## NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2

Here, you will find NCERT Solutions for Class 11 MathsÂ Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2 that will be useful in solving problems in which your have difficulty. These NCERT Solutions for Class 11 Maths will guide you to achieve better marks in the examination. The solutions are accurate and to the point which can save your precious time and also understanding the basic concepts.

1.Â Find the modulus and the argument of the complex number z = -1-iâˆš3

Let z = -1 â€“ iâˆš3

Here x = -1, y = -âˆš3

âˆ´Â Î± = tan

= tan

sience x < 0, and y< 0

âˆ´Â argÂ |i| = -(Ï€Â -Â Î±Â ) = -(Ï€Â -Â Ï€/3) = -2Ï€/ 3

|z| =Â âˆš(-1)

2.Â Find the modulus and the argument of the complex number z = - âˆš3 + i

We know that the polar form of

z = r(cosÎ¸Â + isinÎ¸)

âˆ´Â Let -Â âˆš3 + i = r(cosÎ¸Â + isinÎ¸)

=> rcosÎ¸Â = -âˆš3 and r sinÎ¸Â = 1

By squaring and adding, we get

r

r

by dividing, rsinÎ¸/rcosÎ¸Â = -1/Â âˆš3

i.e.Â Î¸Â lies in second quadrant

=> tanÎ¸Â = 1/Â âˆš3

=>Â Î¸Â = 180 â€“ 30 = 150 =>Â Î¸= 5Ï€/6

âˆ´Â |z|= 2 and arg z = 5Ï€/6

3.Â Convert (1 - i) in the polar form.

Answer

We have 1 â€“ i = r(cosÎ¸Â + i sinÎ¸)

=> rcosÎ¸Â = 1, rsinÎ¸Â = -1

By squaring and adding, we get

r

=> r

âˆ´Â r =Â âˆš2, by dividing r sinÂ Î¸/r cosÂ Î¸Â = -1/1 = -1

tanÂ Î¸Â = -1 i.e.,Â Î¸Â lies in fourth quadrant.

=>Â Î¸ = -45 =>Â Î¸ = -(Ï€/4)

âˆ´Â Polar form of 1 â€“ i

=Â

4.Â Convert (-1 + i) in the polar form.

We have -1 + i = r(cosÂ Î¸Â + i sinÂ Î¸)

=> r cosÂ Î¸Â = -1 and r sinÂ Î¸Â = 1

By squaring and adding, we get

r

âˆ´Â r

by dividing, r sinÂ Î¸/r cosÂ Î¸ = 1/-1 = -1 => tanÂ Î¸Â = - 1

âˆ´Â Î¸Â lies in second quadrant;

Î¸Â = 180Â°Â â€“ 45Â°Â = 135Â°Â i.e.Â Î¸ = 3Ï€/4

âˆ´Â Polar form of -1 + i

=Â âˆš2(cos(3Ï€/4) + i sin(3Ï€/4))

5.Â Convert (-1 â€“ i) in the polar form.

We have -1 â€“ I = r(cosÂ Î¸Â + i sinÂ Î¸)

=> r cosÂ Î¸Â = -1 and r sinÂ Î¸Â = -1

By squaring and adding, we get

r

=> r

âˆ´Â r =Â âˆš2

by dividing r sinÂ Î¸/r cosÂ Î¸Â = -1/-1 = 1 => tanÂ Î¸Â = 1

âˆ´Â Î¸Â lies in IIIrd quadrant.

Î¸Â = -180 + 45 = -135 orÂ Î¸ = - 3Ï€/4

âˆ´Â Polar form of -1 â€“ i

=Â âˆš2(cos (- 3Ï€/4) + i sin(- 3Ï€/4))

6. Convert each of the complex numbers given in Exercise 6 to 8 in the polar form.

-3

z = - 3 = r(cosÂ Î¸Â + i sinÂ Î¸)

âˆ´Â r cosÂ Î¸Â = - 3, r sinÂ Î¸Â = 0

squaring and adding r

tanÂ Î¸Â = 0Â âˆ´Â Î¸Â =Â Ï€ âˆµÂ cosÂ Ï€Â = 0

âˆ´Â -3 = 3 (cosÏ€Â + i sinÏ€)

7.Â âˆš3 + i

r =Â âˆš3 + I = r(cosÂ Î¸Â + i sinÂ Î¸)

âˆ´Â r cosÂ Î¸Â =Â âˆš3, r sinÂ Î¸Â = 1

squaring and adding r

also tanÂ Î¸Â = 1/âˆš3 , sinÂ Î¸Â and cosÂ Î¸Â both are positive

âˆ´Â Î¸Â lies in the Ist quadrant

âˆ´Â Î¸Â = 30Â°Â =Â Ï€/6

âˆ´Â Polar form of z is 2(cos (Ï€/6) + i sin(Ï€/6))

8.Â i

z = i = r(cosÂ Î¸Â + i sinÂ Î¸)

âˆ´Â r cosÂ Î¸Â = 0 , r sinÂ Î¸Â = 1

squaring and adding r

now sinÂ Î¸Â = 1, cosÂ Î¸Â = 0 atÂ Î¸Â =Â Ï€/2

âˆ´Â Polar form of z is cos(Ï€/2) + i sin(Ï€/2)

1.Â Find the modulus and the argument of the complex number z = -1-iâˆš3

**Answer**Let z = -1 â€“ iâˆš3

Here x = -1, y = -âˆš3

âˆ´Â Î± = tan

^{-1}Â |y/x| = tan^{-1}Â |-3/-1|= tan

^{-1}Â tanÂ Ï€/3 =Â Ï€/3sience x < 0, and y< 0

âˆ´Â argÂ |i| = -(Ï€Â -Â Î±Â ) = -(Ï€Â -Â Ï€/3) = -2Ï€/ 3

|z| =Â âˆš(-1)

