NCERT Solutions Class 11 Maths Chapter 5

Q.1 Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

$\begin{array}{l}1.\text{„}\left(5\text{i}\right)\left(-\frac{3}{5}\text{i}\right)\\ \text{2.„}{\mathrm{i}}^9+{\mathrm{i}}^{19}\\ \text{3.„}{\mathrm{i}}^{-39}\\ \text{4.„}3(7+\mathrm{i}7)+\mathrm{i}(7+\mathrm{i}7)\\ \text{5.„}(1-\mathrm{i})-(-1+\mathrm{i}6)\\ \text{6.„}\left(\frac{1}{5}+\mathrm{i}\frac{2}{5}\right)-\left(4+\mathrm{i}\frac{5}{2}\right)\\ \text{7.„}\left[\left(\frac{1}{3}+\mathrm{i}\frac{7}{3}\right)+\left(4+\mathrm{i}\frac{1}{3}\right)\right]-\left(-\frac{4}{3}+\mathrm{i}\right)\\ \text{8.„}{(1-\mathrm{i})}^4\\ 9.\text{„}{\left(\frac{1}{3}+3\text{i}\right)}^3\\ 10.\text{„}{\left(-2-\frac{1}{3}\text{i}\right)}^3\end{array}$

Ans

1.

$\begin{array}{l}\left(5\mathrm{i}\right)(-\frac{3}{5}\mathrm{i})=5\times -\frac{3}{5}{\mathrm{i}}^2\\ =-3(-1)[âˆ\mu {\mathrm{i}}^2=-1]\\ =3+0\mathrm{i},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}which is in the form of a}+\text{ib.}\end{array}$

2.

$\begin{array}{l}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}i}}^{\text{9}}+{\text{i}}^{\text{19}}={\mathrm{i}}^{4\times 2+1}+{\mathrm{i}}^{4\times 4+3}\\ ={\left({\mathrm{i}}^4\right)}^2\mathrm{i}+{\left({\mathrm{i}}^4\right)}^4{\mathrm{i}}^3\\ ={\left(1\right)}^2\times \mathrm{i}+{\left(1\right)}^4(-\mathrm{i})[âˆ\mu {\mathrm{i}}^4=1{\text{and i}}^{\text{3}}=-\text{\hspace{0.17em}i}]\\ =\mathrm{i}-\mathrm{i}\\ =0+0\mathrm{i},\text{which is in the form of a}+\text{ib.}\end{array}$

3.

${\text{\hspace{0.17em}\hspace{0.17em}i}}^{–\text{39}}=\frac{1}{{\mathrm{i}}^{39}}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{1}{{\mathrm{i}}^{4\times 9+3}}\\ =\frac{1}{{\left({\mathrm{i}}^4\right)}^9.{\mathrm{i}}^3}\\ =\frac{1}{{\left(1\right)}^9.(-\mathrm{i})}\\ =-\frac{1}{\mathrm{i}}\times \frac{\mathrm{i}}{\mathrm{i}}\\ =-\frac{\mathrm{i}}{{\mathrm{i}}^2}\\ =-\frac{\mathrm{i}}{-1}\\ =0+\mathrm{i},\text{which is in the form of a}+\text{ib.}\end{array}$

4.
3(7 + i 7)+ i(7 + i 7) = 21 + 21i + 7i + 7i²
= 21 + 28i + 7(– 1)
= 21 + 28i – 7
= 14 + 28i, which is in the form of a + ib.

5.
(1 – i) – (–1 + i 6) = 1 – i + 1 – i 6
= 2 – 7i, which is in the form of a + ib.

6.

$\begin{array}{l}(\frac{1}{5}+\mathrm{i}\frac{2}{5})-(4+\mathrm{i}\frac{5}{2})=(\frac{1}{5}-4)+\mathrm{i}(\frac{2}{5}-\frac{5}{2})\\ =-\frac{19}{5}+\mathrm{i}\left(\frac{4-25}{10}\right)\\ =-\frac{19}{5}-\mathrm{i}\frac{21}{10},\\ \mathrm{which}\text{is in the form of a}+\text{ib.}\end{array}$

7.

$\begin{array}{l}[(\frac{1}{3}+\mathrm{i}\frac{7}{3})+(4+\mathrm{i}\frac{1}{3})]-(-\frac{4}{3}+\mathrm{i})=(\frac{1}{3}+4+\frac{4}{3})+\mathrm{i}(\frac{7}{3}+\frac{1}{3}-1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left(\frac{1+12+4}{3}\right)+\mathrm{i}\left(\frac{7+1-3}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left(\frac{17}{3}\right)+\mathrm{i}\left(\frac{5}{3}\right),\\ \mathrm{Which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}+\mathrm{ib}\mathrm{.}\end{array}$

8.

$\begin{array}{l}{(1–\mathrm{i})}^4={\left\{{(1–\mathrm{i})}^2\right\}}^2\\ ={(1-2\mathrm{i}+{\mathrm{i}}^2)}^2\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}={(1-2\mathrm{i}-1)}^2\\ ={(-2\mathrm{i})}^2\\ =4{\mathrm{i}}^2\\ =4(-1)\\ =-4+\mathrm{i}0,\\ \mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}+\mathrm{ib}.\end{array}$

9.

