Chapter 5: Complex Numbers and Quadratic Equations

Ans: Given, $\operatorname{Re} \left( {{z^2}} \right) = 0,|z| = 2$

Let $z = x + iy$ and use the formula $|z| = \sqrt {{x^2} + {y^2}}$.

Given that, $ \operatorname{Re} \left( {{z^2}} \right) = 0,|z| = 2$

Let $ z = x + iy$

$|z| = \sqrt {{x^2} + {y^2}}$

$\therefore$$\sqrt {{x^2} + {y^2}}  = 2$                                        (On squaring both sides)

$\Rightarrow{x^2} + {y^2} = 4.....\left( i \right)$

and $\operatorname{Re} (z) = x$

Also, $ z = x + iy$

$\Rightarrow{z^2} = {x^2} + 2ixy - {y^2}$                          (On squaring both sides)

$\Rightarrow{z^2} = \left( {{x^2} - {y^2}} \right) + 2ixy$

$\Rightarrow\operatorname{Re} \left( {{z^2}} \right) = {x^2} - {y^2}\left[ {\because \operatorname{Re} \left( {{z^2}} \right) = 0} \right]$

$\Rightarrow{x^2} - {y^2} = 0.....\left( {ii} \right)$

On adding Eq. (i) and (ii), 

${x^2} + {y^2} + {x^2} - {y^2} = 4 + 0$

 $\Rightarrow 2{x^2} = 4$

 $\Rightarrow {x^2} = 2$

 $\Rightarrow x =  \pm \sqrt 2$ 

Putting the value of $x$ in equation (i), we get

$y =  \pm \sqrt 2$

$\because$ $z = x + iy$

$\Rightarrow z = \sqrt 2  \pm i\sqrt 2 , - \sqrt 2  \pm i\sqrt 2$

Ans: Given, $z + \sqrt 2 |(z + 1)| + i = 0$

Let $z = x + iy$ and use the formula $|z| = \sqrt {{x^2} + {y^2}}$. Compare the real and imaginary parts of LHS and RHS.

Given equation is $ z + \sqrt 2 |(z + 1)| + i = 0$                       …..(i)

Let $ z = x + iy$

On substituting the value of $z$ in equation (i), we get

$\Rightarrow  x + iy + \sqrt 2 |x + \dot y + 1| + i = 0$

$\Rightarrow  x + i(1 + y) + \sqrt 2 \left[ {\sqrt {{{(x + 1)}^2} + {y^2}} } \right] = 0$

$\Rightarrow  x + i(1 + y) + \sqrt 2 \sqrt {\left( {{x^2} + 2x + 1 + {y^2}} \right)}  = 0$

On comparing real and imaginary parts of LHS with RHS, we get

$\Rightarrow  x + \sqrt 2 \sqrt {{x^2} + 2x + 1 + {y^2}}  = 0$       (real part)

 $\Rightarrow {x^2} = 2\left( {{x^2} + 2x + 1 + {y^2}} \right)$

 $\Rightarrow {x^2} + 4x + 2{y^2} + 2 = 0.....\left( {ii} \right)$ 

$1 + y = 0$                                                 (imaginary part)

$\Rightarrow y =  - 1$

For $y =  - 1,$

$\Rightarrow {x^2} + 4x + 2 + 2 = 0\left[ {{\text{Using eq}}{\text{. }}\left( {ii} \right)} \right]$

 $\Rightarrow {x^2} + 4x + 4 = 0$

 $\Rightarrow {(x + 2)^2} = 0$

 $\Rightarrow x + 2 = 0$

 $\Rightarrow x =  - 2$

$\therefore z = x + iy =  - 2 - i$ 

Ans: Given, $z = \dfrac{{1 - i}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}$

Use trigonometrically identity to convert complex number into polar form.

