Complex Numbers and Quadratic Equations

Class 11 NCERT

NCERT

1   Express the given complex number in the form $a+ib:(5i)(-\dfrac{3}{5}i)$

Solution :

$ ( 5 i ) ( - \dfrac {3} {5} i ) \\ = - 5 * \dfrac {3} {5} * i * i \\ = - 3 i^2 \\ = - 3 (-1) \ \ \ \ [ i^2 = - 1 ] \\ = 3 \\ = 3 + i(0)$

2   Express the given complex number in the form $a+ib:i^9+i^{19}$

Solution :

$i^9+i^{19}\\ =i^{4*2+1}+i^{4*4+3}\\ =(i^4)^2.i+(i^4)^4.i^3\\ =1*i+1*(-i) \ \ \ \ \ \ \ [i^4=1,i^3=-i]\\ =i+(-i)\\ =0$

3   Express the given complex number in the form $a+ib:i^{-39}$

Solution :

$i^{-39}=i^{-4*9-3}\\ =(i^4)^{-9}.i^{-3}\\ =(1)^{-9}.i^{-3} \ \ \ [i^4=1]\\ =\dfrac{1}{i^3}=\dfrac{1}{-i} \ \ \ \ \ \ \ \ [i^3=-i]\\ =\dfrac{-1}{i}*\dfrac{i}{i}\\ =\dfrac{-i}{i^2}\\ =\dfrac{-i}{-1}=i \ \ \ \ \ \ [i^2=-1]$

4   Express the given complex number in the form a $A+ib:3(7+i7)+i(7+i7)$

Solution :

$ 3 ( 7 + i 7 ) + i ( 7 + i 7 ) \\ = 21 + 21 i + 7 i + 7 i^2 \\ = 21 + 28 i + 7 * ( -1 ) \ \ \ \ \ \ [ \therefore i^2 = - 1 ] \\ = 14 + 28 i $

5   Express the given complex number in the form $a+ib:(1-i)-(-1+i6).$

Solution :

$ ( 1 - i ) - ( - 1 + i 6 ) \\ = 1 - i + 1 - 6 i \\ = 2 - 7 i $

6   Express the given complex number in the form $a + ib :(\dfrac{1}{5}+i\dfrac{2}{5})-(4+i\dfrac{5}{2})$

Solution :

$(\dfrac{1}{5}+i\dfrac{2}{5})-(4+i\dfrac{5}{2})\\ =\dfrac{1}{5}+\dfrac{2}{5}i-4-\dfrac{5}{2}i\\ =(\dfrac{1}{5-4})+i(\dfrac{2}{5}-\dfrac{5}{2})\\ =\dfrac{-19}{5}+i(\dfrac{-21}{10})\\ =\dfrac{-19}{5}-\dfrac{21}{10}i$

7   Express the given complex number in the form $a + ib : [(\dfrac{1}{3}+i\dfrac{7}{3})+(4+i\dfrac{1}{3})]-(-\dfrac{4}{3}+i)$$\\$

Solution :

$[(\dfrac{1}{3}+i\dfrac{7}{3})+(4+i\dfrac{1}{3})]-(-\dfrac{4}{3}+i)\\ =\dfrac{1}{3}+\dfrac{7}{3}i+4+\dfrac{1}{3}i+\dfrac{4}{3}-i\\ =(\dfrac{1}{3}+4+\dfrac{4}{3})+i(\dfrac{7}{3}+\dfrac{1}{3}-1)\\ =\dfrac{17}{3}+i\dfrac{5}{3}$

8   Express the given complex number in the form $a + ib : ( 1 - i )^4$

Solution :

$(1-i)^4=[(1-i)^2]^2\\ =[1^2+i^2-2i]^2\\ =[1-1-2i]^2\\ =(2i)^2\\ =(-2i)*(-2i)\\ =4i^2=-4 \quad [i^2=-1]$$\\$

9   Express the given complex number in the form $a + ib : (\dfrac{1}{3}+3i)^3$

Solution :

$(\dfrac{1}{3}+3i)^3=(\dfrac{1}{3})^3+(3i)^3+3(\dfrac{1}{3})(3i)(\dfrac{1}{3}+3i)\\ =\dfrac{1}{27}+27i^3+3i(\dfrac{1}{3}+3i)\\ =\dfrac{1}{27}+27(-i)+i+9i^2 \quad [i^3=-i]\\ =\dfrac{1}{27} -27i+i-9 \quad [i^2=-1]\\ =(\dfrac{1}{27}-9)+i(-27+1)\\ =\dfrac{-242}{27}-26i$

10   Express the given complex number in the form $a + ib :(-2-\dfrac{1}{3}i)^3$

Solution :

$(-2-\dfrac{1}{3}i)^3=(-1)^3(2+\dfrac{1}{3}i)^3\\ =-[2^3+(\dfrac{i}{3})^3+3(2)(\dfrac{i}{3})(2+\dfrac{i}{3})]\\ =-[8+\dfrac{i^3}{27}+2i(2+\dfrac{i}{3})]\\ =-[8-\dfrac{i}{27}+4i+(2i^2){3}]\quad [i^3=-i]\\ =-[8-\dfrac{i}{27}+4i-\dfrac{2}{3}] \quad [i^2=-1]\\ =-[\dfrac{22}{3}+\dfrac{107i}{27}]\\ =-\dfrac{22}{3}-\dfrac{107}{27}i$

11   Find the multiplicative inverse of the complex number $4- 3i .$

Solution :

Let $z =4 - 3 i$ $\\$ Then,$\\$ $\bar{z} = 4 + 3 i$ and $|\bar{z}| = 4 ^2 +(- 3 )^2= 16 + 9 = 25$ $\\$ Therefore, the multiplicative inverse of $4 - 3i$ is given by $\\$ $ z^{-1}=\dfrac{\bar{z}}{|\bar{z}|^2}=\dfrac{4+3i}{25}=\dfrac{4}{25}+\dfrac{3}{25}i$

12   Find the multiplicative inverse of the complex number $\sqrt{5} + 3i$

Solution :