$\begin{array}{l}{(\frac{1}{3}+3\mathrm{i})}^3={\left(\frac{1}{3}\right)}^3+3{\left(\frac{1}{3}\right)}^2\left(3\mathrm{i}\right)+3\left(\frac{1}{3}\right){\left(3\mathrm{i}\right)}^2+{\left(3\mathrm{i}\right)}^3\\ [âˆ\mu {(\mathrm{a}+\mathrm{b})}^3={\mathrm{a}}^3+3{\mathrm{a}}^2\mathrm{b}+3{\mathrm{ab}}^2+{\mathrm{b}}^3]\\ =\frac{1}{27}+\mathrm{i}+{\mathrm{i}}^{2}9+{\mathrm{i}}^{3}27\\ =\frac{1}{27}+\mathrm{i}-9-\mathrm{i}27\\ =(\frac{1}{27}-9)+\mathrm{i}(1-27)\\ =\left(\frac{1-243}{27}\right)+\mathrm{i}(-26)\\ =\left(\frac{-242}{27}\right)-\mathrm{i}26\\ \mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}+\mathrm{ib}.\end{array}$

10.

$\begin{array}{l}{(-2-\frac{1}{3}\mathrm{i})}^3={(-2)}^3+3{(-2)}^2(-\frac{1}{3}\mathrm{i})+3(-2){(-\frac{1}{3}\mathrm{i})}^2+{(-\frac{1}{3}\mathrm{i})}^3\\ [âˆ\mu {(\mathrm{a}+\mathrm{b})}^3={\mathrm{a}}^3+3{\mathrm{a}}^2\mathrm{b}+3{\mathrm{ab}}^2+{\mathrm{b}}^3]\\ =-8-4\mathrm{i}-\frac{2}{3}{\mathrm{i}}^2-\frac{1}{27}{\mathrm{i}}^3\\ =-8-4\mathrm{i}-\frac{2}{3}(-1)-\frac{1}{27}(-\mathrm{i})[âˆ\mu {\mathrm{i}}^2=-1{\text{and i}}^{\text{3}}=-\mathrm{i}]\\ =-8+\frac{2}{3}-4\mathrm{i}+\frac{1}{27}\mathrm{i}\\ =-\frac{22}{3}-\left(\frac{108-1}{27}\right)\mathrm{i}\\ =-\frac{22}{3}-\left(\frac{107}{27}\right)\mathrm{i},\\ \mathrm{which}\mathrm{is}\mathrm{}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}+\mathrm{ib}.\end{array}$

Q.2 Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

11. 4 – 3i
12.

$\sqrt{5}+3\mathrm{i}$

13. – i

Ans

$\begin{array}{l}\mathrm{Multiplicative}\text{inverse of}(\text{4}-\text{3i})=\frac{1}{(\text{4}-\text{3i})}\times \frac{(\text{4}+\text{3i})}{(\text{4}+\text{3i})}\\ =\frac{(\text{4}+\text{3i})}{({\text{4}}^{\text{2}}-{\text{9i}}^{\text{2}})}\end{array}$ $\begin{array}{l}[âˆ\mu (\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})={\mathrm{a}}^2-{\mathrm{b}}^2]\\ =\frac{(\text{4}+\text{3i})}{(\text{16}+\text{9})}\\ =\frac{\text{4}}{25}+\frac{\text{3}}{25}\text{i}\\ \mathrm{Thus},\text{the multiplicative inverse of}(4-3\mathrm{i})\text{is}(\frac{\text{4}}{25}+\frac{\text{3}}{25}\text{i}).\end{array}$

$\begin{array}{l}\mathrm{Multiplicative}\text{inverse of}(\sqrt{5}+3\mathrm{i})=\frac{1}{(\sqrt{5}+3\mathrm{i})}\times \frac{(\sqrt{5}-3\mathrm{i})}{(\sqrt{5}-3\mathrm{i})}\\ =\frac{(\sqrt{5}-3\mathrm{i})}{\{{\left(\sqrt{5}\right)}^2-{\left(3\mathrm{i}\right)}^2\}}\\ =\frac{(\sqrt{5}-3\mathrm{i})}{\{5-9{\mathrm{i}}^2\}}\\ =\frac{(\sqrt{5}-3\mathrm{i})}{(5+9)}\\ =\frac{\sqrt{5}}{14}-\mathrm{i}\frac{3}{14}\\ \mathrm{Thus},\text{the multiplicative inverse of}(\sqrt{5}+3\mathrm{i})\text{is}(\frac{\sqrt{5}}{14}-\mathrm{i}\frac{3}{14}).\end{array}$

$\begin{array}{l}\mathrm{Multiplicative}\text{inverse of}(–\text{i})=\frac{1}{(–\text{i})}\times \frac{\mathrm{i}}{\mathrm{i}}\\ =\frac{\mathrm{i}}{-{\mathrm{i}}^2}\\ =\frac{\mathrm{i}}{-(-1)}\\ =\frac{\mathrm{i}}{1}\\ =\mathrm{i},\\ \mathrm{Thus},\text{the multiplicative inverse of}-\text{i is i.}\end{array}$

Q.3

$\begin{array}{l}\text{Express the following expression in the form of a+ib:}\\ \frac{\left(\text{3+i}\sqrt{\text{5}}\right)\left(\text{3-i}\sqrt{\text{5}}\right)}{\left(\sqrt{\text{3}}\text{+}\sqrt{\text{2}}\text{i}\right)\text{–}\left(\sqrt{\text{3}}\text{–}\sqrt{\text{2}}\text{i}\right)}\end{array}$

