Rd sharma solution of chapter 13 complex number class 11

Exercise 13.2

QUESTION 1

Express the following complex numbers in the standard form a+ib :

(i) (1+i)(1+2i)

Sol :

=1+2i+i+2i^2

=1+2i+i-2 [\because~(i^2=-1)]

=-1+3i

 

(ii) \dfrac{3+2i}{-2+i}

Sol :

=\dfrac{3+2i}{-2+i}\times\dfrac{-2-i}{-2-i}

=\dfrac{-6-3i-4i-2i^2}{4-i^2}

=\dfrac{-6-7i+2}{4+1} [\because~(i^2=-1)]

=\dfrac{-4-7i}{5}

=\dfrac{-4}{5}-\dfrac{7i}{5}

 

(iii) \dfrac{1}{(2+i)^2}

Sol :

=\dfrac{1}{4+i^2+4i}

=\dfrac{1}{4-1+4i} \left[\because~(i^2=-1)\right]

=\dfrac{1}{3+4i}

=\dfrac{1}{3+4i}\times\dfrac{3-4i}{3-4i}

=\dfrac{3-4i}{(3)^2-(4i)^2} 

=\dfrac{3-4i}{9+16} [\because~(i^2=-1)]

=\dfrac{3}{25}-\dfrac{4i}{25}

 

(iv) \dfrac{1-i}{1+i}

Sol :

=\dfrac{1-i}{1+i}\times \dfrac{1-i}{1-i}

=\dfrac{(1-i)^2}{1-i^2}

=\dfrac{1+i^2-2i}{1-i^2} 

=\dfrac{1-1-2i}{2} [\because~(i^2=-1)]

=\dfrac{-2i}{2}+\dfrac{2i}{2}

=-i

 

(v) \dfrac{(2+i)^3}{2+3i}

Sol :

=\dfrac{8+i^3+6i(2+i)}{2+3i}

=\dfrac{8+i^3+12i+6i^2}{2+3i} 

=\dfrac{8-i+12i-6}{2+3i} [\because~(i^2=-1)~(i^3=-i)]

=\dfrac{2+11i}{2+3i}\times\dfrac{2-3i}{2-3i}

=\dfrac{(2+11i)(2-3i)}{4+9}

=\dfrac{4-6i+22i-33i^2}{13}

=\dfrac{4+16i+33}{13} [\because~(i^2=-1)]

=\dfrac{37+16i}{13}

=\dfrac{37}{13}+\dfrac{16i}{13}

 

(vi) \dfrac{(1+i)(1+\sqrt{3}i)}{1-i}

Sol :

=\dfrac{(1+i)(1+\sqrt{3}i)}{1-i}\times \dfrac{1+i}{1+i}

=\dfrac{(1+i)^2(1+\sqrt{3}i)}{(1-i)(1+i)}

=\dfrac{(1+i^2+2i)(1+\sqrt{3}i)}{(1-i^2)}

=\dfrac{(1-1+2i)(1+\sqrt{3}i)}{1+1} [\because~(i^2=-1)]

=\dfrac{2i+2\sqrt{3}i^2}{2}

=\dfrac{2i-2\sqrt{3}}{2}

=-\sqrt{3}+i

 

(vii) \dfrac{2+3i}{4+5i}

Sol :

=\dfrac{2+3i}{4+5i}\times \dfrac{4-5i}{4-5i}

=\dfrac{(2+3i)(4-5i)}{16-25i^2} 

=\dfrac{(8-10i+12i-15i^2)}{16-25i^2} 

=\dfrac{8+2i+15}{16+25} [\because~(i^2=-1)]

=\dfrac{23+2i}{41}

=\dfrac{23}{41}+\dfrac{2i}{41}

 

(viii) \dfrac{(1-i)^3}{1-i^3}

Sol :

=\dfrac{1-i^3+3i(1-i)}{1-i^3}

=\dfrac{1-i+3i-3i^2)}{1-i} [\because~(i^3=-i)~(i^2=-1)]

=\dfrac{1-i+3i+3)}{1-i}\times \dfrac{1+i}{1+i}

=\dfrac{4+2i}{1-i}\times \dfrac{1+i}{1+i}

=\dfrac{(4+2i)(1+i)}{1-i^2}

=\dfrac{4+4i+2i+2i^2}{1-i^2}

=\dfrac{4+6i-2}{1+1} [\because~(i^2=-1)~(i^3=-i)]

=\dfrac{2+6i}{2}

=1+3i

 

