Class 11 Maths Ncert Solutions Chapter 5 Ex 5.3 Complex Numbers & Quadratic Equations PDF

# Class 11 Maths Ncert Solutions Ex 5.3

## Class 11 Maths Ncert Solutions Chapter 5 Ex 5.3

Q-1: Evaluate the following:

[i18+(1i)25]3$[i^{18} + (\frac{1}{i})^{25}]^{3}$

Sol:

[i18+(1i)25]3$[i^{18} + (\frac{1}{i})^{25}]^{3}$  =  [i4×4+2+1i4×6+1]3$[i^{4\times4 + 2} + \frac{1}{i^{4\times6 + 1}}]^{3}$  =  [(i4)4i2+1(i6)4i]3$[(i^{4})^{4}\cdot i^{2} + \frac{1}{(i^{6})^{4}\cdot i}]^{3}\\$  =  [i2+1i]3$[i^{2} + \frac{1}{i}]^{3}$   [Since, i4 = 1]

=[1+1i×ii]3=[1+ii2]3=[1i]3$[-1 + \frac{1}{i}\times\frac{i}{i}]^{3} = [-1 + \frac{i}{i^{2}}]^{3} = [-1 – i]^{3}$

=(1)3[1+i]3=[1+i3+3i(i+1)]$(-1)^{3}[1 + i]^{3} = -[1 + i^{3} + 3i(i + 1)]$

=[1i3+3i+3i2]$-[1 – i^{3} + 3i + 3i^{2}]$

= – [1 – i + 3i – 3] = – [-2 + 2i]

= 2 – 2i

Q-2: For the two given complex number z1 and z2 below prove that

Re(z1z2) = Re z1Re z2 – Im z1 Im z2

Sol:

Assuming z1 = a1 + ib1 and z2 = a2 + ib2

Therefore, z1z2 = (a1 + ib1) ( a2 + ib2)

= a1(a2 + ib2) + ib2(a2 + ib2)

= a1a2 + ia1b2 + ib1a2 + i2b1b2

= a1a2 + ia1b2 + ib1a2 – b1b2

= (a1a2 – b1b2) + i(a1b2 + b1a2)

Re(z1z2) = a1a2 – b1b2

Re(z1z2) = Rez1 Rez2  – Imz1 Imz2

Hence, Proved

Q-3: Express (114i21+i)(34i5+i)$\left ( \frac{1}{1 – 4i} – \frac{2}{1 + i} \right )\;\left ( \frac{3 – 4i}{5 + i} \right )$ in the standard form i.e. a + ib.

Sol:

[114i21+i][34i5+i]=[(1+i)2(14i)(14i)(1+i)][34i5+i]$\left [ \frac{1}{1 – 4i} – \frac{2}{1 + i} \right ]\;\left [ \frac{3 – 4i}{5 + i} \right ] = \left [ \frac{(1 + i) -2(1 – 4i)}{(1- 4i)(1 + i)} \right ]\left [ \frac{3 – 4i}{5 + i} \right ]$ =[1+i2+8i1+i4i4i2][34i5+i]=[1+9i53i][34i5+i]$= \left [ \frac{1 + i – 2 + 8i}{1 + i – 4i -4i^{2}} \right ]\left [ \frac{3 – 4i}{5 + i} \right ] = \left [ \frac{-1 + 9i}{5 – 3i} \right ]\left [ \frac{3 – 4i}{5 + i} \right ]$ =[3+4i+27i36i225+5i15i3i2]=33+31i2810i=33+31i2(145i)$= \left [ \frac{-3 + 4i + 27i – 36i^{2}}{25 + 5i – 15i – 3i^{2}} \right ] = \frac{33 + 31i}{28 – 10i} = \frac{33 + 31i}{2(14 – 5i)}$ =33+31i2(145i)×14+5i14+5i$= \frac{33 + 31i}{2(14 – 5i)}\times\frac{14 + 5i}{14 + 5i}$

Now, on multiplying and dividing by (14 + 5i) we will get:

=462+165i+434i+155i22[(14)2(5i)2]=307+599i2(19625i2)$=\frac{462 + 165i + 434i + 155i^{2}}{2[(14)^{2} – (5i)^{2}]} = \frac{307 + 599i}{2(196 – 25i^{2})}$ =307+599i2(221)=307+599i442$=\frac{307 + 599i}{2(221)} = \frac{307 + 599i}{442}$ =307442+i599442$=\frac{307}{442} + i\frac{599}{442}$

Q-4: If aib=xiyuiv$a- ib =\sqrt{\frac{x \;-\; iy}{u \;- \;iv}}$ then prove that (a2+b2)2=x2+y2u2+v2$(a^{2} + b^{2})^{2} = \frac{x^{2} \;+ \;y^{2}}{u^{2} \;+\; v^{2}}$

Sol:

aib=xiyuiv$a – ib =\sqrt{\frac{x \;-\; iy}{u \;- \;iv}}$ =xiyuiv×x+iyu+iv$= \sqrt{\frac{x – iy}{u – iv}\times\frac{x + iy}{u + iv}}\\$

Now, on multiplying and dividing by (u + iv) we will get:

=(xu+yv)+i(xvyu)u2+v2$\\\sqrt{\frac{(xu\; +\; yv) \;+\; i\;(xv \;- \;yu)}{u^{2} \;+\; v^{2}}}$

Therefore, (aib)2=(xu+yv)+i(xvyu)u2+v2$(a – ib)^{2} = \frac{(xu \;+\; yv) \;+ \;i(xv\; – \;yu)}{u^{2}\; + \;v^{2}}\\$

a22iabb2=(xu+yv)+i(xvyu)u2+v2$\\\Rightarrow a^{2} – 2iab – b^{2} = \frac{(xu \;+ \;yv) \;+\;i(xv \;-\; yu)}{u^{2} \;+\; v^{2}}$

On comparing both sides, we will get:

a2b2=xu+yvu2+v2and2ab=xvyuu2+v2$a^{2} – b^{2} = \frac{xu \;+\; yv}{u^{2} \;+\; v^{2}}\;and\; -2ab = \frac{xv\; -\; yu}{u^{2} + v^{2}}$……………..(a)

(a2 + b2)2 = (a2 – b2)2 + 4a2b2

=(xu+yvu2+v2)2+(xvuu2+v2)2$\\(\frac{xu \;+\; yv}{u^{2} \;+\; v^{2}})^{2} +(\frac{xv \;- \;u}{u^{2}\; + \;v^{2}})^{2}\\$

Now, by using equation (a) we will get:

=x2u2+y2v2+2xyuv+x2v2+y2u22xyuv(u2+v2)2$= \frac{x^{2}u^{2}\; +\; y^{2}v^{2} \;+\; 2xyuv \;+\; x^{2}v^{2}\; +\; y^{2}u^{2} \;- \;2xyuv}{(u^{2} \;+\; v^{2})^{2}}$ =x2u2+y2v2+x2v2+y2u2(u2+v2)2$=\frac{x^{2}u^{2}\; +\; y^{2}v^{2}\; +\; x^{2}v^{2} \;+\; y^{2}u^{2}}{(u^{2} \;+\; v^{2})^{2}}$ =x2(u2+v2)+y2(u2+v2)2(u2+v2)2$= \frac{x^{2}(u^{2}\; +\; v^{2}) \;+ \;y^{2}(u^{2}\; +\; v^{2})^{2}}{(u^{2}\; + \;v^{2})^{2}}$ =(x2+y2)(u2+v2)(u2+v2)2=x2+y2u2+v2$= \frac{(x^{2} \;+ \;y^{2})\;(u^{2} \;+\; v^{2})}{(u^{2} \;+ \;v^{2})^{2}} = \frac{x^{2} \;+ \;y^{2}}{u^{2}\; +\; v^{2}}\\$

Hence, proved

Q-5) Convert the following complex number in polar form:

1: 1+7i(2i)2$\frac{1 + 7i}{(2 – i)^{2}}\\$

2:1+3i12i$\\\frac{1 + 3i}{1 – 2i}$

Sol:

1: Here, z = 1+7i(2i)2$\frac{1 + 7i}{(2 – i)^{2}}$

1+7i(2i)2=1+7i4+i24i=1+7i414i$\frac{1 + 7i}{(2 – i)^{2}} = \frac{1 + 7i}{4 + i^{2} – 4i} = \frac{1 + 7i}{4 – 1 – 4i}\\$

=1+7i34i×3+4i3+4i=3+4i+21i+28i232+42$\\\frac{1 + 7i}{3 – 4i}\times\frac{3 + 4i}{3 + 4i} = \frac{3 + 4i + 21i + 28i^{2}}{3^{2} + 4^{2}}\\$

=3+4i+21i2832+42=25+25i25$\\\frac{3 + 4i + 21i – 28}{3^{2} + 4^{2}} = \frac{-25 + 25i}{25}$ = -i + 1

Assuming rcosΘ=1andrsinΘ=1$r\cos\Theta = -1 \;and\; r\sin\Theta = 1$

On squaring on both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$

Therefore, Modulus = 2$\sqrt{2}$        [ Since, r > 0]

Now, 2cosΘ=1and2sinΘ=1cosΘ=12andsinΘ=12$\sqrt{2}\cos\Theta = -1 \;and\; \sqrt{2}\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}$

Here, Θ$\Theta$ lies in 2nd quadrant.

Therefore, Θ$\Theta$ = (ππ4)=3π4$(\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$

Therefore, the required polar form is:

1+i=rcosΘ+irsinΘ=2cos(3π4)+i2sin(3π4)$-1 + i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(\frac{3\pi}{4}) + i\sqrt{2}\sin(\frac{3\pi}{4})\\$

=2[cos(3π4)+isin(3π4)]$\\ \sqrt{2}[\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})]$

2: Here, z = 1+3i12i$\frac{1 \;+ \;3i}{1\; -\; 2i}$

1+3i12i=1+3i12i×1+2i1+2i$\frac{1 \;+ \;3i}{1 \;- \;2i} = \frac{1 \;+ \;3i}{1 \;- \;2i}\times\frac{1 \;+\; 2i}{1\; + \;2i}\\$

=1+2i+3i61+4=5+5i5$\\ \frac{1 \;+ 2i \;+ \;3i\; – 6}{1\; + \;4} = \frac{-5 \;+\; 5i}{5}$ = -i + 1

Assuming rcosΘ=1andrsinΘ=1$r\cos\Theta = -1 \;and\; r\sin\Theta = 1$

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1)2+(1)2r2(cos2Θ+sin2Θ)=1+1r2(1)=2r=2$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$

Therefore, Modulus = 2$\sqrt{2}$        [ Since, r > 0]

Now,2cosΘ=1and2sinΘ=1$\sqrt{2}\cos\Theta = -1 \;and\; \sqrt{2}\sin\Theta = 1$

cosΘ=12andsinΘ=12$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}$

Here, Θ$\Theta$ lies in 2nd quadrant.

Now, Θ$\Theta$ = (ππ4)=3π4$(\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$

Therefore, the required polar form is:

1+i=rcosΘ+irsinΘ=2cos(3π4)+i2sin(3π4)$-1 + i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(\frac{3\pi}{4}) + i\sqrt{2}\sin(\frac{3\pi}{4})\\$

=2[cos(3π4)+isin(3π4)]$\\ \sqrt{2}\;[\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})]$

Q-6: Solve the equation given below:

3y2 – 4y + 203$\frac{20}{3}$ = 0

Sol:

Here, 3y2 – 4y + 203$\frac{20}{3}$ = 0

The given equation can be also be written as,

9y2 – 12y + 20 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 9, b = -12 and c = 20

Now, the discriminant is:

D=b24ac=(12)24×9×20=144720=576$D = b^{2} – 4ac = (-12)^{2} – 4\times 9\times 20 = 144 – 720 = -576$

Now, the solution is:

y=b±D2a$y=\frac{-b \;\pm \;\sqrt{D}}{2a}$ =(12)±5762(9)$= \frac{-(12) \;\pm \;\sqrt{-576}}{2(9)}$ =12±576i218$= \frac{12\; \pm \;\sqrt{576i^{2}}}{18}\\$