Ans

$\begin{array}{l}\frac{(3+\mathrm{i}\sqrt{5})(3-\mathrm{i}\sqrt{5})}{(\sqrt{3}+\sqrt{2}\mathrm{i})-(\sqrt{3}-\sqrt{2}\mathrm{i})}=\frac{3^2-{\left(\mathrm{i}\sqrt{5}\right)}^2}{\sqrt{3}+\sqrt{2}\mathrm{i}-\sqrt{3}+\sqrt{2}\mathrm{i}}\left[\begin{array}{l}âˆ\mu (\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})\\ ={\mathrm{a}}^2-{\mathrm{b}}^2\end{array}\right]\\ =\frac{(9-{\mathrm{i}}^{2}5)}{2\sqrt{2}\mathrm{i}}\\ =\frac{\{9-(-1)5\}}{2\sqrt{2}\mathrm{i}}\end{array}$ $\begin{array}{l}=\frac{14}{2\sqrt{2}\mathrm{i}}\\ =\frac{7}{\sqrt{2}\mathrm{i}}\times \frac{\sqrt{2}\mathrm{i}}{\sqrt{2}\mathrm{i}}\\ =\frac{7\sqrt{2}\mathrm{i}}{{\mathrm{i}}^2\left(2\right)}\\ =\frac{7\sqrt{2}\mathrm{i}}{(-1)2}[âˆ\mu {\mathrm{i}}^2=-1]\\ =0-\frac{7\sqrt{2}\mathrm{i}}{2},\mathrm{which}\text{in the form of a + ib.}\end{array}$

Q.4 Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

$\begin{array}{l}1.\text{„}\mathrm{z}=-1-\mathrm{i}\sqrt{3}\\ 2.\text{„z}=-\sqrt{3}+\text{i}\end{array}$

Ans

$\begin{array}{l}\mathrm{z}=-1-\mathrm{i}\sqrt{3}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=-\text{1 and rsin}\mathrm{\theta }=-\sqrt{3}\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={(-1)}^2+{(-\sqrt{3})}^2\\ {\mathrm{r}}^2=1+3\\ \Rightarrow \mathrm{r}=2\\ \mathrm{So},\text{‹ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}modulus}=\text{2}\end{array}$

$\begin{array}{l}\therefore \mathrm{cos\theta }=\frac{-1}{2}\text{and sin}\mathrm{\theta }=-\frac{\sqrt{3}}{2}\\ \Rightarrow \mathrm{cos\theta }=-\cos \frac{\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=-\sin \frac{\mathrm{\pi }}{3}\\ \mathrm{Since},\text{sin}\mathrm{\theta }\text{and cos}\mathrm{\theta }\text{are negative in III quadrant.}\\ \mathrm{So},\mathrm{\theta }=-(\mathrm{\pi }-\frac{\mathrm{\pi }}{3})=-\frac{2\mathrm{\pi }}{3}\\ \mathrm{Thus},\text{modulus and argument of}(-1-\mathrm{i}\sqrt{3})\text{is 2 and}-\frac{2\mathrm{\pi }}{3}\\ \text{respectively.}\end{array}$

$\begin{array}{l}\mathrm{z}=-\sqrt{3}+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=-\sqrt{3}\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squarring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={(-\sqrt{3})}^2+{\left(1\right)}^2\\ {\mathrm{r}}^2=3+1=4\\ \Rightarrow \mathrm{r}=2\\ \mathrm{So},\text{‹ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}modulus}=\text{2}\\ \therefore \mathrm{cos\theta }=\frac{-\sqrt{3}}{2}\text{and sin}\mathrm{\theta }=\frac{1}{2}\\ \Rightarrow \mathrm{cos\theta }=-\cos \frac{\mathrm{\pi }}{6}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\sin \frac{\mathrm{\pi }}{6}\\ \mathrm{Since},\text{sin}\mathrm{\theta }\text{is positive and cos}\mathrm{\theta }\text{is negative in II quadrant.}\end{array}$

$\begin{array}{l}\mathrm{So},\mathrm{\theta }=(\mathrm{\pi }-\frac{\mathrm{\pi }}{6})=\frac{5\mathrm{\pi }}{6}\\ \mathrm{Thus},\text{modulus and argument of}(-\sqrt{3}+\mathrm{i})\text{is 2 and}\frac{5\mathrm{\pi }}{6}\\ \text{respectively.}\end{array}$

Q.5 Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

3. 1 – i
4. – 1 + i
5. – 1 – i
6. – 3
7.

$\sqrt{3}+\mathrm{i}$

8. i

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{z}=1-\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=1\text{and rsin}\mathrm{\theta }=-1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={\left(1\right)}^2+{(-1)}^2\\ {\mathrm{r}}^2=1+1\\ \Rightarrow \mathrm{r}=\sqrt{2}\\ \therefore \mathrm{cos\theta }=\frac{1}{\sqrt{2}}\text{and sin}\mathrm{\theta }=\frac{-1}{\sqrt{2}}\\ \Rightarrow \mathrm{cos\theta }=\cos \frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=-\sin \frac{\mathrm{\pi }}{4}\\ \mathrm{Since},\text{cos}\mathrm{\theta }\text{is positive and sin}\mathrm{\theta }\text{is negative in IV quadrant.}\\ \mathrm{So},\mathrm{\theta }=-\frac{\mathrm{\pi }}{4}[âˆ\mu -\mathrm{\pi }<\mathrm{\theta }\le \mathrm{\pi }]\\ \mathrm{Thus},\text{polar form of z}=\sqrt{2}\{\cos (-\frac{\mathrm{\pi }}{4})+\mathrm{isin}(-\frac{\mathrm{\pi }}{4})\}.\end{array}$