(ix) (1+2i)^{-3}

Sol :

=\dfrac{1}{(1+2i)^{3}}

=\dfrac{1}{[1+8i^3+6i(1+2i)]}

=\dfrac{1}{1+8i^3+6i+12i^2}

=\dfrac{1}{1-8i+6i-12} [\because~(i^2=-1)~(i^3=-i)]

=\dfrac{1}{-11-2i}\times \dfrac{-11+2i}{-11+2i}

=\dfrac{-11+2i}{(-11+2i)(-11-2i)}

=\dfrac{-11+2i}{(121-4i^2)}

=\dfrac{-11+2i}{(121+4}

=\dfrac{-11}{125}+\dfrac{2i}{125}

 

(x) \dfrac{3-4i}{(4-2i)(1+i)}

Sol :

=\dfrac{3-4i}{4+4i-2i-2i^2}

=\dfrac{3-4i}{6+2i} [\because~(i^2=1)]

=\dfrac{3-4i}{6+2i}\times \dfrac{6-2i}{6-2i}

=\dfrac{(3-4i)(6-2i)}{36-4i^2}

=\dfrac{18-6i-24i+8i^2}{36-4i^2}

=\dfrac{18-30i-8}{36+4} [\because~(i^2=-1)]

=\dfrac{10-30i}{40}

=\dfrac{1}{4}-\dfrac{3i}{4}

 

(xi) \left(\dfrac{1}{1-4i}-\dfrac{2}{1+i}\right)\left(\dfrac{3-4i}{5+i}\right)

Sol :

=\left(\dfrac{1+i-2+8i}{(1-4i)(1+i)}\right)\left(\dfrac{3-4i}{5+i}\right)

=\left(\dfrac{1+i-2+8i}{1+i-4i-4i^2}\right)\left(\dfrac{3-4i}{5+i}\right)

=\left(\dfrac{-1+9i}{1-3i+4}\right)\left(\dfrac{3-4i}{5+i}\right) [\because~(i^2=-1)]

=\left(\dfrac{-1+9i}{5-3i}\right)\left(\dfrac{3-4i}{5+i}\right)

=\dfrac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}

=\dfrac{-3+31i+36}{25+25i+3} [\because~(i^2=-1)]

=\dfrac{33+31i}{28+25i}

=\dfrac{33+31i}{28+25i}\times\dfrac{28-25i}{28-25i}

=\dfrac{924+330i+868i+310i^2}{(28)^2-(25i)^2}

=\dfrac{924+1198i-310}{784+100} [\because~(i^2=-1)]

=\dfrac{614+1198i}{884}

=\dfrac{614}{884}+\dfrac{1198i}{884}

=\dfrac{307}{442}+\dfrac{599i}{442}

 

(xii) \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}

Sool :

=\left(\dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}\right)\times\left(\dfrac{1+\sqrt{2}i}{1+\sqrt{2}i}\right)

=\dfrac{(5+\sqrt{2}i)(1+\sqrt{2}i)}{1^2-(\sqrt{2}i)^2}

=\dfrac{5+5\sqrt{2}i+\sqrt{2}i+2i^2}{1-2i^2}

=\dfrac{5+6\sqrt{2}i-2}{1+2} [\because~(i^2=-1)]

=\dfrac{3+6\sqrt{2}i}{3}

=\dfrac{3}{3}+\dfrac{6\sqrt{2}i}{3}

=1+2\sqrt{2}i

 

QUESTION 2

Find the real values of x and y, if 

(i) (x+iy)(2-3i)=4+i

Sol :

=2x-3xi+2yi-3yi^2=4+i

=2x-i(3x+2y)+3y=4+i

=2x+3y-i(3x+2y)=4+i

Comparing both the sides :

2x+3y=4\dots(1)

-3x+2y=1\dots(2)

Multiplying equation (1) by 3 and equation (2) by 2 :

6x+9y=12\dots(3)

-6x+4y=2\dots(4)

Adding equations (3) and (4) :

13y=14

y=\dfrac{14}{13}

Substituting the value of y in equation (1) :

2x+3\times\dfrac{14}{13}=4

2x=4-\dfrac{42}{13}

2x=\dfrac{10}{13}

x=\dfrac{5}{13}

\therefore~x=\dfrac{5}{13}\text{ and } y=\dfrac{14}{13}

 

(ii) (3x-2iy)(2+i)^2=10(1+i)

Sol :

