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NCERT Solution Complex Number part -1




Question 1:

Express the given complex number in the form a + ib: (5i)(-3i/5)

Question 2:

Express the given complex number in the form a + ib:

i9 +i19

Question 3:

Express the given complex number in the form a + ib: i-39

Question 4:

Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)

Question 5:

Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)

Question 6:

Express the given complex number in the form a + ib:

(1/5+2i/5)-(4+i5/2)

Question 7:

Express the given complex number in the form a + ib:

Question 8

Express the given complex number in the form a + ib: (1 – i)4

Question 9:

Express the given complex number in the form a + ib: (1/3+3i)3

Question 10

Express the given complex number in the form a + ib: (-2-1i/3)3

Question 11:

Find the multiplicative inverse of the complex number 4 – 3i.

 

Question 12:

Find the multiplicative inverse of the complex number

Question 13

Find the multiplicative inverse of the complex number –i

 

Question 14

Express the following expression in the form of a + ib.

Solution 1

(5i)(-3i/5)

Multiplying

=-(15/5)i2

Now we know that

i2=-1

so

=3

 

Solution 2

 i9 +i19

 

=i2X4 +1 +i4X4 +3

=(i2)4 I + (i4)4 i3

Now i2=-1 so i4=1  and i3=-i

=i+(-i)

=i-i=0

 

 

Solution 3

 

Solution 4:

 3(7 + i7) + i(7 + i7)

=21+21i+7i+7i2

Now i2=-1

So

=21+28i-7

=14+28i

 

Solution  5:

(1 – i) – (–1 + i6)

=2-7i

Solution 6:

(1/5+2i/5)-(4+i5/2)

=[(1/5)-4] +i[(2/5) –(5/2)]

=(-19/5) +(-21/10)i

 

Solution 7:

 

 

Solution 8

(1 – i)4

Can be written as

=[(1-i)2]2

=(1+i2-2i)2

=(1-1-2i)2

=4i2=-4

 

Solution 9

 

Solution 10

 

Solution 11

Let z=4-3i

Conjugate is given by

  =4+3i

Modulus is given by

|z|=5

Multiple inverse of any complex number z  is given by

 

Solution 12

Let z= √5+3i

Conjugate is given by

  = √5-3i

Modulus is given by

|z|=5+9=14

Multiple inverse of any complex number z  is given by

 

Solution 13

Let z=-i

Conjugate is given by

  =i

Modulus is given by

|z|=1

Multiple inverse of any complex number z  is given by

 

Solution 14

 

 

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