=12±24i18=2±4i3$\\ \frac{12 \;\pm \;24\;i}{18}= \frac{2 \;\pm \; 4\;i}{3}$

=23±43i$=\frac{2}{3}\; \pm \;\frac{4}{3}\;i$

Therefore, y = =23±43i$\boldsymbol{= \frac{2}{3}\; \pm \;\frac{4}{3}\;i}$

Q-7: Solve the equation given below:

y2 – 2y + 32$\frac{3}{2}$ = 0

Sol:

Here, y2 – 2y + 32$\frac{3}{2}$= 0

The given equation can be also be written as:

2y2 – 4y + 3 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 2, b = -4 and c = 3

Now, the discriminant is:

D=b24ac=(4)24×(2)×(3)=1624=8$D = b^{2} – 4ac = (-4)^{2} – 4\times (2)\times (3) = 16 – 24 = -8$

Now, the solution is:

y=b±D2a$y = \frac{-b \;\pm \;\sqrt{D}}{2a}$ =(4)±82(2)$= \frac{-(-4) \;\pm\;\sqrt{-8}}{2(2)}$ =4±8i24$= \frac{4 \;\pm \;\sqrt{8i^{2}}}{4}\\$

=4±22i4=2±2i2$\\\frac{4 \;\pm \;2\sqrt{2}\;i}{4}= \frac{2 \;\pm \;\sqrt{2}\;i}{2}$

=1±22i$= 1 \;\pm \;\frac{\sqrt{2}}{2}i$

Therefore, y =1±22i$\boldsymbol{= 1 \;\pm \;\frac{\sqrt{2}}{2}i}$

Q-8: Solve the equation given below:

27y2 – 10y + 1 = 0

Sol:

Here, 27y2 – 10y + 1 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 27, b = -10 and c = 1

Now, the discriminant is:

D=b24ac=(10)24×(27)×(1)=100108=8$D = b^{2} – 4ac = (-10)^{2} – 4\times (27)\times (1) = 100 – 108 = -8$

Now, the solution is:

y=b±D2a$y = \frac{-b \;\pm \;\sqrt{D}}{2a}$ =(10)±82(27)$= \frac{-(-10)\; \pm \;\sqrt{-8}}{2(27)}$ =10±8i254$= \frac{10 \;\pm \;\sqrt{8i^{2}}}{54}$

=10±22i54=5±2i27$\\\frac{10 \;\pm \;2\sqrt{2}\;i}{54} = \frac{5 \pm \sqrt{2}\;i}{27}$

=57±227i$= \frac{5}{7}\; \pm \;\frac{\sqrt{2}}{27}\;i\\$

Therefore, y =57±227i$\boldsymbol{= \frac{5}{7} \;\pm \;\frac{\sqrt{2}}{27}\;i}$

Q-9: Solve the equation given below:

21y2 – 28y + 10 = 0

Sol:

Here, 21y2 – 28y + 10 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 21, b = -28 and c = 10

Now, the discriminant is:

D=b24ac=(28)24×(21)×(10)=784840=56$D = b^{2} – 4ac = (-28)^{2} – 4\times (21)\times (10) = 784 – 840 = -56$

Now, the solution is:

Since, y=b±D2a$y = \frac{-b \;\pm \;\sqrt{D}}{2a}$

=(28)±562(21)$= \frac{-(-28) \;\pm \;\sqrt{-56}}{2(21)}$ =28±56i242$= \frac{28 \;\pm \;\sqrt{56i^{2}}}{42}\\$

=28±214i42=14±14i21$\\\frac{28 \;\pm \;2\sqrt{14}\;i}{42} = \frac{14 \;\pm \;\sqrt{14}\;i}{21}$=32±1421i$= \frac{3}{2} \;\pm \;\frac{\sqrt{14}}{21}\;i$

Therefore, y =32±1421i$\boldsymbol{= \frac{3}{2} \;\pm \;\frac{\sqrt{14}}{21}\;i}$

