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 \[r\left( {\cos \theta + i\sin \theta } \right)\] \[ = \left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)\]
Exercise 5.1 ⇐
⇒Exercise 5.3