=6x+3xi-4yi-2yi^2=10+10i

=6x+3xi-4yi-2yi^2=10+10i

=6x+3xi-4yi+2y=10+10i [\because~(i^2=-1)]

=6x+2y+i(3x-4y)=10+10i

Comparing both sides:

2x+3y=4\dots(1)

-3x+2y=1\dots(2)

Multiplying equation (1) with 3 and equation (2) with 2 :

6x+9y=12\dots(3)

-6x+4y=2\dots(4)

Adding equations (3) and (4) :

13y=14

y=\dfrac{14}{13}

Substituting the value of y in equation (1):

2x+3\times\dfrac{14}{13}=4

2x=4-\dfrac{42}{13}

2x=\dfrac{10}{13}

x=\dfrac{5}{13}

[\therefore~x=\dfrac{5}{13}\text{ and } y=\dfrac{14}{13}]

 

(ii) (3x-2yi)(2+i)^2=10(1+i)

Sol :

(3x-2yi)(4+i^2+4i)^2=10+10i)

(3x-2yi)(4+i^2+4i)=10+10i [\because~(i^2=-1)]

(3x-2yi)(3+4i)=10+10i

9x+12xi-6yi-8yi^2=10+10i [\because~(i^2=-1)]

9x+12xi-6yi+8y=10+10i [\because~(i^2=-1)]

(9x+8y)+i(12x-6y)=10+10i

Comparing both sides :

9x+8y=10\dots(1)

 

12x-6y=10

6x-3y=5\dots(2)

Multiplying equation (1) by 3 and equation (2) by 8 ,

27x+24y=30\dots(3)

48x-24y=40\dots(4)

Adding equations (3) and (4):

75x=70

\therefore~x=\dfrac{14}{15}

Substituting the value of x in equation (1) :

9\times\dfrac{14}{15}+8y=10

\dfrac{126}{15}+8y=10

8y=10-\dfrac{126}{15}

8y=\dfrac{24}{15}

y=\dfrac{1}{5}

 

(iii) \dfrac{(1+i)x-2i}{3+i}+\dfrac{(2-3i)y+i}{3-i}=i

Sol :

\dfrac{x+xi-2i}{3+i}+\dfrac{2y-3yi+i}{3-i}=i

\dfrac{x+xi+2i}{3+i}+\dfrac{2y-3yi+i}{3-i}=i [\because~(i^2=-1)]

\dfrac{(x+xi+2i)(3-i)+(2y-3yi+i)(3+i)}{(3+i)(3-i)}=i

\dfrac{3x+3xi-6i-xi-xi^2+2i^2+6y-9yi+3i+2yi-3yi^2+i^2}{10}=i

3x+3xi-6i-xi+x-2+6y-9yi+3i+2yi+3y-1=10i [\because~(i^2=-1)]

4x+9y-3+2xi-7yi-3i=10i

(4x+9y-3)+i(2x-7y-3)=10i

Comparing both the sides :

4x+9y-3=0

4x+9y=3\dots(1)

2x-7y-3=10

2x-7y=13\dots(2)

Multiplying equation (2) by 2 :

4x-14y=26\dots(3)

Subtracting equation (3) from 1 :

\begin{matrix}4x+99y=\phantom{0}3\\4x-14y=26\\\underline{~-\phantom{4x}+\phantom{14y}\phantom{=}-}\\\underline{\qquad 23y=-23}\end{matrix}

[\therefore~y=-1]

Substituting the value of y in equation (1) :

4x-9=3

4x=3+9

x=3

 

(iv) (1+i)(x+iy)=2-5i

Sol :

x+xi+yi+yi^2=2-5i

(x-y)+i(x+y)=2-5i [\because~(i^2=-1)]

Comparing both the sides

x-y=2\dots(1)

x+y=-5\dots(2)

Adding equations (1) and (2) 

2x=-3

x=\dfrac{-3}{2}

Substituting the value of x in equation (1)

\dfrac{-3}{2}-y=2

y=\dfrac{-3}{\phantom{-}2}-2

y=\dfrac{-7}{\phantom{-}2}

 

QUESTION 3

Find the conjugates of the following complex numbers:

(i) 4-5i

Sol :

\text{ if }z=a+ib~,~\text{ then } \overline{z}=a-ib

\overline{z}=4+5i

 

(ii) \dfrac{1}{3+5i}

Sol :

=\left(\dfrac{1}{3+5i}\right)\times\left(\dfrac{3-5i}{3-5i}\right)