Let $z=\sqrt{5}+3i$ $\\$ Then,$ \bar{z}=\sqrt{5}-3i$ and $|z|^2=(\sqrt{5})^2+3^2=5+9=14$ $\\$ Therefore, the multiplicative inverse of $\sqrt{5}+3i$ $\\$ $ z^{-1}=\dfrac{\bar{z}}{|z|^2}=\dfrac{\sqrt{5}-3i}{14}=\dfrac{\sqrt{5}}{14}-\dfrac{3i}{14}$

13   Find the multiplicative inverse of the complex number $- i$

Solution :

Let $z =- i$ $\\$ Then, $\bar{z} = i$ and $|z^2| = 1 ^2 =1$ $\\$ Therefore, the multiplicative inverse of $- i$ is given by$\\$ $z^{-1}=\dfrac{\bar{z}}{|z|^2}=\dfrac{i}{1}=i$

14   Express the following expression in the form of $a + ib .$$\\$ $\dfrac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2i})-(\sqrt{3}-i\sqrt{2})}$

Solution :

$\dfrac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2i})-(\sqrt{3}-i\sqrt{2})}$$\\$ $=\dfrac{(3)^2-(i\sqrt{5})^2}{\sqrt{3}+\sqrt{2i}-\sqrt{3}+\sqrt{2i}} \quad [(a+b)(a-b)=a^2-b^2]\\ =\dfrac{9-5i^2}{2\sqrt{2i}}\\ =\dfrac{9-5(-1)}{2\sqrt{2i}} \quad [i^2=-1]\\ =\dfrac{9+5}{2\sqrt{2i}}*\dfrac{i}{i}\\ =\dfrac{14i}{2\sqrt{2i^2}}\\ =\dfrac{14i}{2\sqrt{2(-1)}}\\ =\dfrac{-7i}{\sqrt{2}}*\dfrac{\sqrt{2}}{\sqrt{2}}\\ =\dfrac{-7\sqrt{2i}}{2}$

15   Find the modulus and the argument of the complex number $z =- 1 - i \sqrt{3}$

Solution :

$z =- 1 - i \sqrt{3}$$\\$ Let $ r \cos \theta =-1$ and $ r \sin \theta =-\sqrt{3}$ $\\$ On squaring and adding, we obtain $\\$ $(r \cos \theta)^2+(r \sin \theta )^2=(-1)^2+(-\sqrt{3})^2\\ \implies r^2 (\cos^2 \theta+\sin^2 \theta )=1+3\\ \implies r^2 4 \quad [\cos^2 \theta +\sin ^2 \theta =1]\\ \implies r=\sqrt{4}=2 \quad [\text{Conventionally, } r > 0]\\ \therefore \text{Modulus}=2\\ \therefore 2 \cos \theta =-1 \text{and} 2 \sin \theta =-\sqrt{3}\\ \implies \cos \theta =\dfrac{-1}{2} \text{and} \sin \theta =\dfrac{-\sqrt{3}}{2}$ $\\$ Since both the values of $\sin \theta $ and $\cos \theta $ negative and $\sin \theta $ and $\cos \theta $ are negative in III quadrant,$\\$ Argument =$ -(\pi-\dfrac{\pi}{3})=\dfrac{-2 \pi}{3}$ $\\$ Thus, the modulus and argument of the complex number $-1-\sqrt{3i}$ are $ 2 $ and $-\dfrac{2\pi}{3}$ respectively.

16   Find the modulus and the argument of the complex number $z=-\sqrt{ 3} + i$

Solution :

$z=-\sqrt{ 3} + i$ $\\$ Let $ r \cos \theta =-\sqrt{3} \text{and} r \sin \theta =1$ $\\$ On squaring and adding, we obtain $\\$ $ r^2 \cos^2 \theta + r^2 \sin^2 \theta =(-\sqrt{3})^2 + 1^2\\ \implies r^2= 3+1=4 \quad [\cos^2 \theta +\sin^2 \theta =1]\\ \implies r=\sqrt{4} =2 \quad [\text{Conventionally,} r > 0] \\ \therefore \text{Modulus}=2\\ \therefore 2 \cos \theta =-\sqrt{3} \text{and} 2 \sin \theta =1\\ \implies \cos \theta =\dfrac{-\sqrt{3}}{2} \text{and} \sin \theta =\dfrac{1}{2} \\ \therefore \theta =\pi -\dfrac{\pi}{6}=\dfrac{5 \pi}{6 } \quad \quad [\text{As} \theta \text{lies in the II quadrant}]$ $\\$ Thus, the modulus an argument of the complex number$-\sqrt{3}+i $ are $ 2$ and $ \dfrac{5\pi}{6}$ respectively.

17   Convert the given complex number in polar form: $1 - i$

Solution :

$ 1-i $ $\\$ Let $ r \cos \theta =1 $ and $ r \sin \theta =-1 $ $\\$ On squaring and adding,we obtain $\\$ $ r^2 \cos^2 \theta +r^2 \sin^2 \theta =1^2+(-1)^2\\ \implies r^2(\cos^2 \theta + \sin^2 \theta) =1+1\\ \implies r^2=2\\ \implies r=\sqrt{2} \qquad [\text{Conventionally,} r > 0]\\ \therefore \sqrt{2} \cos \theta =1 $ and $ \sqrt{2} \sin \theta =-1\\ \implies \cos \theta =\dfrac{1}{\sqrt{2}} $ and $ \sin \theta =-\dfrac{1}{\sqrt{2}}\\ \therefore \theta =-\dfrac{\pi}{4} \qquad \quad [\text{As } \theta \text{lies in the IV quadrant}] \\ \therefore 1-i=r \cos \theta + i r \sin \theta =\sqrt{2} \cos (-\dfrac{\pi}{4})+i\sqrt{2}\sin (-\dfrac{\pi}{4}) =\sqrt{2}[\cos(-\dfrac{\pi}{4})+ i \sin(-\dfrac{\pi}{4})]$ $\\$ This is the required polar form.

