# Free Objective Test 01 Practice Test - 11th and 12th

For positive integers n1,n2 the value of the expression  (1+i)n1+(1+i3)n1+(1+i5)n2 + (1+i7)n2 where i=21 is a real number if and only if

A.

n1=n2+1

B.

n1=n2-1

C.

n1=n2

D.

n1>0,n2>0

#### SOLUTION

Solution : D

Using i3=i,i5=i and i7=i, we can write the given expression as
(1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2 where i=21=2[n1C0+n1C2(i)2+n1C4(i)4+n1C6(i)6+]+2[n2C0+n2C2(i)2+n2C4(i)4+n2C6(i)6+]=2[n1C0n1C2+n1C4n1C6+]+2[n2C0n2C2+n2C4n2C6+]
This is a real number irrespective of values of n1 and n2.

The conjugate of (2+i)23+i , in the form of a+ib, is

A.

132+i(152)

B.

1310+i(152)

C.

1310+i(910)

D.

1310+i(910)

#### SOLUTION

Solution : C

z = (2+i)23+i=3+4i3+i×3i3i=1310+i910

Conjugate = 1310i910.

If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+n2) is equal to

A. a2-b2
B. a2+b2
C. a2+b2
D. a2b2

#### 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

If x=-5+4, then the value of the expression x4+9x3+35x2-x+4 is

A.

160

B.

-160

C.

60

D.

-60

#### SOLUTION

Solution : B

x+5 =4i x2+10x+25=-16

Now, x4+9x3+35x2-x+4

=(x2+10x+41)(x2-x+4)-160=-160

1i1+i is equal to

A.

cosπ2+isinπ2

B.

cosπ2-isinπ2

C.

sinπ2+icosπ2

D.

None of these

#### SOLUTION

Solution : B

1i1+i=(1i)(1i)(1+i)(1i)=1+(i)22i1+1=-i

Which can be written as cosπ2-isinπ2

The values of x and y for which the numbers 3+ix2y and x2 +y+4i are conjugate complex are

A.

(-2,-1) or (2,-1)

B.

(1,-2) or (-2,1)

C.

(1,2) or (-1,-2)

D.

None of these

#### SOLUTION

Solution : A

According to condition, 3-ix2 y = x2  + y + 4i

⇒ x2 + y = 3 And x2y = -4 ⇒ x = ±2, y = -1

⇒  (x,y) = (2,-1) or (-2,-1)

For the complex number z, which of the following is true?

A.

z+¯z is real and z¯z is imaginary

B.

z+¯z is imaginary and  z¯z is real

C.

Both z+¯z and z¯z are real.

D.

Both z+¯z and z¯z are imaginary numbers

#### SOLUTION

Solution : C

Let z=x+iy

Here,  z+¯z=(x+iy)+(xiy)=2x (Real)

And z¯z=(x+iy)(xiy)=x2+y2 (Real).

The number of solutions of the equation z2¯z = 0 is

A.

1

B.

2

C.

3

D.

4

#### SOLUTION

Solution : D

Let z = x+iy, so that ¯z = x - iy, therefore

z2+¯z=0(x2y2+x)+i(2xyy) = 0

Equating real and imaginary parts , we get

x2y2+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 x2y2+x=0y2=14+12=34y=±32

Hence, there are four solutions in all.

If  z  is a complex number, then z.¯z  = 0 if and only if

A.

z=0

B.

Re(z)=0

C.

Im(z)=0

D.

None of these

#### SOLUTION

Solution : 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

If z is a complex number such that z2 = (¯z)2,then

A.

z is purely real

B.

z is purely imaginary

C.

Either z is purely real or purely imaginary

D.

None of these

#### SOLUTION

Solution : C

Let z = x+iy, then its coonjucate ¯z = x-iy

Given that z2=(¯z)2

⇒  x2y2 + 2ixy = x2y2 - 2ixy ⇒ 4ixy = 0

if x ≠ 0 then y = 0 and if y ≠ 0 then x = 0

If -1+3=reiθ, then θ is equal to

A.

π3

B.

-π3

C.

2π3

D.

-2π3

#### SOLUTION

Solution : C

Here -1+3=reiθ -1+i3=reiθ

=rcosθ=-1 And rsinθ=3

Hence tanθ=-3  tanθ=tan2π3.Hence θ=2π3

If z=reiθ,then |eiz|=

A.

ersinθ

B.

ersinθ

C.

ercosθ

D.

ercosθ

#### SOLUTION

Solution : B

if z=reiθ=r(cosθ + isinθ)

iz=ir(cosθ + isinθ)=-rcosθ + irsinθ

or eiz = ercosθ+irsinθ=esinθericosθ

or |eiz|=|ersinθ||eircosθ|

=ersinθ[cos2(rcosθ)+sin2(rcosθ)]2=ersinθ

Real part of eeiθ is

A.

ecosθ [cos(sinθ)]

B.

ecosθ [cos(cosθ)]

C.

esinθ [cos(sinθ)]

D.

esinθ [sin(sinθ)]

#### SOLUTION

Solution : A

eeiθ=ecosθ+isinθ=ecosθeisinθ=ecosθ[cos(sinθ)+isin(sinθ)]Real part ofeeiθ is ecosθ[cos(sinθ)]

If z1,z2 and z3 are complex numbers such that

|z1|=|z2|=|z3|=1z1+1z2+1z3=1,

then |z1+z2+z3|

A.

Equal to 1

B.

Less than 1

C.

Greater than 3

D.

Equal to 3

#### SOLUTION

Solution : A

1=1z1+1z2+1z3=z1¯z1z1+z2¯z2z2+z3¯z3z3

(hence, |z1|2=1=z1¯¯¯¯¯z1, etc)

|z1+z2+z3|=|¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯z1+z2+z3|= |z1+z2+z3|

(hence,|¯¯¯¯¯z1|=|z1|)

Find the complex number z satisfying the equations z12z8i=53,  z4z8=1

A.

6

B.

6±8i

C.

6+8i, 6+17i

D.

None of these

#### SOLUTION

Solution : C

We have z12z8i=53,  z4z8=1

Let z=x+iy, then

z12z8i=53 3|z-12| = 5|z-8i|

9(x12)2+9y2 = 25x2+25(y8)2     (i)

z4z8=1 |z-4| = |z-8|

(x4)2+y2 = y2+(x8)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.