$\begin{array}{l}\mathrm{Let}\mathrm{z}=-1+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=-1\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={(-1)}^2+{\left(1\right)}^2\end{array}$ $\begin{array}{l}{\mathrm{r}}^2=1+1\\ \Rightarrow \mathrm{r}=\sqrt{2}\\ \therefore \mathrm{cos\theta }=\frac{-1}{\sqrt{2}}\text{and sin}\mathrm{\theta }=\frac{1}{\sqrt{2}}\\ \Rightarrow \mathrm{cos\theta }=-\cos \frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\sin \frac{\mathrm{\pi }}{4}\\ \mathrm{Since},\text{sin\hspace{0.17em}}\mathrm{\theta }\text{is positive and cos\hspace{0.17em}}\mathrm{\theta }\text{is negative in II quadrant.}\\ \mathrm{So},\mathrm{\theta }=\mathrm{\pi }-\frac{\mathrm{\pi }}{4}=\frac{3\mathrm{\pi }}{4}\\ \mathrm{Thus},\text{polar form of z}=\sqrt{2}\{\cos \left(\frac{3\mathrm{\pi }}{4}\right)+\mathrm{isin}\left(\frac{3\mathrm{\pi }}{4}\right)\}.\end{array}$

$\begin{array}{l}\mathrm{Let}\mathrm{z}=-1-\text{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=-1\text{and rsin}\mathrm{\theta }=-1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={(-1)}^2+{(-1)}^2\\ {\mathrm{r}}^2=1+1\\ \Rightarrow \mathrm{r}=\sqrt{2}\\ \therefore \mathrm{cos\theta }=\frac{-1}{\sqrt{2}}\text{and sin}\mathrm{\theta }=\frac{-1}{\sqrt{2}}\\ \Rightarrow \mathrm{cos\theta }=-\cos \frac{\mathrm{\pi }}{4}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=-\sin \frac{\mathrm{\pi }}{4}\\ \mathrm{Since},\text{sin\hspace{0.17em}}\mathrm{\theta }\text{and cos\hspace{0.17em}}\mathrm{\theta }\text{is negative in III quadrant.}\end{array}$ $\begin{array}{l}\mathrm{So},\mathrm{\theta }=-(\mathrm{\pi }-\frac{\mathrm{\pi }}{4})=-\frac{3\mathrm{\pi }}{4}[âˆ\mu -\mathrm{\pi }<\mathrm{\theta }\le \mathrm{\pi }]\\ \mathrm{Thus},\text{polar form of z}=\sqrt{2}\{\cos (-\frac{3\mathrm{\pi }}{4})+\mathrm{isin}(-\frac{3\mathrm{\pi }}{4})\}.\end{array}$

$\begin{array}{l}\mathrm{Let}\mathrm{z}=-3+0\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=-3\text{and rsin}\mathrm{\theta }=0\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={(-3)}^2+{\left(0\right)}^2\\ {\mathrm{r}}^2=9\\ \Rightarrow \mathrm{r}=3\\ \therefore \mathrm{cos\theta }=\frac{-3}{3}=-\text{1 and sin}\mathrm{\theta }=0\\ \Rightarrow \mathrm{cos\theta }=-\cos 0\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\sin 0\\ \mathrm{Since},\text{sin\hspace{0.17em}}\mathrm{\theta }\text{is positive and cos\hspace{0.17em}}\mathrm{\theta }\text{is negative in II quadrant.}\\ \mathrm{So},\mathrm{\theta }=(\mathrm{\pi }-0)=\mathrm{\pi }\\ \mathrm{Thus},\text{polar form of z}=3\{\mathrm{cos\pi }+\mathrm{isin\pi }\}.\end{array}$

$\begin{array}{l}\mathrm{Let}\mathrm{z}=\sqrt{3}+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=\sqrt{3}\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squaring}\text{and adding, we get}\end{array}$ $\begin{array}{l}{\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={\left(\sqrt{3}\right)}^2+{\left(1\right)}^2\\ {\mathrm{r}}^2=4\\ \Rightarrow \mathrm{r}=2\\ \therefore \mathrm{cos\theta }=\frac{\sqrt{3}}{2}\text{and sin}\mathrm{\theta }=\frac{1}{2}\\ \Rightarrow \mathrm{cos\theta }=\cos \frac{\mathrm{\pi }}{6}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\sin \frac{\mathrm{\pi }}{6}\\ \mathrm{So},\mathrm{\theta }=\frac{\mathrm{\pi }}{6}\\ \mathrm{Thus},\text{polar form of z}=2\{\cos \frac{\mathrm{\pi }}{6}+\mathrm{isin}\frac{\mathrm{\pi }}{6}\}.\end{array}$

$\begin{array}{l}\mathrm{Let}\mathrm{z}=0+\mathrm{i}\\ \mathrm{Comparing}\text{with z}=\mathrm{r}(\mathrm{cos\theta }+\mathrm{isin\theta }),\text{we get}\\ \text{rcos}\mathrm{\theta }=0\text{and rsin}\mathrm{\theta }=1\\ \mathrm{Squaring}\text{and adding, we get}\\ {\text{r}}^{\text{2}}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })={\left(0\right)}^2+{\left(1\right)}^2\\ {\mathrm{r}}^2=1\\ \Rightarrow \mathrm{r}=1\\ \therefore \mathrm{cos\theta }=0\text{and sin}\mathrm{\theta }=1\\ \Rightarrow \mathrm{cos\theta }=\cos \frac{\mathrm{\pi }}{2}\text{\hspace{0.17em}\hspace{0.17em}and sin}\mathrm{\theta }=\sin \frac{\mathrm{\pi }}{2}\\ \mathrm{So},\mathrm{\theta }=\frac{\mathrm{\pi }}{2}\end{array}$ $\mathrm{Thus},\text{polar form of z}=1\{\cos \frac{\mathrm{\pi }}{2}+\mathrm{isin}\frac{\mathrm{\pi }}{2}\}.$