Given that,

$z = \dfrac{{1 - i}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}\left( {{\text{Multiplying and dividing numerator by }}\sqrt 2 } \right)$

 $\Rightarrow \dfrac{{ - \sqrt 2 \left[ {\dfrac{{ - 1}}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right]}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}} = \dfrac{{ - \sqrt 2 \left[ {\cos \left( {\pi  - \dfrac{\pi }{4}} \right) + i\sin \left( {\pi  - \dfrac{\pi }{4}} \right)} \right]}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}$

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

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

$\Rightarrow  - \sqrt 2 \left[ {\cos \dfrac{{5\pi }}{{12}} + i\sin \dfrac{{5\pi }}{{12}}} \right]$

Ans: Given, $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $\arg (z) - \arg (w) = \dfrac{\pi }{2}$.

Let $z = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and $w = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$and  $|zw| = |z||w|$.

Let $ z = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and $w = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

Also, $ |zw| = |z||w| = {r_1}{r_2} = 1$

$\therefore {r_1}{r_2} = 1$

Further, $ \arg (z) = {\theta _1}$ and $\arg (w) = {\theta _2}$

${\text{But,}}$

  $\arg (z) - \arg (w) = \dfrac{\pi }{2}$

 $\Rightarrow {\theta _1} - {\theta _2} = \dfrac{\pi }{2}$

 $\Rightarrow \arg \left( {\dfrac{z}{w}} \right) = \dfrac{\pi }{2}$ 

Now, 

to prove $\bar zw =  - i$

${\text{LHS}} = \bar zw$

$= {r_1}\left( {\cos {\theta _1} - i\sin {\theta _1}} \right){r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

$= {r_1}{r_2}\left[ {\cos \left( {{\theta _2} - {\theta _1}} \right) + i\sin \left( {{\theta _2} - {\theta _1}} \right)} \right]$

 $= \left[ {{r_1}{r_2}\cos \left( {\dfrac{{ - \pi }}{2}} \right) + i\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right]$

$= 1[0 - i] =  - i = {\text{RHS}}$ 

Ans: ${\left| {a{z_1} - b{z_2}} \right|^2} + {\left| {b{z_1} + a{z_2}} \right|^2}$

$= {\left| {a{z_1}} \right|^2} + {\left| {b{z_2}} \right|^2} - 2\operatorname{Re} \left( {a{z_1}.b{{\bar z}_2}} \right) + {\left| {b{z_1}} \right|^2} + {\left| {a{z_2}} \right|^2} + 2\operatorname{Re} \left( {a{z_1}.b{{\bar z}_2}} \right)$

 $= \left( {{a^2} + {b^2}} \right){\left| {{z_1}} \right|^2} + \left( {{a^2} + {b^2}} \right){\left| {{z_2}} \right|^2}$

 $= \left( {{a^2} + {b^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)$ 

Ans:  ${\text{ }}\sqrt { - 25}  \times \sqrt { - 9}$

 $\Rightarrow i\sqrt {25}  \times i\sqrt 9  = {i^2}(5 \times 3)$

 $\Rightarrow  - 15$ 

Ans: ${\text{ }}\dfrac{{{{(1 - i)}^3}}}{{1 - {i^3}}} = \dfrac{{{{(1 - i)}^3}}}{{(1 - i)\left( {1 + i + {i^2}} \right)}}$

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

 $\Rightarrow \dfrac{{ - 2i}}{i} =  - 2$ 

Ans: $i + {i^2} + {i^3} +  \ldots$ upto $1000$ terms,

$= i + {i^2} + {i^3} + {i^4} +  \ldots {i^{1000}} = 0$

$\left[ {\because {i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}} = 0} \right.$, where $n \in N$ i.e., $\left. {\sum\limits_{n = 1}^{1000} {{i^n}}  = 0} \right]$

Ans: Multiplicative inverse of $1 + i$

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

 $\Rightarrow \dfrac{1}{2}(1 - i)$ 

Ans: Let ${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$

${z_1} + {z_2} = \left( {{x_1} + {x_2}} \right) + i\left( {{y_1} + {y_2}} \right)$ which is real.