=\dfrac{3-5i}{9-25i^2}

=\dfrac{3-5i}{9+25} [\because~(i^2=-1)]

=\dfrac{3}{34}-\dfrac{5i}{34}

 

(iii) \dfrac{1}{1+i}

Sol :

=\left(\dfrac{1}{1+i}\right)\times\left(\dfrac{1+i}{1+i}\right)

=\dfrac{1+i}{1-i^2}

=\dfrac{1+i}{2} [\because~(i^2=-1)]

=\dfrac{1}{2}+\dfrac{i}{5}

 

(iv) \dfrac{(3-i)^2}{2+i}

Sol :

=\dfrac{9+i^2-6i}{2+i}

=\dfrac{8-6i}{2+i} [\because~(i^2=-1)]

=\left(\dfrac{8-6i}{2+i}\right)\times\left(\dfrac{2-i}{2-i}\right)

=\dfrac{(8-6i)(2-i)}{4-i^2}

=\dfrac{16-8i-12i+6i^2}{5} [\because~(i^2=-1)]

=\dfrac{16-20i-6}{5}

=\dfrac{10-20i}{5}

=\dfrac{10}{5}-\dfrac{20i}{5}

=2+4i

 

(v) \dfrac{(1+i)(2+i)}{3+i}

Sol :

=\dfrac{2+i+2i+i^2}{3+i}

=\dfrac{2+i+2i-1}{3+i} [\because~(i^2=-1)]

=\dfrac{1+3i}{3+i}\times\dfrac{3-i}{3-i}

=\dfrac{(1+3i)(3-i)}{9-i^2}

=\dfrac{3-i+9i-3i^2}{9-i^2}

=\dfrac{3+8i+3}{9+1} [\because~(i^2=-1)]

=\dfrac{6+8i}{10}

=\dfrac{3+4i}{5}

 

(vi) \dfrac{(3-2i)(2+3i)}{(1+2i)(2-i)}

Sol :

=\dfrac{6+9i-4i-6i^2}{2-i+4i-2i^2}

=\dfrac{6+5i-6i^2}{2+3i-2i^2}

=\dfrac{6+5i+6}{2+3i+2} [\because~(i^2=-1)]

=\dfrac{12+5i}{4+3i}\times\dfrac{4-3i}{4-3i}

=\dfrac{(12+5i)(4-3i)}{16-9i^2} 

=\dfrac{48+20i-36i-15i^2}{16+9}

=\dfrac{63-16i}{25} [\because~(i^2=-1)]

 

QUESTION 4

Find the multiplicative inverse of the following complex numbers:

(i) 1-i

Sol :

Let z=1-i then,

\dfrac{1}{z}=\dfrac{1}{1-i}

=\dfrac{1}{1-i}\times\dfrac{1+i}{1+i}

=\dfrac{1+i}{1-i^2}

=\dfrac{1+i}{1-i^2} [\because~(i^2=-1)]

=\dfrac{1+i}{1+1}

=\dfrac{1}{2}+\dfrac{i}{2}

 

(ii) (1+i\sqrt{3})^2

Sol :

=1+3i^2+2\sqrt{3}i

=1-3+2\sqrt{3}i [\because~(i^2=-1)]

z=-2+2\sqrt{3}i

\dfrac{1}{z}={1}{-2+2\sqrt{3}i}

={1}{-2+2\sqrt{3}i}\times\dfrac{-2-2\sqrt{3}i}{-2-2\sqrt{3}i}

={-2-2\sqrt{3}i}{(-2)^2-(2\sqrt{3}i)^2}

={-2-2\sqrt{3}i}{4-12i^2}

={-2-2\sqrt{3}i}{4+12} [\because~(i^2=-1)]

=\dfrac{-2}{16}-\dfrac{2\sqrt{3}i}{16}

=\dfrac{-1}{8}-\dfrac{\sqrt{3}i}{8}

 

(iii) 4-3i

Sol :

z=4-3i

\dfrac{1}{z}={1}{4-3i}

={1}{4-3i}\times\dfrac{4+3i}{4+3i}

={4+3i}{16-9i^2}

={4+3i}{16+9} [\because~(i^2=-1)]

={4}{25}+\dfrac{3i}{25}

 

(iv) \sqrt{5}+3i

Sol :

z=\sqrt{5}+3i

\dfrac{1}{z}={1}{\sqrt{5}+3i}

={1}{\sqrt{5}+3i}\times\dfrac{\sqrt{5}-3i}{\sqrt{5}-3i}

={\sqrt{5}-3i}{(\sqrt{5})^2-(3i)^2}

={\sqrt{5}-3i}{5-9i^2}

={\sqrt{5}}{14}-\dfrac{3i}{14} [\because~(i^2=-1)

