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 4
th 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 2
nd 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 3
rd 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 1
st 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
\[r\left( {\cos \theta + i\sin \theta } \right)\]
\[ = \left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)\]