11th NCERT Complex Numbers and Quadratic Equations
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Exercise 5.2 Questions 8

Question (1)

Find the modulus and the argument of the complex number $Z = - 1 - i\sqrt 3$

Solution

$Z = - 1 - i\sqrt 3$ Comparing to standard form we get a = -1, b = - √3
|Z| = r
$= \sqrt {{a^2} + {b^2}}$ $= \sqrt {1 + 3} = \sqrt 4$ $= 2$ $a = r\cos \theta ,\quad b = r\sin \theta$ $- 1 = 2\cos \theta ,\quad - \sqrt 3 = 2\sin \theta$ $\cos \theta = \frac{{ - 1}}{2},\quad \sin \theta = \frac{{ - \sqrt 3 }}{2}$ θ lies in 3rd quadrant. $\theta = \pi + \frac{\pi }{3} = \frac{{4\pi }}{3} = - \frac{{2\pi }}{3}$ So priciple argument is the value of θ
= $\frac{{4\pi }}{3} = - \frac{{2\pi }}{3}$

Question (2)

Find the modulus and the argument of the complex number $- \sqrt 3 + i$

Solution

$- \sqrt 3 + i$ Comparing to standard form we get a = -√3, b = 1
|Z| = r
$= \sqrt {{a^2} + {b^2}}$ $= \sqrt {1 + 3} = \sqrt 4$ $= 2$ $a = r\cos \theta ,\quad b = r\sin \theta$ $- \sqrt 3 = 2\cos \theta ,\quad 1 = 2\sin \theta$ $\cos \theta = \frac{{ - \sqrt 3 }}{2},\quad \sin \theta = \frac{1}{2}$ θ lies in 2nd quadrant. $\theta = \pi - \frac{\pi }{6} = \frac{{5\pi }}{6}$ So priciple argument is the value of θ
= $\frac{{5\pi }}{6}$

Question (3)

Convert the given complex number in polar form: 1 – i

Solution

Z = 1 - i
Comparing to standard form we get, a = 1 , b = - 1
$r = \sqrt {{a^2} + {b^2}}$ $= \sqrt {1 + 1} = \sqrt 2$ $a = r\cos \theta ,\quad b = r\sin \theta$ $1 = \sqrt 2 \cos \theta ,\quad - 1 = \sqrt 2 \sin \theta$ $\cos \theta = \frac{1}{{\sqrt 2 }},\quad \sin \theta = - \frac{1}{{\sqrt 2 }}$ So θ lies in 4th quadrant .
$\theta = - \frac{\pi }{4}$ So polar form of complex number is given by $r\left( {\cos \theta + i\sin \theta } \right)$ $= \sqrt 2 \left( {\cos - \frac{\pi }{4} + i\sin - \frac{\pi }{4}} \right)$

Question (4)

Convert the given complex number in polar form: – 1 + i

Solution

Z = -1 + i
Comparing to standard form we get, a = -1 , b = 1
$r = \sqrt {{a^2} + {b^2}}$ $= \sqrt {1 + 1} = \sqrt 2$ $a = r\cos \theta ,\quad b = r\sin \theta$ $-1 = \sqrt 2 \cos \theta ,\quad 1 = \sqrt 2 \sin \theta$ $\cos \theta = \frac{-1}{{\sqrt 2 }},\quad \sin \theta = \frac{1}{{\sqrt 2 }}$ So θ lies in 2nd quadrant .
$\theta = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}$ So polar form of complex number is given by $r\left( {\cos \theta + i\sin \theta } \right)$ $= \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$

Question (5)

Convert the given complex number in polar form: – 1 – i

Solution

Z = - 1 - i
Comparing to standard form we get, a = -1 , b = - 1
$r = \sqrt {{a^2} + {b^2}}$ $= \sqrt {1 + 1} = \sqrt 2$ $a = r\cos \theta ,\quad b = r\sin \theta$ $-1 = \sqrt 2 \cos \theta ,\quad - 1 = \sqrt 2 \sin \theta$ $\cos \theta = \frac{-1}{{\sqrt 2 }},\quad \sin \theta = - \frac{1}{{\sqrt 2 }}$ So θ lies in 3rd quadrant .
$\theta = \pi + \frac{\pi }{4} = \frac{{5\pi }}{4}$ So polar form of complex number is given by $r\left( {\cos \theta + i\sin \theta } \right)$ $= \sqrt 2 \left( {\cos \frac{{5\pi }}{4} + i\sin \frac{{5\pi }}{4}} \right)$

Question (6)

Convert the given complex number in polar form: –3

Solution

Z = -3
Comparing to standard form we get, a = -3 , b = 0
$r = \sqrt {{a^2} + {b^2}}$ $= \sqrt {9 + 0} = 3$ $a = r\cos \theta ,\quad b = r\sin \theta$ $- 3 = 3\cos \theta ,\quad 0 = 3\sin \theta$ $\cos \theta = - 1,\quad \sin \theta = 0$ So θ lies on x axis.
$\theta = \pi$ So polar form of complex number is given by $r\left( {\cos \theta + i\sin \theta } \right)$ $3\left( {\cos \pi + i\sin \pi } \right)$

Question (7)

Convert the given complex number in polar form: $\sqrt 3 + i$

Solution

$\sqrt 3 + i$ Comparing to standard form we get, a = √3 , b = 1
$r = \sqrt {{a^2} + {b^2}}$ $= \sqrt {3 + 1} = \sqrt 4 = 2$ $a = r\cos \theta ,\quad b = r\sin \theta$ $\sqrt 3 = 2\cos \theta ,\quad 1 = 2\sin \theta$ $\cos \theta = \frac{{\sqrt 3 }}{2},\quad \sin \theta = \frac{1}{2}$ So θ lies in 1st quadrant .
$\theta = \frac{\pi }{6}$ So polar form of complex number is given by $r\left( {\cos \theta + i\sin \theta } \right)$ $2\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right)$

Question (8)

Convert the given complex number in polar form: i

Solution

Z = i Comparing to standard form we get, a = 0 , b = 1
$r = \sqrt {{a^2} + {b^2}}$ $= \sqrt {1 + 0} = 1$ $a = r\cos \theta ,\quad b = r\sin \theta$ $0 = \cos \theta ,\quad 1 = \sin \theta$ $\cos \theta = 0,\quad \sin \theta = 1$ So θ lies on y axis. .
$\theta = \frac{\pi }{2}$ So polar form of complex number is given by $r\left( {\cos \theta + i\sin \theta } \right)$ $= \left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)$