NCERT Solutions for Exercise 5.2 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

Access Complex Numbers And Quadratic Equations Class 11 Chapter 5-Exercise: 5.2

Question:1 Find the modulus and the arguments of each of the complex numbers.

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

Answer:

Given the problem is
$z=-1-i\sqrt{3}$
Now, let
$r\cos \theta =-1andr\sin \theta =-\sqrt{3}$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(-1{)}^2+(-\sqrt{3}{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=1+3$
$r^2=4$
$r=2$ $(\because r>0)$
Therefore, the modulus is 2
Now,
$2\cos \theta =-1and2\sin \theta =-\sqrt{3}$
$\cos \theta =-\frac{1}{2}and\sin \theta =-\frac{\sqrt{3}}{2}$
Since, both the values of $\cos \theta and\sin \theta$ is negative and we know that they are negative in III quadrant
Therefore,
Argument = $-(\pi -\frac{\pi }{3})=-\frac{2\pi }{3}$
Therefore, the argument is

$-\frac{2\pi }{3}$

Question:2 Find the modulus and the arguments of each of the complex numbers.

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

Answer:

Given the problem is
$z=-\sqrt{3}+i$
Now, let
$r\cos \theta =-\sqrt{3}andr\sin \theta =1$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(-\sqrt{3}{)}^2+(1{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=1+3$
$r^2=4$
$r=2$ $(\because r>0)$
Therefore, the modulus is 2
Now,
$2\cos \theta =-\sqrt{3}and2\sin \theta =1$
$\cos \theta =-\frac{\sqrt{3}}{2}and\sin \theta =\frac{1}{2}$
Since values of $\cos \theta$ is negative and value $\sin \theta$ is positive and we know that this is the case in II quadrant
Therefore,
Argument = $(\pi -\frac{\pi }{6})=\frac{5\pi }{6}$
Therefore, the argument is

$\frac{5\pi }{6}$

Question:3 Convert each of the complex numbers in the polar form:

$1-i$

Answer:

Given problem is
$z=1-i$
Now, let
$r\cos \theta =1andr\sin \theta =-1$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(1{)}^2+(-1{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=1+1$
$r^2=2$
$r=\sqrt{2}$ $(\because r>0)$
Therefore, the modulus is $\sqrt{2}$
Now,
$\sqrt{2}\cos \theta =1and\sqrt{2}\sin \theta =-1$
$\cos \theta =\frac{1}{\sqrt{2}}and\sin \theta =-\frac{1}{\sqrt{2}}$
Since values of $\sin \theta$ is negative and value $\cos \theta$ is positive and we know that this is the case in the IV quadrant
Therefore,
$\theta =-\frac{\pi }{4}(liesinIVquadrant)$
Therefore,
$1-i=r\cos \theta +ir\sin \theta$
$=\sqrt{2}\cos (-\frac{\pi }{4})+i\sqrt{2}\sin (-\frac{\pi }{4})$
$=\sqrt{2}(\cos (-\frac{\pi }{4})+i\sin (-\frac{\pi }{4}))$

Therefore, the required polar form is $\sqrt{2}(\cos (-\frac{\pi }{4})+i\sin (-\frac{\pi }{4}))$

Question:4 Convert each of the complex numbers in the polar form:

$-1+i$

Answer:

Given the problem is
$z=-1+i$
Now, let
$r\cos \theta =-1andr\sin \theta =1$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(1{)}^2+(-1{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=1+1$
$r^2=2$
$r=\sqrt{2}$ $(\because r>0)$
Therefore, the modulus is $\sqrt{2}$
Now,
$\sqrt{2}\cos \theta =-1and\sqrt{2}\sin \theta =1$
$\cos \theta =-\frac{1}{\sqrt{2}}and\sin \theta =\frac{1}{\sqrt{2}}$
Since values of $\cos \theta$ is negative and value $\sin \theta$ is positive and we know that this is the case in II quadrant
Therefore,
$\theta =(\pi -\frac{\pi }{4})=\frac{3\pi }{4}(liesinIIquadrant)$
Therefore,
$-1+i=r\cos \theta +ir\sin \theta$
$=\sqrt{2}\cos \left(\frac{3\pi }{4}\right)+i\sqrt{2}\sin \left(\frac{3\pi }{4}\right)$
$=\sqrt{2}(\cos \left(\frac{3\pi }{4}\right)+i\sin \left(\frac{3\pi }{4}\right))$