18   Convert the given complex number in polar form: $- 1 + i$

Solution :

$-1+i $ $\\$ Let $ r \cos \theta =-1 $ and $ r \sin \theta =1$ $\\$ On squaring and adding , we obtain $\\$ $ r^2 \cos^2 \theta +r^2 \sin^2 \theta =(-1)^2 +1^2\\ \implies r^2 (\cos^2 \theta + \sin^2 \theta) =1+1\\ \implies r^2=2\\ \implies r=\sqrt{2} \quad \quad [\text{Conventionally,} r > 0]\\ \therefore \sqrt{2}\cos \theta =-1$ and $\sqrt{2} \sin \theta =1 \\ \implies \cos \theta =-\dfrac{1}{\sqrt{2}} $ and $ \sqrt{2} \sin \theta =1\\ \therefore \theta =\pi-\dfrac{\pi}{4}=\dfrac{3 \pi}{4}, \qquad \quad [\text{As} \theta \text{lies in the II quadrant}]$ $\\$ It can be written, $\\$ $ \therefore -1+i=r \cos \theta + i r \sin \theta =\sqrt{2} \cos \dfrac{3 \pi}{4} + i\sqrt{2}\sin \dfrac{3 \pi}{4}=\sqrt{2}(\cos \dfrac{3 \pi}{4}+i \sin \dfrac{3 \pi}{4}) $ $\\$ This is the required polar form.

19   Convert the given complex number in polar form: $- 1 - i$

Solution :

$-1-i$ $\\$ Let $ r \cos \theta =-1 $ and $ r \sin \theta =-1 $ $\\$ On squaring and adding, we obtain $\\$ $r^2 \cos^2 \theta + r^2 \sin^2 \theta =(-1)^2+(-1)^2\\ \implies r^2(\cos^2 \theta +\sin^2 \theta) =1+1\\ \implies r^2 =2 \\ \implies r=\sqrt{2}\qquad \quad [\text{Conventionally,} r > 0]\\ \therefore \sqrt{2} \cos \theta =-1$ and $ \sqrt{2} \sin \theta =-1$ $\\$ $ \implies \cos \theta =-\dfrac{-1}{\sqrt{2}} $ and $ \sin \theta =-\dfrac{1}{\sqrt{2}}\\ \therefore \theta =-(\pi - \dfrac{\pi}{4})=-\dfrac{3 \pi}{4} \quad [\text{As} \theta \text{lies in the III quadrant}] \\ \therefore -1-i= r \cos \theta + i \ r \sin \theta =\sqrt{2} \cos \dfrac{-3 \pi}{4}+i\sqrt{2} \sin \dfrac{-3 \pi}{4}\\ =\sqrt{2}(\cos \dfrac{-3 \pi}{4}+ i \sin \dfrac{-3 \pi} {4})$ $\\$ This is the required polar form.

20   Convert the given complex number in polar form: $- 3$

Solution :

$-3$ $\\$ let $ r \cos \theta =-3 \text{ and \ } r \sin \theta =0 $ $\\$ On squaring and adding , we obtain $\\$ $ r^2 \cos^2 \theta + r^2 \sin^2 \theta =(-3)^2\\ \implies r^2(\cos^2 \theta + \sin^2 \theta ) =9\\ \implies r62=9\\ \implies r=\sqrt{9} =3 \quad \quad [\text{Conventionally,} r > 0] \\ \therefore 3 \cos \theta =-3 \text{and} 3 \sin \theta =0\\ \implies \cos \theta =-1 \text{ and} \sin =0\\ \therefore \theta = \pi\\ \therefore -3=r \cos \theta + i \ r \sin \theta =3 \cos \pi + i 3 \sin \pi =3(\cos \pi + i \sin \pi)$ $\\$ This is the required polar form.

21   Convert the given complex number in polar form: $ \sqrt{3}+i$

Solution :

$\sqrt{3}+i $ $\\$ Let $ r \cos \theta =\sqrt{3} \text{and } r \sin \theta =1 $ $\\$ On squaring and adding, we obtain $\\$ $ r^2 \cos ^2 \theta + r^2 \sin^2 \theta =(\sqrt{3})^2 + 1^2\\ \implies r^2 (\cos^2 \theta +\sin ^2 \theta)=3+1\\ \implies r^2 = 4 \\ \implies r=\sqrt{4} =2 \quad \quad [\text{Conventionally,} r > 0]\\ \therefore 2 \cos \theta =\sqrt{3} \text{and} 2 \sin \theta =1\\ \implies \cos \theta =\dfrac{\sqrt{3}}{2} \text{and } \sin \theta =\dfrac{1}{2} \\ \therefore \theta =\dfrac{\pi}{6} \quad \quad [\text{As} \theta \text{lies in the I quadrant}]\\ \therefore \sqrt{3}+i = r \cos \theta + i \ r \sin \theta =2 \cos \dfrac{\pi}{6}+i \ 2 \sin\dfrac{\pi}{6}=2(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}) $ $\\$ This is the required polar form.

22   Convert the given complex number in polar form:$ i$

Solution :

$i $ $\\$ Let $ r \cos \theta =0 \text{and} r \sin \theta =1$ $\\$ On squaring and adding, we obtain $\\$ $ r^2 \cos^2 \theta + r^2 \sin^2 \theta =0^2 + 1^2\\ \implies r^2(\cos^2 \theta + \sin^2 \theta) =1\\ \implies r^2 =1\\ \implies r=\sqrt{1}=1 \quad [\text{Conventionally,} r > 0]\\ \therefore \cos \theta =0 \text{and } \sin \theta =1\\ \therefore \theta =\dfrac{\pi}{2}\\ \therefore i= r \cos \theta + i \ r \sin \theta =\cos \dfrac{\pi}{2}+i \sin \dfrac{\pi}{2}$ $\\$ This is the required polar form.

