# Class 11 Maths Ncert Solutions Ex 5.2

## Class 11 Maths Ncert Solutions Chapter 5 Ex 5.2

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

z=1i3$z = -1 -\;i\sqrt{3}$

Sol:

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

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

(rcosΘ)2+(rsinΘ)2=(1)2+(3)2r2(cos2Θ+sin2Θ)=1+3r2(1)=4r=4=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (-\sqrt{3})^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 3 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$

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

Now, 2cosΘ=1and2sinΘ=3cosΘ=12andsinΘ=32$2\cos\Theta = -1 \;and\; 2\sin\Theta = \sqrt{3} \\ \Rightarrow \cos\Theta = \frac{-1}{2} \;and\; \sin\Theta = \frac{\sqrt{3}}{2}$

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

Therefore, Argument = (ππ3)=2π3$-(\pi – \frac{\pi}{3}) = \frac{-2\pi}{3}$

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

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

Sol:

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

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

(rcosΘ)2+(rsinΘ)2=(3)2+(1)2r2(cos2Θ+sin2Θ)=3+1r2(1)=4r=4=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-\sqrt{3})^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 3 + 1 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$

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

Now, 2cosΘ=3and2sinΘ=1cosΘ=32andsinΘ=12$2\cos\Theta = -\sqrt{3} \;\;and\;\; 2\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{-\sqrt{3}}{2} \;and\; \sin\Theta = \frac{1}{2}$

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

Therefore, Argument = (ππ6)=5π6$(\pi – \frac{\pi}{6}) = \frac{5\pi}{6}$

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

Sol:

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

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

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (1)^{2} + (-1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}\\$

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

Now, 2cosΘ=1and2sinΘ=1$\sqrt{2}\cos\Theta = 1 \;and\; \sqrt{2}\sin\Theta = -1$

cosΘ=12andsinΘ=12$\Rightarrow \cos\Theta = \frac{1}{\sqrt{2}} \;and\; \sin\Theta = \frac{-1}{\sqrt{2}}\\$

Here, Θ$\Theta$ lies in 4th quadrant.

Therefore, Θ$\Theta$ = (π4)$(- \frac{\pi}{4})$

Therefore, the required polar form is:

1i=rcosΘ+irsinΘ=2cos(π4)+i2sin(π4)$1 – i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(-\frac{\pi}{4}) + i\sqrt{2}\sin(-\frac{\pi}{4})\\$

Therefore, 1i=2[cos(π4)+isin(π4)]$\\\boldsymbol{1-i=\sqrt{2}\left [ \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right ]}$

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

Sol:

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

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

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$

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

Now,2cosΘ=1and2sinΘ=1$\sqrt{2}\cos\Theta = -1 \;and\; \sqrt{2}\sin\Theta = 1$

cosΘ=12andsinΘ=12$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}\\$

Here, Θ$\Theta$ lies in 2nd quadrant.

Therefore, Θ$\Theta$ = (ππ4)=3π4$(\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$

So, the required polar form is:

1+i=rcosΘ+irsinΘ=2cos(3π4)+i2sin(3π4)$-1 + i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(\frac{3\pi}{4}) + i\sqrt{2}\;\sin(\frac{3\pi}{4})\\$

Therefore, -1 + i = 2[cos(3π4)+isin(3π4)]$\\\mathbf{\sqrt{2}\left [ \cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4}) \right ]}$

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

Sol:

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

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

(rcosΘ)2+(rsinΘ)2=(3)2+(0)2r2(cos2Θ+sin2Θ)=9r2(1)=9r=9=3$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-3)^{2} + (0)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 9 \\ \Rightarrow r^{2}(1) = 9 \\ \Rightarrow r = \sqrt{9} = 3\\$

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

Now, 3cosΘ=3and3sinΘ=0cosΘ=1andsinΘ=0$3\cos\Theta = -3 \;and\; 3\sin\Theta = 0 \\ \Rightarrow \cos\Theta = -1 \;and\; \sin\Theta = 0$

Therefore, Θ=π$\Theta = \pi$

So, the required polar form is:

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

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

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

Sol:

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

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

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (-1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$

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

Now, 2cosΘ$\sqrt{2}\cos\Theta$=1and2sinΘ=1$-1 \;\;and\;\; \sqrt{2}\sin\Theta = -1$

cosΘ=12andsinΘ=12$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;\;and\;\; \sin\Theta = \frac{-1}{\sqrt{2}}$

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

Sol:

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

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

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

(rcosΘ)2+(rsinΘ)2=(3)2+(1)2r2(cos2Θ+sin2Θ)=3+1r2(1)=4r=4=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (\sqrt{3})^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 3 + 1 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$

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

Now,

2cosΘ=3and2sinΘ=1cosΘ=32andsinΘ=12$2\cos\Theta = \sqrt{3} \;and\; 2\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{\sqrt{3}}{2} \;and\; \sin\Theta = \frac{1}{2}$

Here, Θ$\Theta$ lies in 1st quadrant.

Therefore, Θ$\Theta$ = (π6)$(\frac{\pi}{6})$

So, the required polar form is:

3+i=rcosΘ+irsinΘ=2cos(π6)+i2sin(π6)$\sqrt{3} + i = r\cos\Theta + ir\sin\Theta = 2\cos(\frac{\pi}{6}) + i2\sin(\frac{\pi}{6})\\$

Therefore, 3+i=2[cos(π6)+isin(π6)]$\\\boldsymbol{\sqrt{3} + i = 2\left [ \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}) \right ]}$

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

Sol:

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

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

(rcosΘ)2+(rsinΘ)2=(0)2+(1)2r2(cos2Θ+sin2Θ)=1r2(1)=1r=1=1$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (0)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 \\ \Rightarrow r^{2}(1) = 1 \\ \Rightarrow r = \sqrt{1} = 1\\$

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

Now, cosΘ=0and3sinΘ=1$\cos\Theta = 0 \;and\; 3\sin\Theta = 1$

Therefore, Θ=π2$\Theta = \frac{\pi}{2}$

So, the required polar form is:

i=cos(π2)+isin(π2)$\boldsymbol{i= \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})}$