Class 11 Maths Ncert Solutions Ex 5.2

Q.1: Find out the modulus and argument of the given complex number:

$z = -1 -\;i\sqrt{3}$

 

Sol:

Assuming $r\cos\Theta = -1 \;and\; r\sin\Theta = -\sqrt{3}$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (-\sqrt{3})^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 3 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$

Therefore, Modulus = 2   [ Since, r > 0 ]

Now, $2\cos\Theta = -1 \;and\; 2\sin\Theta = \sqrt{3} \\ \Rightarrow \cos\Theta = \frac{-1}{2} \;and\; \sin\Theta = \frac{\sqrt{3}}{2}$

Since, the values of both $\cos\Theta \;and\; \sin\Theta$are negative and they both are negative in 3^rd quadrant

Therefore, Argument = $-(\pi – \frac{\pi}{3}) = \frac{-2\pi}{3}$

 

 

Q.2: Find out the modulus and argument of the given complex number

$z = -\sqrt{3} + i$

 

Sol:

Assuming $r\cos\Theta = -\sqrt{3} \;and\; r\sin\Theta = 1$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-\sqrt{3})^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 3 + 1 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$

Therefore, Modulus = 2              [ Since, r > 0 ]

Now, $2\cos\Theta = -\sqrt{3} \;\;and\;\; 2\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{-\sqrt{3}}{2} \;and\; \sin\Theta = \frac{1}{2}$

Here, $\Theta$ lies in the 2^nd quadrant.

Therefore, Argument = $(\pi – \frac{\pi}{6}) = \frac{5\pi}{6}$

 

 

Q.3: Convert the following complex number in polar form: 1 – i

Sol:

Assuming $r\cos\Theta = 1 \;and\; r\sin\Theta = -1$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (1)^{2} + (-1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}\\$

Therefore, Modulus = $\sqrt{2}$         [ Since, r > 0 ]

Now, $\sqrt{2}\cos\Theta = 1 \;and\; \sqrt{2}\sin\Theta = -1$

$\Rightarrow \cos\Theta = \frac{1}{\sqrt{2}} \;and\; \sin\Theta = \frac{-1}{\sqrt{2}}\\$

Here, $\Theta$ lies in 4^th quadrant.

Therefore, $\Theta$ = $(- \frac{\pi}{4})$

Therefore, the required polar form is:

$1 – i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(-\frac{\pi}{4}) + i\sqrt{2}\sin(-\frac{\pi}{4})\\$

Therefore, $\\\boldsymbol{1-i=\sqrt{2}\left [ \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right ]}$

 

 

Q.4: Convert the following complex number in polar form;   -1 + i

 

Sol:

Assuming $r\cos\Theta = -1 \;and\; r\sin\Theta = 1$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$

Therefore, Modulus =  $\sqrt{2}$               [ Since, r > 0 ]

Now,$\sqrt{2}\cos\Theta = -1 \;and\; \sqrt{2}\sin\Theta = 1$

$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}\\$

Here, $\Theta$ lies in 2^nd quadrant.

Therefore, $\Theta$ = $(\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$

So, the required polar form is:

$-1 + i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(\frac{3\pi}{4}) + i\sqrt{2}\;\sin(\frac{3\pi}{4})\\$

Therefore, -1 + i = $\\\mathbf{\sqrt{2}\left [ \cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4}) \right ]}$

 

 

Q.5: Convert the following complex number in polar form;  -3

Sol:

Assuming $r\cos\Theta = -3 \;and\; r\sin\Theta = 0$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-3)^{2} + (0)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 9 \\ \Rightarrow r^{2}(1) = 9 \\ \Rightarrow r = \sqrt{9} = 3\\$

Therefore, Modulus = 3    [ Since, r > 0]

Now, $3\cos\Theta = -3 \;and\; 3\sin\Theta = 0 \\ \Rightarrow \cos\Theta = -1 \;and\; \sin\Theta = 0$

Therefore, $\Theta = \pi$

So, the required polar form is:

$-3 = r\cos\Theta + ir\sin\Theta = 3\cos(\pi) + i.\;3\sin(\pi)\\$

Therefore, $\boldsymbol{-3= 3[\cos(\pi) + i\sin(\pi)]}$

 

 

Q.6: Convert the following complex number in polar form;   -1 – i

Sol:

 

Assuming $r\cos\Theta = -1 \;and\; r\sin\Theta = -1$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (-1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$

Therefore, Modulus = $\sqrt{2}$           [ Since, r > 0 ]

Now, $\sqrt{2}\cos\Theta$=$-1 \;\;and\;\; \sqrt{2}\sin\Theta = -1$

$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;\;and\;\; \sin\Theta = \frac{-1}{\sqrt{2}}$

 

 

Q.7: Convert the following complex number in polar form: $z = \sqrt{3} + i$

 

Sol:

$z = \sqrt{3} + i$

Assuming $r\cos\Theta = \sqrt{3} \;and\; r\sin\Theta = 1$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (\sqrt{3})^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 3 + 1 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$

Therefore, Modulus = 2     [Since, r > 0]

Now,

$2\cos\Theta = \sqrt{3} \;and\; 2\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{\sqrt{3}}{2} \;and\; \sin\Theta = \frac{1}{2}$

Here, $\Theta$ lies in 1^st quadrant.

Therefore, $\Theta$ = $(\frac{\pi}{6})$

So, the required polar form is:

$\sqrt{3} + i = r\cos\Theta + ir\sin\Theta = 2\cos(\frac{\pi}{6}) + i2\sin(\frac{\pi}{6})\\$

Therefore, $\\\boldsymbol{\sqrt{3} + i = 2\left [ \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}) \right ]}$

 

 

Q.8: Convert the following complex number in polar form: i

 

Sol:

Assuming $r\cos\Theta = 0 \;and\; r\sin\Theta = 1$

On squaring both sides and then adding them we will get:

$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (0)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 \\ \Rightarrow r^{2}(1) = 1 \\ \Rightarrow r = \sqrt{1} = 1\\$

Therefore, Modulus = 1     [ Since, as r > 0)]

Now, $\cos\Theta = 0 \;and\; 3\sin\Theta = 1$

Therefore, $\Theta = \frac{\pi}{2}$

So, the required polar form is:

$\boldsymbol{i= \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})}$

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