Therefore, the required polar form is $\sqrt{2}(\cos \left(\frac{3\pi }{4}\right)+i\sin \left(\frac{3\pi }{4}\right))$

Question:5 Convert each of the complex numbers in the polar form:

$-1-i$

Answer:

Given problem is
$z=-1-i$
Now, let
$r\cos \theta =-1andr\sin \theta =-1$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(-1{)}^2+(-1{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=1+1$
$r^2=2$
$r=\sqrt{2}$ &nbsnbsp; $(\because r>0)$
Therefore, the modulus is $\sqrt{2}$
Now,
$\sqrt{2}\cos \theta =-1and\sqrt{2}\sin \theta =-1$
$\cos \theta =-\frac{1}{\sqrt{2}}and\sin \theta =-\frac{1}{\sqrt{2}}$
Since values of both $\cos \theta$ and $\sin \theta$ is negative and we know that this is the case in III quadrant
Therefore,
$\theta =-(\pi -\frac{\pi }{4})=-\frac{3\pi }{4}(liesinIIIquadrant)$
Therefore,
$-1-i=r\cos \theta +ir\sin \theta$
$=\sqrt{2}\cos (-\frac{3\pi }{4})+i\sqrt{2}\sin (-\frac{3\pi }{4})$
$=\sqrt{2}(\cos (-\frac{3\pi }{4})+i\sin (-\frac{3\pi }{4}))$

Therefore, the required polar form is $\sqrt{2}(\cos (-\frac{3\pi }{4})+i\sin (-\frac{3\pi }{4}))$

Question:6 Convert each of the complex numbers in the polar form:

$-3$

Answer:

Given problem is
$z=-3$
Now, let
$r\cos \theta =-3andr\sin \theta =0$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(-3{)}^2+(0{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=9+0$
$r^2=9$
$r=3$ $(\because r>0)$
Therefore, the modulus is 3
Now,
$3\cos \theta =-3and3\sin \theta =0$
$\cos \theta =-1and\sin \theta =0$
Since values of $\cos \theta$ is negative and $\sin \theta$ is Positive and we know that this is the case in II quadrant
Therefore,
$\theta =\pi (liesinIIquadrant)$
Therefore,
$-3=r\cos \theta +ir\sin \theta$
$=3\cos \left(\pi \right)+i3\sin \left(\pi \right)$
$=3(\cos \pi +i\sin \pi )$

Therefore, the required polar form is $3(\cos \pi +i\sin \pi )$

Question:7 Convert each of the complex numbers in the polar form:

$\sqrt{3}+i$

Answer:

Given problem is
$z=\sqrt{3}+i$
Now, let
$r\cos \theta =\sqrt{3}andr\sin \theta =1$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(\sqrt{3}{)}^2+(1{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=3+1$
$r^2=4$
$r=2$ $(\because r>0)$
Therefore, the modulus is 2
Now,
$2\cos \theta =\sqrt{3}and2\sin \theta =1$
$\cos \theta =\frac{\sqrt{3}}{2}and\sin \theta =\frac{1}{2}$
Since values of Both $\cos \theta$ and $\sin \theta$ is Positive and we know that this is the case in I quadrant
Therefore,
$\theta =\frac{\pi }{6}(liesinIquadrant)$
Therefore,
$\sqrt{3}+i=r\cos \theta +ir\sin \theta$
$=2\cos \left(\frac{\pi }{6}\right)+i2\sin \left(\frac{\pi }{6}\right)$
$=2(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$

Therefore, the required polar form is $2(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$

Question:8 Convert each of the complex numbers in the polar form:

$i$

Answer:

Given problem is
$z=i$
Now, let
$r\cos \theta =0andr\sin \theta =1$
Square and add both the sides
$r^2({\cos }^2\theta +{\sin }^2\theta )=(0{)}^2+(1{)}^2$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r^2=0+1$
$r^2=1$
$r=1$ $(\because r>0)$
Therefore, the modulus is 1
Now,
$1\cos \theta =0and1\sin \theta =1$
$\cos \theta =0and\sin \theta =1$
Since values of Both $\cos \theta$ and $\sin \theta$ is Positive and we know that this is the case in I quadrant
Therefore,
$\theta =\frac{\pi }{2}(liesinIquadrant)$
Therefore,
$i=r\cos \theta +ir\sin \theta$
$=1\cos \left(\frac{\pi }{2}\right)+i1\sin \left(\frac{\pi }{2}\right)$
$=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}$

Therefore, the required polar form is $\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}$