Class 11 Maths Ncert Solutions Ex 5.2

Class 11 Maths Ncert Solutions Chapter 5 Ex 5.2

Q.1: Find out the modulus and argument of the given complex number:

z=1i3

 

Sol:

Assuming rcosΘ=1andrsinΘ=3

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(3)2r2(cos2Θ+sin2Θ)=1+3r2(1)=4r=4=2

Therefore, Modulus = 2   [ Since, r > 0 ]

Now, 2cosΘ=1and2sinΘ=3cosΘ=12andsinΘ=32

Since, the values of both cosΘandsinΘare negative and they both are negative in 3rd quadrant

Therefore, Argument = (ππ3)=2π3

 

 

Q.2: Find out the modulus and argument of the given complex number

z=3+i

 

Sol:

Assuming rcosΘ=3andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(3)2+(1)2r2(cos2Θ+sin2Θ)=3+1r2(1)=4r=4=2

Therefore, Modulus = 2              [ Since, r > 0 ]

Now, 2cosΘ=3and2sinΘ=1cosΘ=32andsinΘ=12

Here, Θ lies in the 2nd quadrant.

Therefore, Argument = (ππ6)=5π6

 

 

Q.3: Convert the following complex number in polar form: 1 – i

Sol:

Assuming rcosΘ=1andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2

Therefore, Modulus = 2         [ Since, r > 0 ]

Now, 2cosΘ=1and2sinΘ=1

cosΘ=12andsinΘ=12

Here, Θ lies in 4th quadrant.

Therefore, Θ = (π4)

Therefore, the required polar form is:

1i=rcosΘ+irsinΘ=2cos(π4)+i2sin(π4)

Therefore, 1i=2[cos(π4)+isin(π4)]

 

 

Q.4: Convert the following complex number in polar form;   -1 + i

 

Sol:

Assuming rcosΘ=1andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2

Therefore, Modulus =  2               [ Since, r > 0 ]

Now,2cosΘ=1and2sinΘ=1

cosΘ=12andsinΘ=12

Here, Θ lies in 2nd quadrant.

Therefore, Θ = (ππ4)=3π4

So, the required polar form is:

1+i=rcosΘ+irsinΘ=2cos(3π4)+i2sin(3π4)

Therefore, -1 + i = 2[cos(3π4)+isin(3π4)]

 

 

Q.5: Convert the following complex number in polar form;  -3

Sol:

Assuming rcosΘ=3andrsinΘ=0

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(3)2+(0)2r2(cos2Θ+sin2Θ)=9r2(1)=9r=9=3

Therefore, Modulus = 3    [ Since, r > 0]

Now, 3cosΘ=3and3sinΘ=0cosΘ=1andsinΘ=0

Therefore, Θ=π

So, the required polar form is:

3=rcosΘ+irsinΘ=3cos(π)+i.3sin(π)

Therefore, 3=3[cos(π)+isin(π)]

 

 

Q.6: Convert the following complex number in polar form;   -1 – i

Sol:

 

Assuming rcosΘ=1andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2

Therefore, Modulus = 2           [ Since, r > 0 ]

Now, 2cosΘ=1and2sinΘ=1

cosΘ=12andsinΘ=12

 

 

Q.7: Convert the following complex number in polar form: z=3+i

 

Sol:

z=3+i

Assuming rcosΘ=3andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(3)2+(1)2r2(cos2Θ+sin2Θ)=3+1r2(1)=4r=4=2

Therefore, Modulus = 2     [Since, r > 0]

Now,

2cosΘ=3and2sinΘ=1cosΘ=32andsinΘ=12

Here, Θ lies in 1st quadrant.

Therefore, Θ = (π6)

So, the required polar form is:

3+i=rcosΘ+irsinΘ=2cos(π6)+i2sin(π6)

Therefore, 3+i=2[cos(π6)+isin(π6)]

 

 

Q.8: Convert the following complex number in polar form: i

 

Sol:

Assuming rcosΘ=0andrsinΘ=1

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(0)2+(1)2r2(cos2Θ+sin2Θ)=1r2(1)=1r=1=1

Therefore, Modulus = 1     [ Since, as r > 0)]

Now, cosΘ=0and3sinΘ=1

Therefore, Θ=π2

So, the required polar form is:

i=cos(π2)+isin(π2)