Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.2

Last Updated : 25 Jan, 2021

Question 1.  z = â€“ 1 â€“ i âˆš3

Solution:

We have,

z = -1 – iâˆš3

We know that, z = r (cosÎ¸ + i sinÎ¸)

Therefore,

r cosÎ¸ = -1  —(1)

r sinÎ¸ = -âˆš3  —-(2)

On Squaring and adding (1) and (2), we obtain

r2 (cos 2Î¸ + sin 2Î¸) = (-1)2 + (-âˆš3)2

r2 = 1 + 3

r = âˆš4

Since r has to positive, Therefore r = 2

Putting r = 2 on (1) and (2), we get

cosÎ¸ = -1 / 2 and sinÎ¸ = -âˆš3 / 2

Therefore, Î¸ = – 2Ï€ / 3 (Since cosÎ¸ and sinÎ¸ both are negative, therefore Î¸ lies in third quadrant)

Hence, modulus and argument of z = -1 – iâˆš3 are 2 and – 2Ï€ / 3 respectively.

Question 2. z =  -âˆš3 + i

Solution:

We have,

z = -âˆš3 + i

We know that, z = r (cosÎ¸ + i sinÎ¸)

Therefore,

r cosÎ¸ = -âˆš3  —(1)

r sinÎ¸ = 1  —-(2)

On Squaring and adding (1) and (2), we obtain

r2 (cos 2Î¸ + sin 2Î¸) = (-âˆš3)2 + (1)

r2  = 3 + 1   (Since, cos 2Î¸ + sin 2Î¸ = 1)

r2 = 3 + 1

r = âˆš4

Since r has to positive, Therefore r = 2

Putting r = 2 on (1) and (2), we get

cosÎ¸ = -âˆš3 / 2 and sinÎ¸ = 1 / 2

Therefore, Î¸ = 5Ï€ / 6 (Since cosÎ¸ negative and sinÎ¸ positive, therefore Î¸ lies on second quadrant)

Hence, modulus and argument of z = -âˆš3 + i are 2 and 5Ï€ / 6 respectively.

Question 3. 1 â€“ i

Solution:

We have z = 1 – i,

Let  r cosÎ¸  = 1   —(1)  and,

r sinÎ¸  = -1 —(2)

On Squaring and adding  (1) and (2) , we obtain

r2 (  cos 2Î¸ + sin 2Î¸  )  =  (1)2 + (-1)

r2 =  2

r = âˆš2 ( Since r has to be positive )

Putting r = âˆš2 on (1) and (2) , we get

cosÎ¸ = 1 / âˆš2  and  sinÎ¸ = -1 / âˆš2

Therefore, Î¸ =  – Ï€ / 4 ( Since cosÎ¸ positive and sinÎ¸  negative, therefore Î¸ is negative as it lies on fourth quadrant)

Hence , z in polar form:  z =  r cosÎ¸  + i  r sinÎ¸ = âˆš2 (cos (- Ï€ / 4) + i sin (- Ï€ / 4)).

Question 4. -1 + i

Solution:

We have z = -1 + i,

Let  r cosÎ¸  = -1   —(1)  and,

r sinÎ¸  = 1 —(2)

On Squaring and adding (1) and (2), we obtain

r2 (cos 2Î¸ + sin 2Î¸)  = (-1)2 + (1)2

r2 = 2

r = âˆš2 (Since r has to be positive)

Putting r = âˆš2 on (1) and (2), we get

cosÎ¸ = -1 / âˆš2 and sinÎ¸ = 1 / âˆš2

Therefore, Î¸ = 3Ï€ / 4 (Since cosÎ¸ negative and sinÎ¸ positive, therefore Î¸ is positive as it lies on second quadrant)

Hence, z in polar form:  z = r cosÎ¸ + i  r sinÎ¸ = âˆš2 (cos (3Ï€ / 4) + i sin (3Ï€ / 4)).

Question 5. -1 – i

Solution:

We have z = -1 – i,

Let  r cosÎ¸  = -1   —(1)  and,

r sinÎ¸  = -1 —(2)

On Squaring and adding  (1) and (2), we obtain

r2 (cos 2Î¸ + sin 2Î¸)  =  (-1)2 + (-1)2

r2 =  2

r = âˆš2 ( Since r has to be positive )

Putting r = âˆš2 on (1) and (2) , we get

cosÎ¸ = -1 / âˆš2  and  sinÎ¸ = -1 / âˆš2

Therefore,  Î¸ =  -3Ï€ / 4 (Since  cosÎ¸  negative and sinÎ¸  negative, therefore Î¸ is negative as it lies on third quadrant)

Hence, z in polar form:  z =  r cosÎ¸  + i  r sinÎ¸  = âˆš2 (cos (-3Ï€ / 4) + i sin (-3Ï€ / 4)).

Question 6.  -3

Solution:

We have z = -3,

Let  r cosÎ¸  = -3   —(1)  and,

r sinÎ¸  = 0 —(2)

On Squaring and adding (1) and (2), we obtain

r2 (cos 2Î¸ + sin 2Î¸) =  (-3)2 + (0)

r2 = 9

r = 3 (Since r has to be positive)

Putting r = 3 on (1) and (2), we get

cosÎ¸ = -3 / 3  and  sinÎ¸ = 0 / 3

Therefore,  Î¸ =  Ï€

Hence, z in polar form:  z =  r cosÎ¸  + i  r sinÎ¸ = 3(cos (Ï€) + i sin (Ï€)).

Question 7. âˆš3 + i

Solution:

We have z = âˆš3 + i,

Let  r cosÎ¸  = âˆš3  —(1)  and,

r sinÎ¸  = 1 —(2)

On Squaring and adding  (1) and (2) , we obtain

r2 (cos 2Î¸ + sin 2Î¸) = (âˆš3)2 + (1)2

r2 = 4

r = 2 (Since r has to be positive )

Putting r = 2 on (1) and (2), we get

cosÎ¸ = âˆš3 / 2 and  sinÎ¸ = 1 / 2

Therefore, Î¸ =  Ï€ / 6 ( Since  cosÎ¸  positive and sinÎ¸  positive, therefore Î¸ is positive as it lies on first quadrant)

Hence, z in polar form:  z =  r cosÎ¸  + i  r sinÎ¸  = 2 (cos (Ï€ / 6) + i sin (Ï€ / 6)).

Question 8. i

Solution:

We have z = i,

Let  r cosÎ¸  = 0 —(1)  and,

r sinÎ¸  = 1 —(2)

On Squaring and adding  (1) and (2) , we obtain

r2 (cos 2Î¸ + sin 2Î¸)  =  (0)2 + (1)

r2 =  1

r = 1 (Since r has to be positive)

Putting r = 1 on (1) and (2), we get

cosÎ¸ = 0 / 1 and sinÎ¸ = 1 / 1

Therefore, Î¸ = Ï€ / 2

Hence, z in polar form: z = r cosÎ¸ + i r sinÎ¸ = cos (Ï€ / 2) + i sin (Ï€ / 2)

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