 

QUESTION 5

If z_{1}=2-i,z_{2}=1+i, find \left|\dfrac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|

Sol :

\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}{1-i}\right|

=\dfrac{4}{\left|1-i\right|}

Also , \left|1-i\right|=\sqrt{1^2+i^2} \left[\because~(\left|a+ib\right|=\sqrt{a^2+b^2})\right]

=\sqrt{2}

\therefore \left|\dfrac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=\dfrac{4}{\sqrt{2}}

 

QUESTION 6

If z_1=2-i,z_2=-2+i , find

(i) Re\left(\dfrac{z_1z_2}{\overline{z_1}}\right)

(ii) Im\left(\dfrac{1}{z_1\overline{z_1}}\right)

Sol :

(i)Given z_1=2-i,z_2=-2+i \overline{z_1}=2+i

Re\left(\dfrac{z_1z_2}{\overline{z_1}}\right)= \left(\dfrac{(2-i)(-2+i)}{2+i}\right)

=\left(\dfrac{-4+2i+2i-i^2}{2+i}\right)

=\left(\dfrac{-4+4i+1}{2+i}\right) [\because~(i^2=-1)]

=\left(\dfrac{-3+4i}{2+i}\right)

=\left[\dfrac{-3+4i}{2+i}\times\left(\dfrac{2-i}{2+i}\right)\right]

=\left(\dfrac{-6+3i+8i-4i^2}{2^2-i^2}\right)

=\left(\dfrac{-6+11i+4}{4+1}\right) [\because~(i^2=-1)]

=\dfrac{-2}{5}+\dfrac{11i}{5}

Re\left(\dfrac{z_1z_2}{\overline{z_1}}\right)=\dfrac{-2}{\phantom{-}5}

 

(ii) Given z_1=2-i,z_2=-2+i \overline{z_1}=2+i

Im\left(\dfrac{1}{z_1\overline{z_1}}\right)= \dfrac{1}{(2-i)(2+i)}

=\dfrac{1}{4+2i-2i-i^2}

=\dfrac{1}{5} [\because~(i^2=-1)]

Im\left(\dfrac{1}{z_1\overline{z_1}}\right)=0  Since no term containing is present 

 

 

QUESTION 7

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

Sol :

=\dfrac{(1+i)(1+i)-(1-i)(1-i)}{(1-i)(1+i)}

=\dfrac{1+i+i+i^2-(1-i-i+i^2)}{1+i-i-i^2}

=\dfrac{1+2i-1-(1-2i-1)}{1+1} [\because~(i^2=-1)]

=\dfrac{2i+2i)}{2}

=\dfrac{2i+2i)}{2}

=2i

\therefore~{\left|2i\right|=\sqrt{0^2+2^2}}

=2  \left[\because~{\left|a+bi\right|=\sqrt{a^2+b^2}}\right]

\Rightarrow \left|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\right|=2

 

QUESTION 8

If x+iy=\dfrac{a+ib}{a-ib} , prove that x^2+y^2=1

Sol :

x+iy=\dfrac{a+ib}{a-ib}

Taking mod on the both sides:

\left|x+iy\right|=\left|\dfrac{a+ib}{a-ib}\right|

\sqrt{x^2+y^2}=\dfrac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}

\sqrt{x^2+y^2}=1

x^2+y^2=1 squaring both the sides

hence proved

 

QUESTION 9

Find the least positive integral value of for which \left(\dfrac{1+i}{1-i}\right)^n

Sol :

\left(=\dfrac{1+i}{1-i}\right)^n

=\left[\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i}\right]^n

=\left[\dfrac{1+i^2+2i}{1-i^2}\right]^n [\because~(i^2=-1)]

=\left[\dfrac{1-1+2i}{1+1}\right]^n

=\left[\dfrac{2i}{2}\right]^n

=i^n

For =i^n to be real , the least positive value of will be 2 .

As i^2=-1 

 

QUESTION 10

Find the real values of \theta for which the complex number \dfrac{1+icos~\theta}{1-2icos~\theta} is purely real .

Sol :

=\dfrac{1+icos~\theta}{1-2icos~\theta}\times\dfrac{1+2icos~\theta}{1+2icos~\theta}

=\dfrac{1+2icos~\theta+icos~\theta-2cos~\theta}{1+4icos^2~\theta}

=\dfrac{1-2cos~\theta +i3cos~\theta}{1+4cos^2~\theta}

For it to be purely real , the imaginary part must be zero .