If ${z_1} + {z_2}$ is real, then ${y_1} + {y_2} = 0$

$\Rightarrow  {y_1} =  - {y_2}$

$\because  {z_2} = {x_2} - {\dot i_1}$

$\Rightarrow {z_2} = {\bar z_1}\left[ {{\text{when }}{x_1} = {x_2}} \right]$

Ans: $\arg (z) + \arg (\bar z),(\bar z \ne 0)$

$\Rightarrow  \theta  + ( - \theta ) = 0$

Ans: Given that, $|z + 4| \leqslant 3$ For the greatest value of $|z + 1|$.

 $\Rightarrow |z + 1| = |z + 4 - 3| \leqslant |z + 4| + | - 3|$

 $\Rightarrow |z + 4 - 3| \leqslant 3 + 3$

 $\Rightarrow |z + 4 - 3| \leqslant 6$ 

So, greatest value of $|z + 1| = 6$ For, now, least value of $|z + 1|$, we know that the least value of the modulus of a complex number is zero. So, the least value of $|z + 1|$ is zero.

Ans: Given that, $\left| {\dfrac{{z - 2}}{{z + 2}}} \right| = \dfrac{\pi }{6}$

 $\Rightarrow  \dfrac{{|x + \dot y - 2|}}{{|x + iy + 2|}} = \dfrac{\pi }{6}$

 $\Rightarrow \dfrac{{|x - 2 + \dot iy|}}{{|x + 2 + \dot i|\mid }} = \dfrac{\pi }{6}$ 

$\Rightarrow  6|x - 2 + \mathop {iy}\limits^. | = \pi |x + 2 + iy|$

$\Rightarrow  6\sqrt {{{(x - 2)}^2} + {y^2}}  = \pi \sqrt {{{(x + 2)}^2} + {y^2}}$

$\Rightarrow  36\left[ {{x^2} + 4 - 4x + {y^2}} \right] = {\pi ^2}\left[ {{x^2} + 4x + 4 + {y^2}} \right]$

$\Rightarrow \left( {36 - {\pi ^2}} \right){x^2} + \left( {36 - {\pi ^2}} \right){y^2} - \left( {144 + 4{\pi ^2}} \right)x + 144 + 4{\pi ^2} = 0$, which is a circle.

Ans: Given that, $|z| = 4$ and arg $(z) = \dfrac{{5\pi }}{6}$

${\text{Let,}}$

$z = x + iy = r(\cos \theta  + i\sin \theta )$

$\left| z \right| = r = 4{\text{ and }}\arg (z) = \theta$

 $\because \tan \theta  = \dfrac{{5\pi }}{6}$

 $\Rightarrow z = 4\left( {\cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right)$

 $\Rightarrow 4\left[ {\cos \left( {\pi  - \dfrac{\pi }{6}} \right) + i\sin \left( {\pi  - \dfrac{\pi }{6}} \right)} \right]$

 $\Rightarrow 4\left[ { - \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right]$

 $\Rightarrow 4\left[ {\dfrac{{ - \sqrt 3 }}{2} + i\dfrac{1}{2}} \right] =  - 2\sqrt 3  + 2i$ 

Ans: False

We can compare two complex numbers when they are purely real. Otherwise comparison of complex number is not possible.

Ans: False

Ans: True

 Let, $z = x + iy$

 $\Rightarrow |z| + |z - 1| = \sqrt {{x^2} + {y^2}}  + \sqrt {{{(x - 1)}^2} + {y^2}}$ 

If $x = 0,y = 0$, then the value of $|z| + |z - 1| = 1$.

Ans: True

Let, $z = x + iy$

  $|z - 1| = |z - i|$ 

Then,

 $\Rightarrow |x - 1 + iy| = |x - i(1 - y)|$

 $\Rightarrow {(x - 1)^2} + {y^2} = {x^2} + {(1 - y)^2}$

 $\Rightarrow {x^2} - 2x + 1 + {y^2} = {x^2} + 1 + {y^2} - 2y$

 $\Rightarrow  - 2x + 1 = 1 - 2y$

 $\Rightarrow  - 2x + 2y = 0$

 $\Rightarrow x - y = 0$ 

Equation of a line through the points $(1,0)$ and $(0,1)$,                             …..(i)

$y - 0 = \dfrac{{1 - 0}}{{0 - 1}}(x - 1)$

$\Rightarrow y =  - (x - 1) \Rightarrow x + y = 1$                                                                           …..(ii)

which is perpendicular to the line $x - y = 0$.