Q.6 Solve the following equation:
x² + 3 = 0

Ans

\begin{array}{l}{\text{x}}^{\text{2}}+\text{3}=0\\ comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=1,b=0\text{and c}=3\\ By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-0\pm \sqrt{0^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}\\ =\pm \frac{\sqrt{-12}}{2}\\ =\pm \frac{2i\sqrt{3}}{2}\left[âˆ\mu \sqrt{-1}=i\right]\\ x=\pm i\sqrt{3}\end{array}

Q.7 Solve the following equation:
2x² + x + 1 = 0

Ans

\begin{array}{l}{\text{2x}}^{\text{2}}+\text{x}+\text{1}=0\\ comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=2,b=1\text{and c}=1\end{array} \begin{array}{l}By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-1\pm \sqrt{1^2-4\left(2\right)\left(1\right)}}{2\left(2\right)}\\ =\frac{-1\pm \sqrt{1-8}}{4}\\ =\frac{-1\pm \sqrt{-7}}{4}\\ x=\frac{-1\pm i\sqrt{7}}{4}\left[âˆ\mu \sqrt{-1}=i\right]\end{array}

Q.8 Solve the following equation:
x² + 3x + 9 = 0

Ans

\begin{array}{l}{\text{x}}^{\text{2}}+\text{3x}+\text{9}=0\\ comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=1,b=3\text{and c}=9\\ By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-3\pm \sqrt{3^2-4\left(1\right)\left(9\right)}}{2\left(1\right)}\\ =\frac{-3\pm \sqrt{9-36}}{2}=\frac{-3\pm \sqrt{-27}}{4}\end{array} x=\frac{-3\pm 3\sqrt{3}i}{2}\left[âˆ\mu \sqrt{-1}=i\right]

Q.9 Solve the following equation:
– x² + x – 2 = 0

Ans

\begin{array}{l}-x^2+x-2=0\\ comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=-1,b=1\text{and c}=-2\\ By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-1\pm \sqrt{1^2-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}\\ =\frac{-1\pm \sqrt{1-8}}{-2}=\frac{-1\pm \sqrt{-7}}{-2}\\ x=\frac{-1\pm i\sqrt{7}}{-2}\left[âˆ\mu \sqrt{-1}=i\right]\end{array}

Q.10 Solve the following equation:
x² + 3x + 5 = 0

Ans

\begin{array}{l}{\text{x}}^{\text{2}}+\text{3x}+\text{5}=0\\ comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=1,b=3\text{and c}=5\\ By\text{using quadratic formula,}\end{array} \begin{array}{l}\text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-3\pm \sqrt{3^2-4\left(1\right)\left(5\right)}}{2\left(1\right)}\\ =\frac{-3\pm \sqrt{9-20}}{2}=\frac{-1\pm \sqrt{-11}}{2}\\ x=\frac{-3\pm i\sqrt{11}}{2}\left[âˆ\mu \sqrt{-1}=i\right]\end{array}

Q.11 Solve the following equation:
x² – x + 2 = 0

Ans

\begin{array}{l}{\text{x}}^{\text{2}}–\text{x}+\text{2}=0\\ Comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=1,b=-1\text{and c}=2\\ By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}\\ =\frac{1\pm \sqrt{1-8}}{2}=\frac{1\pm \sqrt{-7}}{2}\\ x=\frac{1\pm \sqrt{7}i}{2}\left[âˆ\mu \sqrt{-1}=i\right]\end{array}

Q.12 Solve the following equation:

\sqrt{2}{x}^2+x+\sqrt{2}=0

Ans

\begin{array}{l}\sqrt{2}{x}^2+x+\sqrt{2}=0\\ Comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=\sqrt{2},b=1\text{and c}=\sqrt{2}\\ By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-1\pm \sqrt{1^2-4\left(\sqrt{2}\right)\left(\sqrt{2}\right)}}{2\left(\sqrt{2}\right)}\\ =\frac{-1\pm \sqrt{1-8}}{2\sqrt{2}}=\frac{-1\pm \sqrt{-7}}{2\sqrt{2}}\\ x=\frac{-1\pm i\sqrt{7}}{2\sqrt{2}}\left[âˆ\mu \sqrt{-1}=i\right]\end{array}

Q.13 Solve the following equation:

\sqrt{3}{x}^2-\sqrt{2}x+3\sqrt{3}=0

Ans

\begin{array}{l}\sqrt{3}{x}^2-\sqrt{2}x+3\sqrt{3}=0\\ Comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=\sqrt{3},b=-\sqrt{2}\text{and c}=3\sqrt{3}\\ By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array} \begin{array}{l}=\frac{-\left(-\sqrt{2}\right)\pm \sqrt{{\left(-\sqrt{2}\right)}^2-4\left(\sqrt{3}\right)\left(3\sqrt{3}\right)}}{2\left(\sqrt{3}\right)}\\ =\frac{\sqrt{2}\pm \sqrt{2-36}}{2\sqrt{3}}=\frac{\sqrt{2}\pm \sqrt{-34}}{2\sqrt{3}}\\ x=\frac{\sqrt{2}\pm \sqrt{34}i}{2\sqrt{3}}\left[âˆ\mu \sqrt{-1}=i\right]\end{array}