23   Solve the equation $x^ 2 + 3 = 0$

Solution :

The given quadratic equation is $x^ 2 + 3 = 0$$\\$ On comparing the given equation with $ax^ 2 + bx + c = 0,$ $\\$ We obtain $a = 1, b = 0,$ and $c = 3$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D = b ^2 - 4 ac = 0^ 2 - 4 * 1 * 3 = - 12$$\\$ Therefore, the required solutions are$\\$ $=\dfrac{-b\pm \sqrt{D}}{2a} =\dfrac{\pm\sqrt{-12}}{2*1}=\dfrac{\pm\sqrt{12i}}{2} \quad \quad [\sqrt{1}=i]\\ =\dfrac{\pm2\sqrt{3i}}{2}=\pm \sqrt{3i}$

24   Solve the equation $ 2 x ^2 + x + 1 = 0$

Solution :

The given quadratic equation is $2 x^ 2 + x + 1 = 0$ $\\$ On comparing the given equation with $ax ^2 + bx + c = 0 ,$ $\\$ We obtain $a = 2, b = 1$ and $c = 1$ $\\$ Therefore, the discriminant of the given equation is $\\$ $D = b^ 2 - 4 ac = 1^ 2 - 4 * 2 * 1 = 1 - 8 =- 7$ $\\$ Therefore, the required solutions are $\\$ $\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-1\pm \sqrt{-7}}{2*2} =\dfrac{-1 \pm \sqrt{7i}}{4} \quad [\sqrt{-1}=i]$ $\\$

25   Solve the equation $x ^2 + 3 x + 9 = 0$

Solution :

The given quadratic equation is $x^ 2 + 3 x + 9 = 0$ $\\$ On comparing the given equation with $ ax ^2 + bx + c = 0 ,$ $\\$ We obtain $ a = 1, b = 3,$ and $c = 9$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D = b^ 2 - 4 ac = 3 ^2 - 4 * 1 * 9 = 9 - 36 =- 27$ $\\$ Therefore, the required solutions are $\\$ $\dfrac{-b \pm\sqrt{D}}{2a}=\dfrac{-3\pm\sqrt{-27}}{2(1)}\\ \dfrac{-3\pm3\sqrt{-3}}{2}=\dfrac{-3\pm 3\sqrt{3i}}{2} \quad [\sqrt{-1}=i]$

26   Solve the equation $- x^ 2 + x - 2 = 0$

Solution :

The given quadratic equation is $-x ^2 + x - 2 = 0$ $\\$ On comparing the given equation with $ax ^2 + bx + c = 0 ,$ $\\$ We obtain $a =- 1, b = 1, $ and $c =- 2$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D =b ^2 - 4 ac = 1^ 2 - 4 *(- 1)*(- 2 )=1 - 8 =- 7$ $\\$ Therefore, the required solutions are$\\$ $\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-1\pm \sqrt{-7}}{2(-1)}=\dfrac{-1\pm\sqrt{7i}}{-2} \quad [\sqrt{-1}=i]$

27   Solve the equation $x ^2 + 3 x + 5 = 0$

Solution :

The given quadratic equation is $x^ 2 + 3 x + 5 = 0$ $\\$ On comparing the given equation with $ax ^2 + bx + c = 0 ,$ $\\$ We obtain $a = 1, b = 3,$ and $c = 5$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D =b ^2 -4 ac = 3 ^2 - 4 * 1 * 5 = 9 - 20 =- 11$ $\\$ Therefore, the required solutions are $\\$ $ \dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-3\pm\sqrt{-11}}{2*1}=\dfrac{-3\pm\sqrt{11i}}{2} \quad [\sqrt{-1}i]$ $\\$

28   Solve the equation $x^ 2 + x + 2 = 0$

Solution :

The given quadratic equation is $x^ 2 - x + 2 = 0$ $\\$ On comparing the given equation with $ax ^2 + bx + c = 0 ,$ $\\$ We obtain $a = 1, b =- 1,$ and $c = 2$ $\\$ Therefore, the discriminant of the given equation is $\\$ $D = b ^2 - 4 ac =(- 1 )^2- 4 * 1 * 2 =1 - 8 =- 7$$\\$ Therefore, the required solutions are$\\$ $\dfrac{-b\pm \sqrt{D}}{2a} =\dfrac{-(-1)\pm\sqrt{-7}}{2*1}=\dfrac{1\pm\sqrt{7 i}}{2} \quad \quad [\sqrt{-1}=i]$

29   Solve the equation $\sqrt{2}x^2+x+\sqrt{2}=0$

Solution :

The given quadratic equation is $\sqrt{2}x^2+x+\sqrt{2}=0$ $\\$ On comparing the given equation with $ax^ 2 + bx + c = 0 ,$ $\\$ We obtain $a = 2, b = 1,$ and $c = 2$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D = b ^2 - 4 ac = 1^ 2 - 4 *\sqrt{ 2} *\sqrt{2 }= 1 - 8 =- 7$ Therefore, the required solutions are $\\$ $\dfrac{-b\pm \sqrt{D}}{2a} =\dfrac{-1\pm\sqrt{-7}}{2* \sqrt{2}} \\ =\dfrac{-1 \pm \sqrt{7i}}{2a}=\dfrac{-1\pm \sqrt{-7}}{2* \sqrt{2}} =\dfrac{-1 \pm \sqrt{7i}}{2 \sqrt{2}} \quad [\sqrt{-1=i}]$

30   Solve the equation $\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0$

Solution :

The given quadratic equation is $\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0$ $\\$ On comparing the given equation with $ax ^2 + bx + c = 0 ,$ $\\$ We obtain $a = 3, b =- 2,$ and $c = 3\sqrt{ 3 }$ $\\$ Therefore, the discriminant of the given equation is $\\$ $D=b^2-4ac=(-\sqrt{2})^2-4(\sqrt{3})(3\sqrt{3})=2-36=-34$ $\\$ Therefore, the required solutions are$\\$ $\dfrac{-b\pm \sqrt{D}}{2a} =\dfrac{-(-\sqrt{2})\pm \sqrt{-34}}{2*\sqrt{3}}=\dfrac{\sqrt{2}\pm \sqrt{34i}}{2 \sqrt{3}} \quad \quad [\sqrt{-1}=i]$

31   Solve the equation $ x^2+x+\dfrac{1}{\sqrt{2}}=0 $

Solution :