Ans: False

Let $z = x + iy,z \ne 0$ and $\operatorname{Re} (z) = 0$

i.e., $ x = 0$

$\therefore  z = iy$

$\operatorname{Im} \left( {{z^2}} \right) = {i^2}{y^2} =  - {y^2}$ which is real.

Ans: True

 Given inequality, $ |z - 4| < |z - 2|$

Let $ z = x + iy$

$\therefore {|x - 4 + \dot y| < |x - 2 + iy|}$ 

 $\Rightarrow {\sqrt {{{(x - 4)}^2} + {y^2}}  < \sqrt {{{(x - 2)}^2} + {y^2}} } $

$\Rightarrow  {(x - 4)^2} + {y^2} < {(x - 2)^2} + {y^2}$

$\Rightarrow  {x^2} - 8x + 16 + {y^2} < {x^2} - 4x + 4 + {y^2}$

 $\Rightarrow  - 8x + 16 <  - 4x + 4$

 $\Rightarrow  - 8x <  - 4x - 12$

 $\Rightarrow  - 4x <  - 12$

 $\Rightarrow 4x > 12$

 $\Rightarrow  x > 3$ 

Ans: False

Let, ${z_1} = {x_1} + i{y_1} {\text{ and }} {z_2} = {x_2} + i{y_2}$

${\text{Given that,}}$

$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

 $\Rightarrow \left| {{x_1} + i{y_1} + {x_2} + i{y_2}} \right| = \left| {{x_1} + i{y_1}} \right| + \left| {{x_2} + i{y_2}} \right|$

 $\Rightarrow \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} + {{\left( {{y_1} + {y_2}} \right)}^2}}  = \sqrt {\left( {{x_1}^2 + {y_1}^2} \right)}  + \sqrt {\left( {{x_2}^2 + {y_2}^2} \right)}$ 

On squaring both sides, we get

${\left( {{x_1} + {x_2}} \right)^2} + {\left( {{y_1} + {y_2}} \right)^2} = \left( {x_1^2 + y_1^2} \right) + \left( {x_2^2 + y_2^2} \right) + 2\sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)}$

$\Rightarrow x_1^2 + x_2^2 + 2{x_1}{x_2} + y_1^2 + y_2^2 + 2{y_1}{y_2} = x_1^2 + y_1^2 + x_2^2 + y_2^2 + 2\sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)}$

 $\Rightarrow {} {2{x_1}{x_2} + 2{y_1}{y_2}}{ = 2\sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)} }$ 

 $\Rightarrow {}{{x_1}{x_2} + {y_1}{y_2}}{ = \sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)} }$ 

On squaring both sides, we get 

  $x_1^2x_2^2 + y_1^2y_2^2 + 2{x_1}{x_2}{y_1}{y_2} = x_1^2x_2^2 + y_1^2x_2^2 + x_1^2y_2^2 + y_1^2y_2^2$

 $\Rightarrow {\left( {{x_1}{y_2} - {x_2}{y_1}} \right)^2} = 0$ 

$\Rightarrow  {x_1}{y_2} = {x_2}{y_1}$

$\Rightarrow  \dfrac{{{y_1}}}{{{x_1}}} = \dfrac{{{y_2}}}{{{x_2}}}$

$\Rightarrow  \left( {\dfrac{{{y_1}}}{{{x_1}}}} \right) - \left( {\dfrac{{{y_2}}}{{{x_2}}}} \right) = 0$

$\Rightarrow  \arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right) = 0$

Ans: True

We know that, 2 is a real number.

Since, 2 is not a complex number.