Q.14 Solve the following equation:

x^2+x+\frac{1}{\sqrt{2}}=0

Ans

\begin{array}{l}{x}^2+x+\frac{1}{\sqrt{2}}=0\\ Comparing{\text{with ax}}^{\text{2}}+bx+c=0,\text{we get}\\ \text{a}=1,b=1\text{and c}=\frac{1}{\sqrt{2}}\\ By\text{using quadratic formula,}\\ \text{x}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^2-4\left(1\right)\left(\frac{1}{\sqrt{2}}\right)}}{2\left(1\right)}\\ =\frac{-1\pm \sqrt{1-2\sqrt{2}}}{2}=\frac{-1\pm \sqrt{1-2\sqrt{2}}}{2}\end{array} x=\frac{-1\pm i\sqrt{2\sqrt{2}-1}}{2}\left[âˆ\mu \sqrt{-1}=i\right]

Q.15 Solve the following equation:

${\mathrm{x}}^2+\frac{\mathrm{x}}{\sqrt{2}}+1=0$

Ans

$\begin{array}{l}{\mathrm{x}}^2+\frac{\mathrm{x}}{\sqrt{2}}+1=0\\ \mathrm{Comparing}{\text{with ax}}^{\text{2}}+\mathrm{bx}+\mathrm{c}=0,\text{we get}\\ \text{a}=1,\mathrm{b}=\frac{1}{\sqrt{2}}\text{and c}=1\\ \mathrm{By}\text{using quadratic formula,}\\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\\ =\frac{-\left(\frac{1}{\sqrt{2}}\right)\pm \sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}\\ =\frac{-\frac{1}{\sqrt{2}}\pm \sqrt{\frac{1}{2}-4}}{2}\\ =\frac{-\frac{1}{\sqrt{2}}\pm \sqrt{-\frac{7}{2}}}{2}\\ \mathrm{x}=\frac{-1\pm \mathrm{i}\sqrt{7}}{2\sqrt{2}}[âˆ\mu \sqrt{-1}=\mathrm{i}]\end{array}$

Q.16 Find the square root of the following: – 15 – 8i

Ans

$\begin{array}{l}\mathrm{Let}\text{‹ \hspace{0.17em}\hspace{0.17em}x}+\mathrm{iy}=\sqrt{-\text{15}-\text{8i}}\\ \mathrm{Squarring}\text{both sides, we get}\\ {(\mathrm{x}+\mathrm{iy})}^2=-\text{15}-\text{8i}\\ {\mathrm{x}}^2+2\mathrm{ixy}+{\left(\mathrm{iy}\right)}^2=-\text{15}-\text{8i}\\ {\mathrm{x}}^2+2\mathrm{ixy}-{\mathrm{y}}^2=-\text{15}-\text{8i}\\ {\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{ixy}=-\text{15}-\text{8i}\\ \mathrm{Comparing}\text{real and imaginary parts from both sides, we get}\\ {\mathrm{x}}^2-{\mathrm{y}}^2=-\text{15 …}\left(\mathrm{i}\right)\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}xy}=-4\\ {({\mathrm{x}}^2+{\mathrm{y}}^2)}^2={({\mathrm{x}}^2-{\mathrm{y}}^2)}^2+4{\left(\mathrm{xy}\right)}^2\\ ={(-15)}^2+4{(-4)}^2\\ =225+64\\ =289\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=17\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} …}\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right)\\ 2{\mathrm{x}}^2=2\Rightarrow \mathrm{x}=\pm 1\\ \mathrm{and}\mathrm{putting}\text{x}=1\text{in xy}=-4,\text{we get}\end{array}$ $\begin{array}{l}\Rightarrow \left(1\right)\text{y}=-4\\ \Rightarrow \mathrm{y}=-4\\ \mathrm{putting}\text{x}=-1\text{in xy}=-4,\text{we get}\\ \Rightarrow (-1)\text{y}=-4\\ \Rightarrow \mathrm{y}=4\\ \therefore \mathrm{x}+\mathrm{iy}=1-4\mathrm{i}\text{or}-1+4\text{i}\\ \Rightarrow \sqrt{-\text{15}-\text{8i}}=1-4\mathrm{i}\text{or}-1+4\text{i}\end{array}$

Q.17 Find the square root of the following: – 8 – 6i.

Ans

$\begin{array}{l}\mathrm{Let}\text{‹ \hspace{0.17em}\hspace{0.17em}x}+\mathrm{iy}=\sqrt{-\text{\hspace{0.17em}8}-\text{6i}}\\ \mathrm{Squaring}\text{both sides, we get}\\ {(\mathrm{x}+\mathrm{iy})}^2=-\text{\hspace{0.17em}8}-\text{6i}\\ {\mathrm{x}}^2+2\mathrm{ixy}+{\left(\mathrm{iy}\right)}^2=-\text{\hspace{0.17em}8}-\text{6i}\\ {\mathrm{x}}^2+2\mathrm{ixy}-{\mathrm{y}}^2=-\text{\hspace{0.17em}8}-\text{6i}\\ {\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{ixy}=-\text{\hspace{0.17em}8}-\text{6i}\\ \mathrm{Comparing}\text{real and imaginary parts from both sides, we get}\\ {\mathrm{x}}^2-{\mathrm{y}}^2=-\text{8}...\left(\mathrm{i}\right)\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}xy}=-3\\ {({\mathrm{x}}^2+{\mathrm{y}}^2)}^2={({\mathrm{x}}^2-{\mathrm{y}}^2)}^2+4{\left(\mathrm{xy}\right)}^2\\ ={(-8)}^2+4{(-3)}^2\\ =64+36\\ =100\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=10...\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}2{\mathrm{x}}^2=2\Rightarrow \mathrm{x}=\pm 1\\ \mathrm{and}\mathrm{putting}\text{x}=1\text{in xy}=-6,\text{we get}\\ \Rightarrow \left(1\right)\text{y}=-3\\ \Rightarrow \mathrm{y}=-3\\ \mathrm{putting}\text{x}=-1\text{in xy}=-3,\text{we get}\\ \Rightarrow (-1)\text{y}=-3\\ \Rightarrow \mathrm{y}=3\\ \therefore \mathrm{x}+\mathrm{iy}=1-3\mathrm{i}\text{or}-1+3\text{i}\\ \Rightarrow \sqrt{-\text{8}-\text{6i}}=1-3\mathrm{i}\text{or}-1+3\text{i}\end{array}$