The given quadratic equation is $ x^2+x+\dfrac{1}{sqrt{2}}=0$ $\\$ This equation can also be written as $ \sqrt{2}x^2 +\sqrt{2}x+1=0 $ $\\$ On comparing the given equation with $ ax^ 2 +bx + c = 0 ,$ we obtain $a =2, b = 2, $and $c = 1$ $\\$ $\therefore \text{Discriminant} (D)=b^2-4 ac =(\sqrt{2})^2-4*(sqrt{2})*1=2-4 sqrt{2}$ $\\$ Therefore, the required solutions are $ \\$ $\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{-\sqrt{2}\pm\sqrt{2-4\sqrt{2}}}{2*\sqrt{2}}=\dfrac{-\sqrt{2}\pm\sqrt{2(1-2\sqrt{2})}}{2 \sqrt{2}}\\ =(\dfrac{-\sqrt{2}\pm \sqrt{2}(\sqrt{2\sqrt{2}-1})i}{2\sqrt{2}}) \quad \quad [\sqrt{-1}=i]\\ =\dfrac{-1\pm(\sqrt{2\sqrt{2}-1})i}{2}$

32   Solve the equation $ x^2+ \dfrac{x}{\sqrt{2}}+1=0$

Solution :

The given quadratic equation is $ x^2+ \dfrac{x}{\sqrt{2}}+1=0$ $\\$ This equation can also be written as $\sqrt{2}x^2+x+\sqrt{2}=0 $ $\\$ On comparing the given equation with $ax ^2 +bx + c = 0 ,$ $\\$ We obtain $ a=\sqrt{2},b=1,$ and $c=\sqrt{2}$ $\\$ $\therefore \text{Discriminant }(D)=b^2-4ac=1^2-4* \sqrt{2}*\sqrt{2}=1-8=-7$ $\\$ Therefore, the required solutions are $\\$ $\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-1\pm\sqrt{-7}}{2 \sqrt{2}}=\dfrac{-1\pm\sqrt{7i}}{2\sqrt{2}} \quad \quad [\sqrt{-1}=i]$

33   Evalute: $ [i^{18}+(\dfrac{1}{i})^25]^3$

Solution :

$ [i^{18}+(\dfrac{1}{i})^25]^3 \\ =[i^{4*4+2}+\dfrac{1}{i^{4*6+1}}]^3\\ =[(i^4).i^2+\dfrac{1}{(i^4)^6.i}]^3\\ =[i^2+\dfrac{1}{i}]^3 \quad [i^4=1]\\ =[-1+\dfrac{1}{i}*\dfrac{i}{i}]^3 \quad [i^2=-1]\\ =[-1+\dfrac{i}{i^2}]^3\\ =[-1-i]^3\\ =(-1)^3[1+i]^3\\ =-[1^3+i^3+3.1.i(1+i)]\\ =-[1+i^3+3i+3i^2]\\ =-[1-i+3i-3]\\ =-[-2+2i]\\ =2-2i$

34   For any two complex numbers$ z_1$ and $ z_2 $ prove that $\\$ $ Re(z_1 z_2)=Re z_1 Rez_2-Imz_1Imz_2$

Solution :

Let $ z_1=x_1+iy_1$ and $ z_2=x_2+iy_2 $ $\\$ $\therefore z_1 z_2 =(x_1+iy_1)(x_2+iy_2)\\ =x_1(x_2+iy_2)+iy_1(x_2+iy_2)\\ =x_1x_2+ix_1y_2+iy_1x_2+i^2y_1y_2\\ =x_1x_2+ix_1y_2+iy_1x_2-y_1y_2 \quad [i^2=-1]\\ =(x_1x_2-y_1y_2)+i(x_1y_2+y_1x_2)\\ \implies Re(z_1z_2)=x_1x_2-y_1y_2\\ \implies Re(z_!z_2)=Rez_1 Rez_2-Imz_1Imz_2$ $\\$ Hence, proved.

35   Reduce $ (\dfrac{1}{1-4i}-\dfrac{2}{1+i})(\dfrac{3-4i}{5+i})$ to the standard form

Solution :

$ (\dfrac{1}{1-4i}-\dfrac{2}{1+i})(\dfrac{3-4i}{5+i})\\ =[\dfrac{(1+i)-2(1-4i)}{(1-4i)(1+i)}][\dfrac{3-4i}{5+i}]\\ =[\dfrac{1+i-2+8i}{1+i-4i-4i^2}][\dfrac{3-4i}{5+i}]\\ =[\dfrac{-1+9i}{5-3i}][\dfrac{3-4i}{5+i}]\\ =[\dfrac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}]=\\ \dfrac{33+31i}{28-10i}=\dfrac{33+31i}{2(14-5i)}\\ =\dfrac{(33+31i)}{2(14-5i)}*\dfrac{14+5i}{14+5i}\\ \quad \quad [\text{On multiplying numerator and denominator by } (14+5i)]\\ \dfrac{462+165i+434i+155i^2}{2[(14)^2-(5i)^2]}\\ =\dfrac{307+599i}{2(196-25i^2)}\\ =\dfrac{307+599i}{2(221)}=\dfrac{307+599i}{442}\\ \dfrac{307}{442}+\dfrac{599i}{442}\\ $ $\\$ This is the required standard form.

36   If $ x-iy=\sqrt{\dfrac{a-ib}{c-id}}$ prove that $(x^2+y^2)^2=\dfrac{a^2+b^2}{c^2+d^2} $

Solution :

$x-iy =\sqrt{\dfrac{a-ib}{c-id}}\\ =\sqrt{\dfrac{a-ib}{c-id}*\dfrac{c+id}{c+id}} \\ [\text{On multiplying numerator and denominator by }(c+id)]$

$=\sqrt{\dfrac{(ac+bd)+i(ad-bc)}{c^2+d^2}}\\ \therefore (x-iy)^2=\dfrac{(ac+bd)+i(ad-bc)}{c^2+d^2}\\ \implies x^2-y^2-2ixy=\dfrac{(ac+bd)+i(ad-bc)}{c^2+d^2}$ $\\$ On comparing real and imaginary parts, we obtain $\\$ $x^2-y^2=\dfrac{ac+bd}{c^2+d^2},-2xy=\dfrac{ad-bc}{c^2+d^2}.....(1)\\ (x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2\\ =(\dfrac{ac+bd}{c^2+d^2})^2+(\dfrac{ad-bc}{c^2+d^2}) \qquad \quad [\text{Using (1)}]\\ =\dfrac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^ 2+d^2)^2}\\ =\dfrac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}\\ =\dfrac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^2}\\ =\dfrac{(c^2+d^2)(a^2+b^2)}{(c62+d^2)^2}\\ =\dfrac{a^2+b^2}{c^2+d^2}$ $\\$ Hence, proved.

