Class 11 Maths Ncert Solutions Ex 5.1

Class 11 Maths Ncert Solutions Chapter 5 Ex 5.1

Q.1: Express the following complex number in x + iy form; (4i)(74i)

Sol:

(4i)(74i)  = -4 × 74 × i × i = -7 i2 = -7(-1)   [ Since, i2 = -1 ]

= 7

Q.2: Express the following complex number in x + iy form; i19+i9 

Sol:

i19+i9=i4×2+1+i4×4+3 = (i4)2i+(i4)4i3

=1×i+1×(i)   [ Since,  i4 = 1, i3 = -1 ]

= i + (-i)

= 0

 

 

Q.3: Express the following complex number in x + iy form; i39

 

Sol:

i39=i4×93 =(i4)9i3

=(1)9i3   [ Since, i4 = 1 ]   

=1i3=1i   [ Since, i3 = -i ]

=1i×ii

=ii2=i1 = i   [ Since, i2 = -1 ]

 

 

Q.4: Express the following complex number in x + iy form; 5(6 + 6i) + i(6 + 6i)

 

Sol:

5(6 + 6i) + i(6 + 6i) = 30 + 30i + 6i + 6i2

= 30 + 36i + 6(-1)    [ Since, i2 = -1 ]

= 30 – 6 + 36i

= 24 + 36i

 

 

Q.5: Express the following complex number in x + iy form; (2 – 3i) – (-4 + 5i)

Sol:

(2 – 3i) – (-4 + 5i) = 2 – 3i + 4 – 5i

= 6 – 8i

 

 

Q.6: Express the following complex number in x + iy form;

(14+i34)(6+i43)

 

Sol:

(14+i34)(6+i43)=14+i346i43

=(146)+i(3443)

=234+i(712)

=234i712

 

 

Q.7: Express the following complex number in x + iy form:

[(13+i73)+(4+i13)](43+i)

 

Sol:

[(13+i73)+(4+i13)](43+i)

=13+i73+4+i13+43i

=(13+43+4)+i(73+131)

=173+i53

 

 

Q.8: Express the following complex number in x + iy form: (1 – i)4

Sol:

 

(1 – i)4 = [(1i)2]2

=[12i+i2]2=[112i]2=(2i)2=(2i)×(2i)=4i2          [ Since i2 = -1 ]

= -4

 

 

Q.9: Express the following complex number in ‘x + iy’ form; (13+3i)3

Sol:

 

=(13+3i)3:

=(13)3+(3i)3+3(13)(3i)(13+3i)

=127+27i3+3i(13+3i)

=127+27i3+i+9i2

=12727i+i9      [Since, i3 = -i and i2 = -1]

=(1279)+i(27i+1) = 2422726i

 

 

Q.10: Express the following complex number in ‘x + iy’ form: (2i13)3

SoL:

 

(2i13)3=(1)3(2+i13)3 = [23+(i3)3+3(2)(i3)(2+i3)] = [8+i327+2i(2+i3)]

=[8+i327+4i+2i23)] = [8i27+4i23]      [ Since, i3 = -i and i2 = -1 ]

=[223+i10727]=221i10727

 

 

Q.11: Find the multiplicative inverse of the given complex number, 7 – 6i.

Sol:

 

Assuming, z = 7 – 6i

Now, z¯¯¯=7+6i

And, |z|2 = (7)2 + (-6)2 = 49 + 36 = 85

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=7+6i85=785+685i

 

 

Q.12: Find the multiplicative inverse of the given complex number, 7+4i

Sol:

Assuming, z = 7+4i

Now, z¯¯¯=7+4i

And, |z|2 = (7)2+(4)2=23

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=74i23=723423i

 

 

Q.13: Find the multiplicative inverse of the given complex number, – i

Sol:

 

Assuming, z = – i

Now, z¯¯¯=i

And, |z|2 = (-i)2 = 1

Therefore, the multiplicative inverse of the given complex number is given by:

z1=z¯¯¯|z|2=i1 = i

 

 

Q.14: Express the following complex number in ‘x + iy’ form: (3+i5)(3i5)(3+i2)(3i2)

Sol:

 

(3+i5)(3i5)(3+i2)(3i2)=(3)2(i5)23+i23+i2

Now, by using (a + b) (a – b) = (a – b)2

=95i222i=95×(1)22i=9+522i×ii=14i22i2

=14i22=7i2×22=72i2