Q.18 Find the square root of the following: 1 – i

Ans

$\begin{array}{l}\mathrm{Let}\text{‹ \hspace{0.17em}\hspace{0.17em}x}+\mathrm{iy}=\sqrt{1-\text{i}}\\ \mathrm{Squarring}\text{both sides, we get}\\ {(\mathrm{x}+\mathrm{iy})}^2=\text{\hspace{0.17em}1}-\text{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}+{\left(\mathrm{iy}\right)}^2=\text{\hspace{0.17em}1}-\text{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}-{\mathrm{y}}^2=\text{1}-\text{i}\\ {\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{ixy}=\text{1}-\text{i}\\ \mathrm{Comparing}\text{real and imaginary parts from both sides, we get}\\ {\mathrm{x}}^2-{\mathrm{y}}^2=\text{1 …}\left(\mathrm{i}\right)\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}xy}=-\frac{1}{2}\\ {({\mathrm{x}}^2+{\mathrm{y}}^2)}^2={({\mathrm{x}}^2-{\mathrm{y}}^2)}^2+4{\left(\mathrm{xy}\right)}^2\\ ={\left(1\right)}^2+4{(-\frac{1}{2})}^2\end{array}$ $\begin{array}{l}=1+1\\ =2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=\sqrt{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}…}\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right)\\ 2{\mathrm{x}}^2=1+\sqrt{2}\Rightarrow \mathrm{x}=\pm \sqrt{\frac{1+\sqrt{2}}{2}}\\ \mathrm{and}\mathrm{putting}\text{x}=\sqrt{\frac{1+\sqrt{2}}{2}}\text{in xy}=-\frac{1}{2},\text{we get}\\ \Rightarrow \left(\sqrt{\frac{1+\sqrt{2}}{2}}\right)\text{y}=-\frac{1}{2}\\ \Rightarrow \mathrm{y}=-\frac{1}{2}\times \sqrt{\frac{2}{\sqrt{2}+1}}\\ \Rightarrow \mathrm{y}=-\sqrt{\frac{\sqrt{2}+1}{2(\sqrt{2}+1)}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}\\ \Rightarrow \mathrm{y}=-\sqrt{\frac{\sqrt{2}+1}{2}}\\ \mathrm{Putting}\text{x}=-\sqrt{\frac{1+\sqrt{2}}{2}}\text{in xy}=-\frac{1}{2},\text{we get}\\ \Rightarrow (-\sqrt{\frac{1+\sqrt{2}}{2}})\text{y}=-\frac{1}{2}\\ \Rightarrow \mathrm{y}=\frac{1}{2}\times \sqrt{\frac{2}{\sqrt{2}+1}}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{y}=\sqrt{\frac{\sqrt{2}+1}{2(\sqrt{2}+1)}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}\\ \Rightarrow \mathrm{y}=\sqrt{\frac{\sqrt{2}+1}{2}}\\ \therefore \mathrm{x}+\mathrm{iy}=\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{\sqrt{2}+1}{2}}\mathrm{i}\text{or}-\sqrt{\frac{1+\sqrt{2}}{2}}+\sqrt{\frac{\sqrt{2}+1}{2}}\mathrm{i}\\ \Rightarrow \sqrt{-\text{8}-\text{6i}}=\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{\sqrt{2}+1}{2}}\mathrm{i}\text{or}-\sqrt{\frac{1+\sqrt{2}}{2}}+\sqrt{\frac{\sqrt{2}+1}{2}}\\ \Rightarrow \sqrt{-\text{8}-\text{6i}}=\pm \sqrt{\frac{1+\sqrt{2}}{2}}∓\sqrt{\frac{\sqrt{2}+1}{2}}\mathrm{i}\end{array}$

Q.19 Find the square root of the following: –i

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}+\mathrm{iy}=\sqrt{-\mathrm{i}}\\ \mathrm{Squaring}\text{both sides, we get}\\ {\left(\mathrm{x}+\mathrm{iy}\right)}^2=0-\mathrm{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}+{\left(\mathrm{iy}\right)}^2=0-\mathrm{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}-{\mathrm{y}}^2=0-\mathrm{i}\\ {\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{ixy}=0-\mathrm{i}\\ \mathrm{Comparing}\text{real and imaginary parts from both sides, we get}\\ {\mathrm{x}}^2-{\mathrm{y}}^2=0...\left(\mathrm{i}\right)\\ \mathrm{and}\mathrm{xy}=-\frac{1}{2}\\ {\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2={\left({\mathrm{x}}^2-{\mathrm{y}}^2\right)}^2+4{\left(\mathrm{xy}\right)}^2\end{array}$ $\begin{array}{l}\Rightarrow \text{‹‹‹‹‹‹‹‹‹‹„„„„„„„‹y}=\sqrt{\frac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}\\ \Rightarrow \text{‹‹‹‹‹‹‹‹‹‹‹„„„„„„„y}=\sqrt{\frac{\sqrt{2}+1}{2}}\\ \therefore \text{„„„x}+\mathrm{iy}=\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{\sqrt{2}+1}{2}}\text{i„}\mathrm{or}-\sqrt{\frac{1+\sqrt{2}}{2}}+\sqrt{\frac{\sqrt{2}+1}{2}}\text{i}\\ \Rightarrow \sqrt{1-\text{i}}=\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{\sqrt{2}+1}{2}}\text{i„}\mathrm{or}-\sqrt{\frac{1+\sqrt{2}}{2}}+\sqrt{\frac{\sqrt{2}+1}{2}}\text{i}\\ \Rightarrow \sqrt{1-\text{i}}=\pm \sqrt{\frac{1+\sqrt{2}}{2}}∓\sqrt{\frac{\sqrt{2}+1}{2}}\text{i}\end{array}$