37   Convert the following in the polar form:$\\$ (i)$\dfrac{1+7i}{(2-i)^2},\ \ \ \ \ \ $ (ii)$\dfrac{1+3i}{1-2i}$

Solution :

(i) Here,$ z=\dfrac{1+7i}{(2-i)^2}\\ =\dfrac{1+7i}{(2-i)^2}=\dfrac{1+7i}{4+i^2-4i}=\dfrac{1+7i}{4-1-4i}\\ =\dfrac{1+7i}{3-4i}*\dfrac{3+4i}{3+4i}\\ =\dfrac{3+4i+21i+28i^2}{3^2+4^2}\\ =\dfrac{3+4i+21i-28}{3^2+4^2}=\dfrac{-25+25i}{25}\\ =-1+i$ $\\$ Let $ r \cos \theta =-1$ and $ r \sin \theta =1$ $\\$ On squaring and adding, we obtain$ r^2(\cos^2 \theta +\sin^2 \theta) =1$$\\$ $\implies r^2(\cos^2 \theta +\sin^2 \theta) =2\\ \implies r^2=2 \quad \quad [\cos^2 \theta +\sin^2 \theta =1] \\ \implies r=\sqrt{2} \quad \quad [\text{Conventionally,}r > 0]\\ \therefore \sqrt{2}\cos \theta =-1 \text{and} \sqrt{2}\sin \theta =1\\ \implies \cos \theta =\dfrac{-1}{\sqrt{2}}\text{and} \sin \theta =\dfrac{1}{\sqrt{2}}\\ \therefore \theta =\pi -\dfrac{\pi}{4}=\dfrac{3\pi}{4} \qquad \quad [\text{As} \theta \text{lies in II quadrant} ]\\ \therefore z= r\cos \theta + i \ r \sin \theta \\ =\sqrt{2}\cos \dfrac{3\pi}{4}+i\sqrt{2}\sin \dfrac{3\pi}{4}=\\ \sqrt{2}(\cos\dfrac{3\pi}{4}+i \sin \dfrac{3 \pi}{4})$ $\\$ This is the required polar form.$\\$ (ii) Here,$ z=\dfrac{1+3i}{1-2i}\\ =\dfrac{1+3i}{1-2i}*\dfrac{1+2i}{1+2i}\\ =\dfrac{1+2i+3i-6}{1+4}\\ =\dfrac{-5+5i}{5}=-1+i$ $\\$ Let $ r \cos \theta =-1$ and $ r \sin \theta $ $\\$ $=1 $ on squaring and adding, we obtain $ r^2(\cos^2\theta +\sin^2 \theta)\\ =1+1\\ \implies r^2(\cos^2 \theta +\sin^2 \theta)=2\\ \implies r^2=2 \qquad \quad [\cos^2 \theta + \sin ^2 \theta =1]\\ \implies r=\sqrt{2} \qquad \quad [\text{Conventionally,} r>0]\\ \therefore \sqrt{2}\cos \theta=-1$ and $ \sqrt{2} \sin \theta =1 \\ \implies \cos \theta =\dfrac{-1}{\sqrt{2}} $ and $ \sin \theta =\dfrac{1}{\sqrt{2}}\\ \therefore \theta=\pi-\dfrac{\pi}{4}=\dfrac{3 \pi}{4} \qquad \quad [\text{As} \theta \text{lies in II quadrant}]\\ \therefore z= r \cos \theta+ i \ r \sin \theta\\ =\sqrt{2 }\cos \dfrac{3\pi}{4}+i\sqrt{2}\sin \dfrac{3 \pi}{4}=\sqrt{2(\cos \dfrac{3\pi}{4}+i \sin \dfrac{3 \pi}{4})}$ $\\$ This is the required polar form.

38   Solve the equation $ 3x^2-4 x+\dfrac{20}{3}=0$

Solution :

The given quadratic equation is $ 3x^2-4x+\dfrac{20}{3}=0$ $\\$ This equation can also be written as $9 x ^2 - 12 x + 20 = 0$$\\$ On comparing this equation with $ax ^2 + bx + c = 0$ , we obtain $a = 9, b=- 12 $ and $c = 20$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D = b^ 2 - 4 ac =(-12 )^2- 4*9 *20 = 144 - 720 =- 576$$\\$ Therefore, the required solutions are$\\$ $\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-(12)\pm\sqrt{-576}i}{18} \quad [\sqrt{-1}=i]\\ =\dfrac{12\pm24i}{18}=\dfrac{6(2\pm 4i)}{18}=\dfrac{2\pm 4i}{3}=\dfrac{2}{3}\pm\dfrac{4}{3}i$

39   Solve the equation $ x^2-2x+\dfrac{3}{2}=0$

Solution :

The given quadratic equation is $ x^2-2x+\dfrac{3}{2}=0$$\\$ This equation can also be written as $2 x^ 2 - 4 x + 3 = 0$ $\\$ On comparing this equation with $ ax ^2 + bx + c = 0 ,$ we obtain $a = 2, b =- 4$ and $c = 3$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D = b ^2 - 4 ac =(- 4 )^2 -4*2* 3 = 16 - 24 =- 8 $ $\\$ Therefore, the required solutions are$\\$ $\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-(-4)\pm\sqrt{8}}{2*2}=\dfrac{4\pm2\sqrt{2i}}{4} \quad [\sqrt{-1}=i]\\ =\dfrac{2\pm\sqrt{2i}}{2}=1\pm\dfrac{\sqrt{2}}{2}i$

40   Solve the equation $27 x^ 2 - 10 x + 1 = 0$

Solution :