3~cos~\theta=0

This is true for odd multiples of \dfrac{\pi}{2}

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QUESTION 11

Find the smallest positive integer value of for which \dfrac{(1+i)^{n\phantom{-2}}}{(1-i)^{n-2}}

Sol :

=\dfrac{(1+i)^{m\phantom{-2}}}{(1-i)^{m-2}}

=\dfrac{(1+i)^{m}}{(1-i)^{m}(1-i)^{-2}}

=\dfrac{(1+i)^{m}~(1-i)^{2}}{(1-i)^{m}}

=\dfrac{(1+i)^{m}\times(1+i)^{m}\times(1+i^2-2i)}{(1-i)^{m}\times(1+i)^{m}}

=\left(\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i}\right)^m\times(1+i^2-2i)

=\left(\dfrac{1+i^2+2i}{1-i^2}\right)^m\times(1+i^2-2i)

=\left(\dfrac{1-1+2i}{1+1}\right)^m\times(1-1-2i) [\because~(i^2=-1)]

=\left(\dfrac{2i}{2}\right)^m\times(-2i)

=(i^m)\times(-2i)

=-2(i)^{m+1}

For this to be real , the smallest positive value of will be 1 

Thus, i^{1+1}=i^2=-1 , which is real

 

QUESTION 12

If \left(\dfrac{1+i}{1-i}\right)^3-\left(\dfrac{1-i}{1+i}\right)^3=x+iy , find (x,y)

Sol :

\left(\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i}\right)

=\dfrac{(1+i)^2}{1^2-i^2}

=\dfrac{(1+i^2+2i}{1-i^2}

=\dfrac{(1-1+2i}{1+1} [\because~(i^2=-1)]

=\dfrac{2i}{2}

=i\dots(1)

 

Also,

\left(\dfrac{1-i}{1+i}=\dfrac{1-i}{1+i}\times\dfrac{1-i}{1-i}\right)

=\dfrac{(1-i)^2}{1^2-i^2}

=\dfrac{1+i^2-2i}{1^2-i^2}

=\dfrac{1-1-2i}{1+1} [\because~(i^2=-1)]

=\dfrac{-2i}{2}

=-i\dots(2)

It is given that ,

\left(\dfrac{1+i}{1-i}\right)^3-\left(\dfrac{1-i}{1+i}\right)^3=x+iy

\Rightarrow \left(i\right)^3-\left(-i\right)^3=x+iy [from (1) and (2)]

\Rightarrow i^3+i^3=x+iy

\Rightarrow 2i^3=x+iy

\Rightarrow 0-2i=x+iy  [\because~(i^3=-i)]

\Rightarrow x=0 \text{ and } y=-2

Thus ,(x,y)=(0,-2)

 

QUESTION 13

If \dfrac{(1+i)^2}{2-i}=x+iy find (x+y)

Sol :

\dfrac{(1+i)^2}{2-i}=\dfrac{1+i^2+2i}{2-i}

=\dfrac{1+i^2+2i}{2-i}\times\dfrac{2+i}{2+i}

=\dfrac{(1+i^2+2i)(2+i)}{(2-i)(2+i)}

=\dfrac{(1+i^2+2i)(2+i)}{(2^2-i^2)}

=\dfrac{(1+i^2+2i)(2+i)}{(2^2-i^2)} [\because~(i^2=-1)]

=\dfrac{(1-1+2i)(2+i)}{(4+1)}

=\dfrac{4i+2i^2}{5}

=\dfrac{-2+4i}{5} [\because~(i^2=-1)]

=\dfrac{-2}{\phantom{-}5}+\dfrac{4}{5i}\dots(1)

It is given that

\dfrac{(1+i)^2}{2-i}=x+iy

\dfrac{-2}{\phantom{-}5}+\dfrac{4}{5i}=x+iy [from (1)]

x=-\dfrac{2}{5}\text{ and } y=\dfrac{4}{5}

\therefore~(x+y)=-\dfrac{2}{5}+\dfrac{4}{5}

Thus (x+y)=\dfrac{2}{5}

 

QUESTION 14

If \left(\dfrac{1-i}{1+i}\right)^100=a+ib find (a+b)

Sol :