Q.20 Find the square root of the following: i

Ans

$\begin{array}{l}\mathrm{Let}\text{‹ \hspace{0.17em}\hspace{0.17em}x}+\mathrm{iy}=\sqrt{\mathrm{i}}\\ \mathrm{Squaring}\text{both sides, we get}\\ {(\mathrm{x}+\mathrm{iy})}^2=\text{\hspace{0.17em}0}+\text{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}+{\left(\mathrm{iy}\right)}^2=\text{\hspace{0.17em}0}+\text{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}-{\mathrm{y}}^2=\text{0}+\text{i}\\ {\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{ixy}=\text{0}+\text{i}\\ \mathrm{Comparing}\text{real and imaginary parts from both sides, we get}\\ {\mathrm{x}}^2-{\mathrm{y}}^2=\text{0 …}\left(\mathrm{i}\right)\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}xy}=\frac{1}{2}\\ {({\mathrm{x}}^2+{\mathrm{y}}^2)}^2={\left(0\right)}^2+4{\left(\frac{1}{2}\right)}^2\\ =1\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} …}\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right)\\ 2{\mathrm{x}}^2=1\Rightarrow \mathrm{x}=\pm \frac{1}{\sqrt{2}}\\ \mathrm{and}\mathrm{putting}\text{x}=\frac{1}{\sqrt{2}}\text{in xy}=\frac{1}{2},\text{we get}\\ \Rightarrow \frac{1}{\sqrt{2}}\text{y}=\frac{1}{2}\\ \Rightarrow \mathrm{y}=\frac{1}{\sqrt{2}}\end{array}$ $\begin{array}{l}\mathrm{Putting}\text{x}=-\frac{1}{\sqrt{2}}\text{in xy}=\frac{1}{2},\text{we get}\\ \Rightarrow (-\frac{1}{\sqrt{2}})\text{y}=\frac{1}{2}\\ \Rightarrow \mathrm{y}=-\frac{1}{\sqrt{2}}\\ \therefore \mathrm{x}+\mathrm{iy}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\mathrm{i}\text{or}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\mathrm{i}\\ \Rightarrow \sqrt{\text{i}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\mathrm{i}\text{or}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\mathrm{i}\\ \Rightarrow \sqrt{\text{i}}=\pm \frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}\mathrm{i}\end{array}$

Q.21 Find the square root of the following: 1 + i

Ans

$\begin{array}{l}\mathrm{Let}\text{‹ \hspace{0.17em}\hspace{0.17em}x}+\mathrm{iy}=\sqrt{1+\text{i}}\\ \mathrm{Squaring}\text{both sides, we get}\\ {(\mathrm{x}+\mathrm{iy})}^2=\text{\hspace{0.17em}1}+\text{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}+{\left(\mathrm{iy}\right)}^2=\text{\hspace{0.17em}1}+\text{i}\\ {\mathrm{x}}^2+2\mathrm{ixy}-{\mathrm{y}}^2=\text{1}+\text{i}\\ {\mathrm{x}}^2-{\mathrm{y}}^2+2\mathrm{ixy}=\text{1}+\text{i}\\ \mathrm{Comparing}\text{real and imaginary parts from both sides, we get}\\ {\mathrm{x}}^2-{\mathrm{y}}^2=\text{1 …}\left(\mathrm{i}\right)\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}xy}=\frac{1}{2}\\ {({\mathrm{x}}^2+{\mathrm{y}}^2)}^2={({\mathrm{x}}^2-{\mathrm{y}}^2)}^2+4{\left(\mathrm{xy}\right)}^2\end{array}$ $\begin{array}{l}{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2={\left(1\right)}^2+4{\left(\frac{1}{2}\right)}^2\\ =1+1\\ {\mathrm{x}}^2+{\mathrm{y}}^2=\sqrt{2}...\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ {\text{2x}}^{\text{2}}=1+\sqrt{2}\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{2}}=\frac{1+\sqrt{2}}{2}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=\pm \sqrt{\frac{\sqrt{2}+1}{2}}\\ \mathrm{Putting}\text{value of x in xy}=\frac{1}{2},\text{we get}\\ \text{y}=\frac{1}{2\mathrm{x}}\\ =\frac{1}{2}\times \pm \sqrt{\frac{2}{\sqrt{2}+1}}\\ =\pm \sqrt{\frac{1}{2\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}\\ =\pm \sqrt{\frac{\sqrt{2}-1}{2(2-1)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=\pm \sqrt{\frac{\sqrt{2}-1}{2}}\\ \therefore \mathrm{x}+\mathrm{iy}=\pm \sqrt{\frac{\sqrt{2}+1}{2}}\pm \sqrt{\frac{\sqrt{2}-1}{2}}\mathrm{i}\\ \Rightarrow \sqrt{1+\mathrm{i}}=\pm \sqrt{\frac{\sqrt{2}+1}{2}}\pm \sqrt{\frac{\sqrt{2}-1}{2}}\mathrm{i}\end{array}$