The given quadratic equation is $27 x^ 2 - 10 x + 1 = 0$ $\\$ On comparing this equation with $ax ^2 + bx + c = 0 $, we obtain $a = 27, b =-10 $ and $c = 1$ $\\$ Therefore, the discriminant of the given equation is$\\$ $D=b^2-4ac=(-10)^2-4827*1=100-108=-8$ $\\$ Therefore, the required solutions are$\\$ $\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-(-10)\pm \sqrt{-8}}{2*27}=\dfrac{10\pm2\sqrt{2i}}{54} \qquad \quad [\sqrt{-1}=i]\\ =\dfrac{5\pm\sqrt{2i}}{27}=\dfrac{5}{27}\pm \dfrac{\sqrt{2}}{27}i$

41   Solve the equation $21 x ^2 -28 x + 10 = 0$

Solution :

The given quadratic equation is $21 x^ 2 - 28 x + 10 = 0$ $\\$ On comparing this equation with $ax ^2 + bx + c = 0 $, we obtain $a = 21, b =- 28$ and $c = 10$ $\\$ Therefore, the discriminant of the given equation is $\\$ $D = b^2 - 4 ac =(- 28 )^2- 4 * 21 * 10 = 784 - 840 =- 56 $ Therefore, the required solutions are$\\$ $\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{-(-28)\pm\sqrt{56i}}{2*21}\\ =\dfrac{28\pm\sqrt{56i}}{42} \quad \quad [\sqrt{-1}=i]\\ =\dfrac{28\pm 2\sqrt{14i}}{42}=\dfrac{28}{42\pm}\pm \dfrac{2\sqrt{14}}{42}i=\dfrac{2}{3}\pm\dfrac{14}{21}i$

42   If $ z_1=2-i,z_2=1+i$ find $\left|\dfrac{z_1+z_2+1}{z_1-z_2+1}\right|$

Solution :

$z_1=2-i,z_2=1+i $ $ \\$ $\therefore \left| \dfrac{z_1+z_2+1}{z_1-z_2+1} \right| \\ \left| \dfrac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1} \right| $$ \\$ $= \left| \dfrac{4}{2-2i} \right| = \left| \dfrac{4}{2(1-i)} \right| $$ \\$ $=\left| \dfrac{2}{1-i} * \dfrac{1+i}{1+i} \right| = \left| \dfrac{2(1+i)}{(1^2-i^2)} \right|$ $\\$ $=\left| \dfrac{2(1+i)}{1+1} \right| \quad [i^2=-1]$$\\$ $=\left| \dfrac{2(1+i)}{2} \right| $ $ \\$ $=\left| 1+i \right| = \sqrt{1^2+1^2}= \sqrt{2}$ $\\$ Thus, the value of $ \left| \dfrac{z_1+z_2+1}{z_1-z_2+1} \right| $ is $ \sqrt{2}.$

43   If $a+ib=\dfrac{(x+i)^2}{2x^2+1},$ prove that $ a^2+b^2=\dfrac{(x^2+1)^2}{(2x^2+1)^2}$

Solution :

$a+ib=\dfrac{(x+i)^2}{2x^2+1}\\ =\dfrac{x^2+i^2+2xi}{2x^2+1}\\ =\dfrac{x^2-1+i2x}{2x^2+1}\\ =\dfrac{x^2-1}{2x^2+1}+i(\dfrac{2x}{2x^2+1})$ $\\$ On comparing real and imaginary parts,we obtain $\\$ $a=\dfrac{x^2-1}{2x^2+1} \text{and} b=(\dfrac{2x}{2x^2+1})\\ \therefore a^2+b^2=(\dfrac{x^2-1}{2x62+1})^2+(\dfrac{2x}{2x^2+1})^2\\ =\dfrac{x^4+1-2x^2+4x^2}{(2x+1)^2}\\ =\dfrac{x^4+1+2x^2}{(2x^2+1)^2}\\ =\dfrac{(x^2+1)^2}{(2x^2+1)^2}\\ \therefore a^2+ b^2=\dfrac{(x^2+1)^2}{(2x^2+1)^2}$ Hence, proved.

44   Let $ z_1=2-i, z_2=-2+i.$ $\\$ Find (i) $ Re(\dfrac{z_1z_2}{z_1}), \ \ \ \ \ \ $ (ii) $ Im(\dfrac{1}{z_1\bar{z_2}})$

Solution :

$ z_1=2-i,z_2=-2+i\\ (i) z_1z_2=(2-i)(-2+i)=-4+2i+2i-i^2=-4+4i-(-1)=-3+4i\\ \bar{z_1}=2+i\\ \therefore \dfrac{z_1z_2}{z_1}=\dfrac{-3+4i}{2+i}$ $\\$ On multiplying numerator and denominator by $(2-i),$ we obtain$\\$ $\dfrac{z_1z_2}{z_1}=\dfrac{(-3+4i)(2-i)}{(2+i)(2-i)}=\dfrac{-6+3i+8i-4i^2}{2^2+1^2}\\ =\dfrac{-6+11i-4(-1)}{2^2+1^2}\\ =\dfrac{-2+11i}{5}=\dfrac{-2}{5}+\dfrac{11}{5}i$ $\\$ On comparing real parts, we obtain $\\$ $ Re(\dfrac{z_1z_2}{\bar{z_1}})=\dfrac{-2}{5}$ $\\$ (iii)$\dfrac{1}{z_1 \bar{z_1}}=\dfrac{1}{(2-i)(2+i)}=\dfrac{1}{(2)^2+(1)^2}=\dfrac{1}{5}$ $\\$ On comparing imaginary parts, we obtain $\\$ $Im(\dfrac{1}{z_1 \bar{z_1}})=0$

45   Find the modulus and argument of the complex number $\dfrac{1+2i}{1-3i}$

Solution :