\dfrac{1-i}{1+i}=\dfrac{1-i}{1+i}\times\dfrac{1-i}{1-i}

=\dfrac{(1-i)^2}{1^2-i^2}

=\dfrac{1+i^2-2i}{1-i^2} 

=\dfrac{1-1-2i}{1+1} [\because~(i^2=-1)]

=\dfrac{-2i}{2}

=-i\dots(1)

It is given that

\left(\dfrac{1-i}{1+i}\right)^100=a+ib

\Rightarrow \left(-i\right)^100=a+ib [from (i)]

 \Rightarrow i^100=a+ib [\because~(-1)^n=1 \text{ if n is even }]

\Rightarrow i^{4\times 25}=a+ib

\Rightarrow 1+0i=a+ib [\because~(i4=1)\text{ and } 1^n=1]

a=1\text{ and } b=0

Thus , (a+b)=(1,0)

 

QUESTION 15

If a=cos~\theta+isin~\theta , find the value of \dfrac{1+a}{1-a}

Sol :

\dfrac{1+a}{1-a}=\dfrac{1+a}{1-a}\times\dfrac{1+a}{1+a}

=\dfrac{(1+a)^2}{1^2-a^2}

=\dfrac{1+a^2+2a}{1-a^2}

on putting value of a=cos~\theta+isin~\theta

=\dfrac{1+(cos~\theta+i~sin~\theta)^2+2(cos~\theta+i~sin~\theta)}{1-(cos~\theta+isin~\theta)^2}

=\dfrac{1+cos^2~\theta+i^2~sin^2~\theta+2i~cos~\theta ~sin~\theta+2cos~\theta+2i~sin~\theta}{1-cos^2~\theta-i^2sin^2~\theta-2icos~\theta~sin~\theta}

=\dfrac{1+cos^2~\theta-~sin^2~\theta+2i~cos~\theta ~sin~\theta+2cos~\theta+2i~sin~\theta}{1-cos^2~\theta+sin^2~\theta-2icos~\theta~sin~\theta}  [\because~(i^2=-1)]

 

QUESTION 16

Evaluate the following:

(i) 2x^3+2x^2-7x+72 when x=\dfrac{3-5i}{2}

Sol :

x=\dfrac{3-5i}{2}

\Rightarrow x^2=\left(\dfrac{3-5i}{2}\right)^2 [squaring both the sides]

x^2=\dfrac{9+25i^2-30i}{4}

x^2=\dfrac{-16-30i}{4} [\because~(i^2=-1)]

 

\Rightarrow x^3=x^2\times x^1

\Rightarrow x^3=\dfrac{-16-30i}{4}\times\dfrac{3-5i}{2}  [putting value of  x^2\text{ and } x]

\Rightarrow x^3=\dfrac{-48+80i-90i+150i^2}{8}

\Rightarrow x^3=\dfrac{-198-10i}{8}

 

\therefore 2x^3+2x^2-7x+72

\therefore 2\left(\dfrac{-198-10i}{8}\right)+2\left(\dfrac{-16-30i}{4}\right)-7\left(\dfrac{3-5i}{2}\right)+72

\therefore \left(\dfrac{-198-10i}{4}\right)+\left(\dfrac{-16-30i}{2}\right)-\left(\dfrac{21-35i}{2}\right)+72

=\dfrac{-198-10i-32-60i-42+70i+288}{4}

=\dfrac{16}{4}

= 4

 

(ii) x^4+4x^3+4x^2+8x+44 when x=3+2i

Sol :

x=3+2i

\Rightarrow x^2=(3+2i)^2

\Rightarrow x^2=9+4i^2+12i

\Rightarrow x^2=9-4+12i [\because~(i^2=-1)]

\Rightarrow x^2=5+12i

 

\Rightarrow x^3=x^2\times x 

putting the values of x^2 \text{ and } x

\Rightarrow x^3=(5+12i)\times(3+2i) 

\Rightarrow x^3=(15+10i+36i-24)

\Rightarrow x^3=-9+46i

 

\Rightarrow x^4=(x^2)^2

\Rightarrow x^4=(5+12i)^2

\Rightarrow x^4=25+144i^2+120i

\Rightarrow x^4=-119+120i

 

\therefore x^4+4x^3+4x^2+8x+44

=-119+120i-4(-9+46i)+4(5+12i)+8(3+2i)+44

=-119+120i+36-184i+20+48i+24+16i+44

= 5

 