Q-10:  z1 = 2 – i, z2 = 1 + i, find z1+z2+1z1z2+i$\left | \frac{z_{1} + z_{2} + 1}{z_{1} – z_{2} + i} \right |$

Sol:

Here, z1 = 2 – i, z2 = 1 + i

z1+z2+1z1z2+i=(2i)+(1+i)+1(2i)(1+i)+1$\left | \frac{z_{1} + z_{2} + 1}{z_{1} – z_{2} + i} \right | = \left | \frac{(2 – i) + ( 1 + i) + 1}{(2 – i) – ( 1 + i) + 1} \right |\\$

=422i=21i=21i×1+i1+i=2(1+i)1i2$\\ \left | \frac{4}{2 – 2i} \right | = \left | \frac{2}{1 – i} \right |= \left | \frac{2}{1 – i}\times\frac{1 + i}{1 + i} \right | = \left | \frac{2(1 + i)}{1 – i^{2}} \right |\\$

=2(1+i)1+1=2(1+i)2=|1+i|=12+12=2$\\ \left | \frac{2(1 + i)}{1 + 1} \right | = \left | \frac{2(1 + i)}{2} \right |= \left | 1 + i \right | = \sqrt{1^{2} + 1^{2} } = \sqrt{2}$

Q-11:  If x + iy = (a+i)22a2+1$\frac{(a + i)^{2}}{2a^{2} + 1}$ then prove that

x2 + y2(a2+1)2(2a+1)2$\frac{(a^{2} + 1)^{2}}{(2a + 1)^{2}}$

Sol:

Here, x + iy = (a+i)22a2+1$\frac{(a + i)^{2}}{2a^{2} + 1}\\$

=a2+i2+2ai2a2+1=a21+2ai2a2+1=a212a2+1+i2a2a2+1$\\\frac{a^{2} + i^{2} + 2ai}{2a^{2} + 1} = \frac{a^{2} – 1 + 2ai}{2a^{2} + 1} = \frac{a^{2} – 1}{2a^{2} + 1} + i\frac{2a}{2a^{2} + 1}\\$

On comparing both sides, we will get:

x=a212a2+1andy=2a2a2+1$x = \frac{a^{2} – 1}{2a^{2} + 1}\;and\; y = \frac{2a}{2a^{2} + 1}\\$

Therefore, x2+y2=(a212a2+1)2+(2a2a2+1)2$x^{2} + y^{2} = (\frac{a^{2} – 1}{2a^{2} + 1})^{2} + (\frac{2a}{2a^{2} + 1})^{2}\\$

=a4+12a2+4a2(2a+1)2=a4+1+2a2(2a2+1)2$= \frac{a^{4} + 1 – 2a^{2} + 4a^{2}}{(2a + 1)^{2}} = \frac{a^{4} + 1 + 2a^{2}}{(2a^{2} + 1)^{2}}$ =(a2+1)2(2a2+1)2$= \frac{(a^{2} + 1)^{2}}{(2a^{2} + 1)^{2}}\\$

Hence, proved

Q-12: If z1 = 2 – i and z2 = -2 + i is given find out,

1: Im(1z1z1¯¯¯¯¯)$Im(\frac{1}{z_{1}\overline{z_{1}}})$

2:Re(z1z2z1)$Re(\frac{z_{1}z_{2}}{z_{1}})$

Sol:

Here, z1 = 2 – i and z2 = -2 + i

1. Im(1z1z1¯¯¯¯¯)$Im(\frac{1}{z_{1}\overline{z_{1}}})$

=1(2+i)(2+i)=122i2=14+1=15$= \frac{1}{(2 + i)(-2 + i)} = \frac{1}{2^{2} – i^{2}} \\ = \frac{1}{4 + 1} = \frac{1}{5}$

On comparing both sides, we will get:

Im(1z1z1¯¯¯¯¯)=0$Im(\frac{1}{z_{1}\overline{z_{1}}}) = 0\\$

2. Re(z1z2z1)$Re(\frac{z_{1}z_{2}}{z_{1}})$

z1z2 = (2 – i)(- 2 + i) = -4 + 2i + 2i – i2 = -4 + 4i – (-1) = -3 + 4i

z1¯¯¯¯¯=2+i$\overline{z_{1}} = 2 + i\\$ z1z2z1¯¯¯¯¯=3+4i2+i$\frac{z_{1}z_{2}}{\overline{z_{1}}} = \frac{-3 + 4i}{2 + i}$

Now multiplying and dividing it by (2 – i), we will get:

=3+4i2+i×2i2i$\frac{-3 + 4i}{2 + i}\times\frac{2 – i}{2 – i}$

=6+3i+8i4i222i2$\frac{-6 + 3i + 8i – 4i^{2}}{2^{2} – i^{2}}$

=6+11i4(1)4+1$\frac{-6 + 11i – 4(-1)}{4 + 1}$

=2+11i5=25+115i$\frac{2 + 11i}{5} = \frac{-2}{5} + \frac{11}{5}i$

On comparing both the sides, we will get:

Re(z1z2z1¯¯¯¯¯)=25$\boldsymbol{Re(\frac{z_{1}\;z_{2}}{\overline{z_{1}}}) = \frac{-2}{5}}$

Q-13: Find out the modulus and argument of the given complex number

1+2i13i$\frac{1 \;+\;2i}{1\; -\; 3i}$

Sol:

Here, z = 1+2i13i$\frac{1 + 2i}{1 – 3i}$

1+2i13i×1+3i1+3i=1+3i+2i+6i2129i2$\frac{1\; +\; 2i}{1\; -\; 3i}\times\frac{1\; +\; 3i}{1 \;+ \;3i} = \frac{1 \;+\; 3i \;+\; 2i \;+ \;6i^{2}}{1^{2} \;- \;9i^{2}}\\$

=1+3i+2i+6i212+9=5+5i10$\\\frac{1\; +\; 3i \;+\; 2i \;+ \;6i^{2}}{1^{2} +\; 9} = \frac{-5\; +\; 5i}{10}$

=510+510i=12+12i$\frac{-5}{10}\; +\; \frac{5}{10}i = \frac{-1}{2}\; +\; \frac{1}{2}i$

Assuming rcosΘ=1/2andrsinΘ=1/2$r\cos\Theta = -1/2 \;and\; r\sin\Theta = 1/2$

On squaring both sides and then adding them we will get:

(rcosΘ)2+(rsinΘ)2=(1/2)2+(1/2)2r2(cos2Θ+sin2Θ)=1/4+1/4r2(1)=1/2r=12$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1/2)^{2} + (1/2)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1/4 + 1/4 \\ \Rightarrow r^{2}(1) = 1/2 \\ \Rightarrow r = \frac{1}{\sqrt{2}}\\$

Therefore, Modulus = 12$\frac{1}{\sqrt{2}}$          [ Since, r > 0]

Now, 12cosΘ=12and12sinΘ=12$\frac{1}{\sqrt{2}}\cos\Theta = \frac{-1}{2} \;and\; \frac{1}{\sqrt{2}}\sin\Theta = \frac{1}{2}\\$

cosΘ=12andsinΘ=12$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}$

Here, Θ$\Theta$ lies in 2nd  quadrant.

Therefore, Θ=(ππ4)=3π4$\Theta = (\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$

Q-14: Find the real number a and b if (3 + 5i) (a – ib) is the conjugate if -6 – 24i.

Sol:

Assuming z = (3 + 5i) (a – ib)

z = 3a + 5ai – 3bi – 5bi2 = 3a + 5ai – 3bi + 5b = (3a + 5b) + i(5a – 3b)

Therefore, z¯¯¯=(3a+5b)i(5a3b)$\overline{z} = (3a + 5b) – i(5a – 3b)$

Here, given that z¯¯¯=624i$\overline{z} = -6 – 24i$

Therefore, (3a + 5b) – i(5a – 3b) = -6 – 24i

On comparing both sides, we will get:

3a + 5b = -6 . . . . . . . . . . . . (a)

5a – 3b = 24 . . . . . . . . . . . . (b)

On solving these two equations, we will get:

9a + 15b = -18

25a – 15b = 120

34a = 102

Therefore, a = 3

Now, 3(3) + 5b = -6       [From equation (a)]

5b = -15

Therefore, b = -3

Hence, a = 3 and b = -3

Q-15: Find the modulus of 1+i1i1i1+i$\frac{1\; +\; i}{1\; – \;i}\; -\; \frac{1\; -\; i}{1 \;+\; i}$

Sol:

1+i1i1i1+i=(1+i)2(1i)2(1i)(1+i)$\frac{1 \;+\; i}{1\; -\; i} – \frac{1\; -\; i}{1\; +\; i} = \frac{(1 \;+\; i)^{2} – (1\; – \;i)^{2}}{(1 \;- \;i)(1\; +\; i)}\\$

=1+i2+2i1i2+2i1+1=4i2=2i$\\ \frac{1 \;+ \;i^{2} \;+\; 2i \;-\; 1\; -\; i^{2}\; +\;2i}{1 \;+\; 1} = \frac{4i}{2} = 2i\\$

$\\\Rightarrow$   1+i1i1i1+i=|2i|=22=2$\left | \frac{1\; +\; i}{1\; -\; i} \;- \;\frac{1\; -\; i}{1 \;+\; i} \right | = \left | 2i \right | = \sqrt{2^{2}} = 2$

Q-16: If (a + ib)3 = u + iv, then prove that:

ua+vb=4(a2b2)$\frac{u}{a} + \frac{v}{b} = 4(a^{2} – b^{2})$

Sol:

Here, (a + ib)3 = u + iv

a3 + (ib)3 + 3a(ib)(a + ib) = u + iv

a3 + i3b3 + 3a2bi + 3ab2i2 = u + iv

a3 – ib3 + 3a2bi – 3ab2 = u + iv

(a3 – 3ab2) + i(3a2b – b3) = u + iv

On comparing both the sides, we will get:

u = a3 – 3ab2     and      v = 3a2b – b3

Hence, Proved

Q-17: If A and B are two different complex numbers with |B| = 1, then find BA1A¯¯¯¯B$\left | \frac{B – A}{1 – \overline{A}B} \right |$

Sol:

Assuming, A = x + iy and B = u + iv

Here, |B| = 1,

Therefore, u2+v2=1$\sqrt{u^{2} + v^{2}} = 1\\$

$\\\Rightarrow$ u2 + v2 = 1 . . . . . . . . . . . . (a)

BA1A¯¯¯¯B=(x+iy)(u+iv)1(uiv)(x+iy)$\\\left | \frac{B – A}{1\; – \;\overline{A}B} \right | \;= \;\left | \frac{(x \;+\; iy) \;-\; (u \;+\; iv)}{1 – (u\; – \;iv)(x \;+\; iy)} \right |\\$

=(xu)+i(yv)1(xu+uiy+ivx+yv)$\\ \left | \frac{(x \;- \;u) \;+ \;i(y\; -\; v)}{1 \;- \;(xu \;+\; uiy \;+ \;ivx \;+ \;yv)} \right |\\$

=(xu)+i(yv)(1xuyv)+i(xvyu)=|(xv)+i(yv)||(1xuyv)+i(xvyu)|$\\ \left | \frac{(x\; – \;u) + i(y \;- \;v)}{(1 \;- \;xu\; – \;yv) \;+\; i(xv \;-\; yu)} \right | = \frac{\left | (x \;-\; v) \;+\; i(y\; -\; v) \right |}{\left | (1 \;- \;xu \;-\; yv) \;+\; i(xv\; – \;yu) \right |}\\$

Since, z1z2=|z1||z2|$\\\left | \frac{z_{1}}{z_{2}} \right | = \frac{\left | z_{1} \right |}{\left | z_{2} \right |}\\$

Therefore, =(xu)2+(yv)2(1xuyv)+i(xvuy)$\\= \frac{\sqrt{(x \;-\; u)^{2} \;+\; (y\; – \;v)^{2} }}{\sqrt{(1 \;- \;xu\; -\; yv)+ i(xv\; -\; uy)}}\\$

=