Let $ z=\dfrac{1+3i}{1-3i},$ then$\\$ $z=\dfrac{1+2i}{1-3i}*\dfrac{1+3i}{1+3i}=\dfrac{1+3i+2i+6i^2}{1^2+3^2}\\ =\dfrac{1+5i+6(-1)}{1+9}\\ =\dfrac{-5+5i}{10}=\dfrac{-5}{10}+\dfrac{5i}{10}=\dfrac{-1}{2}+\dfrac{1}{2}i\\ Let z=r \cos \theta + i \ r \sin \theta \\ i.e., r \cos \theta =\dfrac{-1}{2}\text{and} r \sin \theta =\dfrac{1}{2} $ $\\$ On squaring and adding, we obtain$\\$ $r^2(\cos^2+ \sin ^2 \theta)=(\dfrac{-1}{2})^2+(\dfrac{1}{2})^2\\ \implies r^2=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\\ \implies r=\dfrac{1}{\sqrt{2}} \qquad \quad [\text{Conventionally,} r > 0]\\ \therefore \dfrac{1}{\sqrt{2}}\cos \theta =\dfrac{-1}{2}\text{and} \dfrac{1}{\sqrt{2}} \sin \theta =\dfrac{1}{2}\\ \implies \cos \theta =\dfrac{-1}{\sqrt{2}} \text{and} \sin \theta =\dfrac{1}{\sqrt{2}} \\ \therefore \theta =\pi -\dfrac{\pi}{4}=\dfrac{3 \pi}{4} \qquad [\text{As} \theta \text{lies in the II quadrant}]$ $\\$ Therefore, the modulus and argument of the given complex number are $ \dfrac{1}{\sqrt{2}} $ and $ \dfrac{3 \pi}{4}$ respectively.

46   Find the real numbers x and y if $( x - iy )( 3 + 5 i )$ is the conjugate of $- 6 - 24i .$

Solution :

Let $ z=(x-iy)(3+5i)\\ z=3x+5xi-3yi-5yi^2=3x+5xi-3yi+5y=(3x+5y)+i(5x-3y)\\ \therefore \bar{z}=(3x+5y)-i(5x-3y)\\ \text{It is given that,} \bar{z}=-6-24i\\ \therefore (3x+5y)-i(5x-3y)=-6-24 i$ $\\$ Equating real and imaginary parts, we obtain $\\$ $3x+5y=-6......(i)\\ 5x-3y=24......(ii)$ $\\$ Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain $\\$ $ 9x+15y=-18\\ \underline{25x-15y=120}\\ \quad 34x=102 \\ \therefore x=\dfrac{102}{34}=3$ $\\$ Putting the value of x in equation (i), we obtain $\\$ $3(3)+5y=-6\\ \implies 5y=-6-9=-15\\ \implies y=-3 $ $\\$ Thus, the values of x and y are 3 and -3 respectively.

47   Find the modulus of $ \dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$

Solution :

$\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}=\dfrac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}\\ =\dfrac{1+i^2+2i-1-i^2+2i}{1^2+1^2}\\ =\dfrac{4i}{2}=2i \\ \therefore \left|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\right|=|2i|=\sqrt{2^2}=2$

48   if $(x+iy)^3=u+iv,$ then show that: $\dfrac{u}{x}+\dfrac{v}{y}=4(x^2-y^2)$

Solution :

$(x+iy)^3=u+iv\\ \implies x^3+(iy)^3+3.x.iy(x+iy)=u+iv\\ \implies x^3 +i^3y^3+3x^2yi+3x^2yi+3xy^2i^2=u+iv\\ \implies x^3-iy^3+3x^2yi-3xy^2=u+iv\\ \implies (x^3-3xy^2)+i(3x^2y-y^3)=u+iv$ $\\$ On equating real and imaginary parts, we obtain $\\$ $ u=x^3-3xy^2,v=3x^2y-y^3\\ \therefore \dfrac{u}{x}+\dfrac{v}{y}=\dfrac{x^3-3xy^2}{x}+\dfrac{3x^2y-y^3}{y}\\ =\dfrac{x(x^2-3y^2)}{x}+\dfrac{y(3x^2-y^2)}{y}\\ =x^2-3y^2+3x^2-y^2\\ =4x^2-4y^2\\ =4(x^2-y^2)\\ \therefore \dfrac{u}{x}+\dfrac{v}{y}=4(x^2-y^2)$ $\\$ Hence,proved.

49   Find the number of non-zero integral solutions of the equation $|1-i|^x=2^x$

Solution :

$|1-i|^x=2^x\\ \implies (\sqrt{1^2+(-1)^2})^x=2^x\\ \implies (\sqrt{2})^x=2^x\\ \implies 2^{x/2}=2^x\\ \implies \dfrac{x}{2}=x\\ \implies x=2x\\ \implies 2x-x=0\\ \implies x=0$ $\\$ Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0.

50   If $(a+ib)(c+id)(e+if)(g+ih)=A+iB,$ then show that:$\\$ $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2$

Solution :

$(a+ib)(c+id)(e+if)(g+ih)=A+iB,\\ \therefore | (a+ib)(c+id)(e+if)(g+ih)|=|A+iB,|\\ |(a+ib)|*|(c+id)|*|(e+if)|*|(g+ih)|=|A+iB,| \quad \because[|z_1z_2|=|z_1||z_2|]\\ \implies \sqrt{a^2+b^2}*\sqrt{c^2+d^2}*\sqrt{e^2+f^2}*\sqrt{g^2+h^2}=\sqrt{A^2+B^2}$ $\\$ On squaring both sides, we obtain $\\$ $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2.$ $\\$ Hence proved.

51   If$(\dfrac{1+i}{1-i})^m=1$ then find the least positive integral value of $m.$

Solution :

$(\dfrac{1+i}{1-i})^m=1\\ \implies (\dfrac{1+i}{1-i}*\dfrac{1+i}{1+i})^m=1\\ \implies (\dfrac{(1+i)^2}{1^2+1^2})^m=1\\ \implies (\dfrac{1^2+i^2+2i}{2})^m=1\\ \implies (\dfrac{1-1+2i}{2})^m=1\\ \implies (\dfrac{2i}{2})^m=1\\ \implies i^m=1\\ \implies i^m=i^{4k}\\ \therefore m=4k,$ where k is some integer.$\\$ Therefore, the least positive integer is 1.$\\$ Thus, the least positive integral value of m is $4(=4*1).$