii)

Let a=cosα+isinα, b=cosβ+isinβ

c=cosγ+isinγ

a+b+c=0 [By (i) and (ii)]....(iii)

1a+1b+1c

=cosα-isinα+cosβ-isinβ+cosγ-isinγ

ab+bc+ca=0 ......(iv) [by (i) and (ii)]

Squaring both sides of equation (iii),

we get a2+b2+c2+2ab+2bc+2ca=0

or a2+b2+c2=0     [by (iv)]

[cosα+isinα]2 + [cosβ+isinβ]2 + [cosγ+isinγ]2 = 0

(cos2α+cos2β+cos2γ) + i(sin2α+sin2β+sin2γ)=0

Separation of real and imaginary part,

cos2α+cos2β+cos2γ=0 .........(v)

And sin2α+sin2β+sin2γ =0 .........(iv)

1-2sin2α+1-2sin2β+1-2sin2γ=0 [By eq.(v)]

Therefore, sin2α+sin2β+sin2γ= 32

The maximum distance from the origin of coordinates to the point z satisfying the equation z+1z=a is

A.

12(2(a2+1)+a)

B.

12(2(a2+2)+a)

C.

12(2(a2+4)+a)

D.

None of these

#### SOLUTION

Solution : C

let z=r (cosθ+isinθ)

Then z+1z=a z+1z2=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+1r22=a2r1r=ar=a+2a2+42

3+i=(a+ib)(c+id),thentan1ba+tan1dc

has the value

A.

2nπ+π3, nϵI

B.

nπ+π6, nϵI

C.

nπ-π3, nϵI

D.

2nπ-π6, nϵI

#### SOLUTION

Solution : B

3+i=(a+ib)(c+id)

Now  tan1(ba)+tan1(dc)

=nπ+π6, nϵI

The number of non-zero integral solutions of the equation  |1i|x=2x is

A.

zero

B.

1

C.

2

D.

None of these

#### SOLUTION

Solution : A

Since1-i = 2 [cosπ4isinπ4], |1-i|

|1i|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.

If c+ici = a+ib, where a,b,c are real, then a2+b2

A.

1

B.

-1

C.

c2

D.

c2

#### SOLUTION

Solution : A

c+ici = a+ib     .........(i)

∴   cic+i = a-ib     .........(ii)

Multiplying (i) and (ii), we get

c2+1c2+1=a2+b2a2+b2=1.

The square root of 3 - 4i is

A.

±(2+i).

B.

±(2-i).

C.

±(1-2i).

D.

±(1+2i).

#### SOLUTION

Solution : B

Let 34i=x+iy 3-4i= x2-y2+2ixy

x2-y2=3, 2xy=-4   ..........(i)

(x2+y2)2=(x2y2)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).