(iii) x^4+4x^+6x^2+4x+9 when x=-1+i\sqrt{2}

Sol :

x=-1+i\sqrt{2}

\Rightarrow x^2=(-1+i\sqrt{2})^2

\Rightarrow x^2=1+2i^2-2\sqrt{2}i

\Rightarrow x^2=1-2-2\sqrt{2}i [\because~(i^2=-1)]

\Rightarrow x^2=-1-2\sqrt{2}i

 

\Rightarrow x^3=x^2\times x

[putting the values of x^2\text{ and } x]

\Rightarrow x^3=\left(-1-2\sqrt{2}i\right)\times \left(-1+i\sqrt{2}\right)

\Rightarrow x^3=1-\sqrt{2}i+2\sqrt{2}i-4i^2

\Rightarrow x^3=5+\sqrt{2}i

 

\Rightarrow x^4=(x^2)^2  [putting the value of x^2]

\Rightarrow x^4=(-1-2\sqrt{2}i)^2

\Rightarrow x^4=1+8i^2+4\sqrt{2}i

\Rightarrow x^4=-7+4\sqrt{2}i

 

\therefore x^4+4x^+6x^2+4x+9

-7+4\sqrt{2}i+20+4\sqrt{2}i-6-12\sqrt{2}i-4+4\sqrt{2}i+9

= 12

 

(iv) x^6+x^4+x^2+1 when x=\dfrac{1+i}{\sqrt{2}}

Sol :

x=\dfrac{1+i}{\sqrt{2}}

\Rightarrow x^2=\left(\dfrac{1+i}{\sqrt{2}}\right)^2 [squaring both the sides]

\Rightarrow x^2=\dfrac{1+i^2+2i}{2}

\Rightarrow x^2=\dfrac{1-1+2i}{2} [\because~(i^2=-1)]

\Rightarrow x^2=i

 

\Rightarrow x^6=(x^2)^3

\Rightarrow x^6=i^3 [putting value of x^2]

\Rightarrow x^6=-i

 

\Rightarrow x^4=(x^2)^2 

\Rightarrow x^4=(i)^2 [putting the value of x^{2}]

\Rightarrow x^4=i^2

\Rightarrow x^4=-1 [\because~(i^2=-1)]

 

\therefore x^6+x^4+x^2+1

=-i-1+i+1

= 0

 

(v) 2x^4+5x^3+7x^2-x+41 when x=-2-\sqrt{3}i

Sol :

x=-2-\sqrt{3}i

\Rightarrow x^2=(-2-\sqrt{3}i)^2

\Rightarrow x^2=4+3i^2+4\sqrt{3}i

\Rightarrow x^2=4-3+4\sqrt{3}i [\because~(i^2=-1)]

\Rightarrow x^2=1+4\sqrt{3}i

 

\Rightarrow x^3=x^2\times x

putting the value of x^2\text{ and } x

\Rightarrow x^3=(1+4\sqrt{3}i)\times(-2-\sqrt{3}i)

\Rightarrow x^3=-2-\sqrt{3}i-8\sqrt{3}i-12i^2

\Rightarrow x^3=-2-9\sqrt{3}i+12 [\because~(i^2=-1)]

\Rightarrow x^3=10-9\sqrt{3}i

 

\Rightarrow x^4=(x^2)^2

on putting the value of x^2

\Rightarrow x^4=(1+4\sqrt{3}i)^2

\Rightarrow x^4=1+48i^2+8\sqrt{3}i

\Rightarrow x^4=1-48+8\sqrt{3}i [\because~(i^2=-1)]

\Rightarrow x^4=-47+8\sqrt{3}i

 

\therefore~2x^4+5x^3+7x^2-x+41

=2(-47)+8\sqrt{3}i+5(10-9\sqrt{3}i)+7(1+4\sqrt{3}i)-(-2-\sqrt{3})+41

=-94+16\sqrt{3}i+50-45\sqrt{3}i+7+28\sqrt{3}i+2+\sqrt{3}+41

= 6

 

QUESTION 17

For a positive integer , find the value of (1-i)^n(1-\dfrac{1}{i})^n

Sol :

(1-i)^n(1-\dfrac{1}{i})^n

=(1-i)^n(1-\dfrac{i^4}{i})^n [\because~(i^4=1)]

=(1-i)^n(1-i^3)^n

=(1-i)^n(1+i)^n [\because~(i^3=-i)]

=[(1-i)(1+i)]^n

=[1-i^2]^n 

=[1+1]^n [\because~(i^2=-1)]

= 2^n

Thus , the value of (1-i)^n(1-\dfrac{1}{i})^n is = 2^n

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