Exercise 5.2 (Complex number)

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.(Exercise 5.2 complex no. ncert math solution class 11)

Question 1: $Z=-1-i \sqrt{3}$

Solution : Let $Z=-1-i \sqrt{3}=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=-1--(i)$

$r \sin \theta=-\sqrt{3}--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1+3$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=4$

$\Rightarrow r^{2}=4 \quad \Rightarrow r=2 \quad[\because r=|Z|>0]$

Therefore, modulus $=2$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{-\sqrt{3}}{-1}$

$\Rightarrow \tan \theta=\sqrt{3}$

From the equations (1), (2) and (3), it is clear that $\sin \theta$ and $\cos \theta$ are negative but $\tan \theta$ is positive. So, $\theta$ lies in III quadrant. Therefore,

Argument $\theta=-\left(\pi-\frac{\pi}{3}\right)=-\frac{2 \pi}{3}$

Question 2: $Z=-\sqrt{3}+i$

Solution : Let $Z=-\sqrt{3}+i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=-\sqrt{3}--(i)$

$r \sin \theta=1--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=3+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=4$

$\Rightarrow r^{2}=4$

$\Rightarrow r=2 \quad[\because r=|Z|>0]$

Therefore, modulus $=2$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{1}{-\sqrt{3}}$

$\Rightarrow \tan \theta=-\frac{1}{\sqrt{3}}--(iii)$

From the equations (1), (2) and (3), it is clear that $\cos \theta$ and $\tan \theta$ are negative but $\sin \theta$ is positive. So, $\theta$ lies in II quadrant. Therefore,

$\text { Argument } \theta=\left(\pi-\frac{\pi}{6}\right)=\frac{5 \pi}{6}$

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

Question 3: $1-i$

Solution : Let $Z=1-i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=1--(i)$

$r \sin \theta=-1--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2$

$\Rightarrow r^{2}=2$

$\Rightarrow r=\sqrt{2} \quad[\because r=|Z|>0]$

Therefore, modulus $=\sqrt{2}$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{-1}{1}$

$\Rightarrow \tan \theta=-1--(iii)$

From the equations (1), (2) and (3), it is clear that $\sin \theta$ and $\tan \theta$ are negative but $\cos \theta$ is positive. So, $\theta$ lies in IV quadrant. Therefore,

$\text { Argument } \theta=-\frac{\pi}{4}$

Therefore, the polar form of $Z=1-i$ is

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

Question 4: $-1+i$

Solution : Let $Z=-1+i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=-1--(i)$

$r \sin \theta=1--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2$

$\Rightarrow r^{2}=2 \quad \Rightarrow r=\sqrt{2} \quad[\because r=|Z|>0]$

Therefore, modulus $=\sqrt{2}$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{1}{-1}$

$\Rightarrow \tan \theta=-1--(iii)$

From the equations (1), (2) and (3), it is clear that $\cos \theta$ and $\tan \theta$ are negative but $\sin \theta$ is positive. So, $\theta$ lies in II quadrant. Therefore,

$\text { Argument } \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$

Therefore, the polar form of $Z=-1+i$ is

$\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$

Question 5: $-1-i$

Solution : Let $Z=-1-i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=-1--(i)$

$r \sin \theta=-1--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2$

$\Rightarrow r^{2}=2$

$\Rightarrow r=\sqrt{2} \quad[\because r=|Z|>0]$

Therefore, modulus $=\sqrt{2}$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{1}{1}$

$\Rightarrow \tan \theta=1$

From the equations (1), (2) and (3), it is clear that $\sin \theta$ and $\cos \theta$ are negative but tan $\theta$ is positive. So, $\theta$ lies in III quadrant. Therefore,

$\text { Argument } \theta=-\left(\pi-\frac{\pi}{4}\right)=-\frac{3 \pi}{4}$

Therefore, the polar form of $Z=-1-i$ is given by

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

Question 6: $-3$

Solution : Let $Z=-3+0 i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=-3--(i)$

$r \sin \theta=0--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=9+0$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=9$

$\Rightarrow r^{2}=9$

$\Rightarrow r=3 \quad[\because r=|Z|>0]$

Therefore, modulus $=3$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{0}{-3}$

$\Rightarrow \tan \theta=0--(iii)$

From the equations (1), (2) and (3), it is clear that $\sin \theta$ and $\tan \theta$ are 0 but $\cos \theta$ is negative. Therefore,

$\text { Argument } \theta=\pi$

Therefore, the polar form of $Z=-3$ is

$3[\cos \pi+i \sin \pi]$

Question 7: $\sqrt{3}+i$

Solution : Let $Z=\sqrt{3}+i=r(\cos \theta+i \sin \theta)$

Comparing the real and imaginary parts, we have

$r \cos \theta=\sqrt{3}--(i)$

$r \sin \theta=1--(ii)$

Squaring and adding equation (1) and (2), we have

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=3+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=4$

$\Rightarrow r^{2}=4$

$\Rightarrow r=2 \quad[\because r=|Z|>0]$

Therefore, modulus $=2$

Now, dividing equation (2) by (1), we have

$\frac{r \sin \theta}{r \cos \theta}=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}---(iii)$

From the equations (1), (2) and (3), it is clear that $\sin \theta, \cos \theta$ and $\tan \theta$ all are positive. So, $\theta$ lies in I quadrant. Therefore,