^{2}Â + (-3)^{2}Â =Â âˆš1+3= 22.Â Find the modulus and the argument of the complex number z = - âˆš3 + i

**Answer**We know that the polar form of

z = r(cosÎ¸Â + isinÎ¸)

âˆ´Â Let -Â âˆš3 + i = r(cosÎ¸Â + isinÎ¸)

=> rcosÎ¸Â = -âˆš3 and r sinÎ¸Â = 1

By squaring and adding, we get

r

^{2}(cos^{2}Î¸Â + sin^{2}Î¸) = (Â âˆš3)^{2}Â + 1r

^{2}.1 = 4 âˆ´Â r = 2by dividing, rsinÎ¸/rcosÎ¸Â = -1/Â âˆš3

i.e.Â Î¸Â lies in second quadrant

=> tanÎ¸Â = 1/Â âˆš3

=>Â Î¸Â = 180 â€“ 30 = 150 =>Â Î¸= 5Ï€/6

âˆ´Â |z|= 2 and arg z = 5Ï€/6

3.Â Convert (1 - i) in the polar form.

Answer

We have 1 â€“ i = r(cosÎ¸Â + i sinÎ¸)

=> rcosÎ¸Â = 1, rsinÎ¸Â = -1

By squaring and adding, we get

r

^{2}(cos^{2}Â Î¸Â + sin^{2}Î¸) = 1^{2}Â + (-1)^{2}=> r

^{2}. 1 = 1 + 1 => r^{2}= 2âˆ´Â r =Â âˆš2, by dividing r sinÂ Î¸/r cosÂ Î¸Â = -1/1 = -1

tanÂ Î¸Â = -1 i.e.,Â Î¸Â lies in fourth quadrant.

=>Â Î¸ = -45 =>Â Î¸ = -(Ï€/4)

âˆ´Â Polar form of 1 â€“ i

=Â

4.Â Convert (-1 + i) in the polar form.

**Answer**We have -1 + i = r(cosÂ Î¸Â + i sinÂ Î¸)

=> r cosÂ Î¸Â = -1 and r sinÂ Î¸Â = 1

By squaring and adding, we get

r

^{2}(cos^{2}Â Î¸Â + sin^{2}Â Î¸) = (-1)^{2}Â + 1^{2}Â => r^{2}Â . 1 = 1 + 1âˆ´Â r

^{2}Â = 2 âˆ´ r =Â âˆš2by dividing, r sinÂ Î¸/r cosÂ Î¸ = 1/-1 = -1 => tanÂ Î¸Â = - 1

âˆ´Â Î¸Â lies in second quadrant;

Î¸Â = 180Â°Â â€“ 45Â°Â = 135Â°Â i.e.Â Î¸ = 3Ï€/4

âˆ´Â Polar form of -1 + i

=Â âˆš2(cos(3Ï€/4) + i sin(3Ï€/4))

5.Â Convert (-1 â€“ i) in the polar form.

**Answer**We have -1 â€“ I = r(cosÂ Î¸Â + i sinÂ Î¸)

=> r cosÂ Î¸Â = -1 and r sinÂ Î¸Â = -1

By squaring and adding, we get

r

^{2}(cos^{2}Â Î¸Â + sin^{2}Â Î¸Â ) = (-1)^{2}Â + (- 1)^{2}=> r

^{2}Â . 1 = 1 + 1 => r^{2}Â = 2âˆ´Â r =Â âˆš2

by dividing r sinÂ Î¸/r cosÂ Î¸Â = -1/-1 = 1 => tanÂ Î¸Â = 1

âˆ´Â Î¸Â lies in IIIrd quadrant.

Î¸Â = -180 + 45 = -135 orÂ Î¸ = - 3Ï€/4

âˆ´Â Polar form of -1 â€“ i

=Â âˆš2(cos (- 3Ï€/4) + i sin(- 3Ï€/4))

6. Convert each of the complex numbers given in Exercise 6 to 8 in the polar form.

-3

**Answer**z = - 3 = r(cosÂ Î¸Â + i sinÂ Î¸)

âˆ´Â r cosÂ Î¸Â = - 3, r sinÂ Î¸Â = 0

squaring and adding r

^{2}Â = (-3)^{2}Â âˆ´ r = 3tanÂ Î¸Â = 0Â âˆ´Â Î¸Â =Â Ï€ âˆµÂ cosÂ Ï€Â = 0

âˆ´Â -3 = 3 (cosÏ€Â + i sinÏ€)

7.Â âˆš3 + i

**Answer**r =Â âˆš3 + I = r(cosÂ Î¸Â + i sinÂ Î¸)

âˆ´Â r cosÂ Î¸Â =Â âˆš3, r sinÂ Î¸Â = 1

squaring and adding r

^{2}Â = 3 + 1 = 4, r = 2also tanÂ Î¸Â = 1/âˆš3 , sinÂ Î¸Â and cosÂ Î¸Â both are positive

âˆ´Â Î¸Â lies in the Ist quadrant

âˆ´Â Î¸Â = 30Â°Â =Â Ï€/6

âˆ´Â Polar form of z is 2(cos (Ï€/6) + i sin(Ï€/6))

8.Â i

**Answer**z = i = r(cosÂ Î¸Â + i sinÂ Î¸)

âˆ´Â r cosÂ Î¸Â = 0 , r sinÂ Î¸Â = 1

squaring and adding r

^{2}Â = 1,Â âˆ´ r = 1now sinÂ Î¸Â = 1, cosÂ Î¸Â = 0 atÂ Î¸Â =Â Ï€/2

âˆ´Â Polar form of z is cos(Ï€/2) + i sin(Ï€/2)