Ans: (i) Given,

$z = i + \sqrt 3  = r(\cos \theta  + \sin \theta )$

$\because r\cos \theta  = \sqrt 3 ,r\sin \theta  = 1$

 $\Rightarrow {r^2} = 1 + 3 = 4\left[ {\because r > 0} \right]$

 $\Rightarrow r = 2$ 

$\Rightarrow  \tan \alpha= \left| {\dfrac{{r\sin \theta }}{{r\cos \theta }}} \right| = \dfrac{1}{{\sqrt 3 }}$

 $\Rightarrow  \tan \alpha  = \dfrac{1}{{\sqrt 3 }}$

 $\Rightarrow \alpha  = \dfrac{\pi }{6}$ 

$\because x > 0, y > 0$

and $ \arg (z) = \theta  = \dfrac{\pi }{6}$

So the polar form of $z$ is $2\left( {\cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right)$.

(ii) Given that,

$z =  - 1 + \sqrt { - 3}  =  - 1 + i\sqrt 3$

$\therefore  \tan \alpha= \left| {\dfrac{{\sqrt 3 }}{{ - 1}}} \right| = \sqrt 3$

\[ \Rightarrow \tan \alpha  = \tan \dfrac{\pi }{3}\]

 $\Rightarrow \alpha  = \dfrac{\pi }{3}$

  $\because x < 0,y > 0$

 $\Rightarrow \theta  = \pi  - \alpha$

$= \pi  - \dfrac{\pi }{3} = \dfrac{{2\pi }}{3}$ 

(iii) Given that,

$|z + 2| = |z - 2|$

$\Rightarrow  |x + 2 + \dot r| = |x - 2 + iy|$

$\Rightarrow  {(x + 2)^2} + {y^2} = {(x - 2)^2} + {y^2}$

 $\Rightarrow  {x^2} + 4x + 4 = {x^2} - 4x + 4$

 $\Rightarrow 8x = 0$

$\therefore x = 0$ 

It is a straight line which is a perpendicular bisector of segment joining the points $( - 2,0)$ and $(2,0)$.

(iv) Given that,

$|z + 2i| = |z - 2i|$

$\Rightarrow  |x + i(y + 2)| = |x + i(y - 2)|$

$\Rightarrow {x^2} + {(y + 2)^2} = {x^2} + {(y - 2)^2}$

 $\Rightarrow  4y = 0$

 $\Rightarrow y = 0$ 

It is a straight line, which is a perpendicular bisector of segment joining $(0, - 2)$ and $(0,2)$.

(v) Given that,

  $|z + 4i| \geqslant 3 = |x + iy + 4i| \geqslant 3$

 $\Rightarrow |x + i(y + 4)| \geqslant 3$

 $\Rightarrow \sqrt {{x^2} + {{(y + 4)}^2}}  \geqslant 3$

 $\Rightarrow {x^2} + {(y + 4)^2} \geqslant 9$

 $\Rightarrow {x^2} + {y^2} + 8y + 16 \geqslant 9$

 $\Rightarrow {x^2} + {y^2} + 8y + 7 \geqslant 0$ 

Which represents a circle. On or outside having center $(0, - 4)$ and radius 3.

(vi) Given that, 

$|z + 4| \leqslant 3$

 $\Rightarrow |x + iy + 4| \leqslant 3$

 $\Rightarrow |x + 4 + iy| \leqslant 3$

 $\Rightarrow \sqrt {{{(x + 4)}^2} + {y^2}}  \leqslant 3$ 

$\Rightarrow {(x + 4)^2} + {y^2} \leqslant 9$

$\Rightarrow  {x^2} + 8x + 16 + {y^2} \leqslant 9$

$\Rightarrow  {x^2} + 8x + {y^2} + 7 \leqslant 0$

It represents the region which is on or inside the circle having the center $( - 4,0)$ and radius 3.

(vii) Given that,

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

 $\Rightarrow \dfrac{{1 + 2i + i + 2{i^2}}}{{1 - {i^2}}} = \dfrac{{1 - 2 + 3i}}{{1 + 1}}$

 $\Rightarrow \dfrac{{ - 1 + 3i}}{2}$

$\therefore \bar z = \dfrac{{ - 1}}{2} - \dfrac{{3i}}{2}$ 

Hence, $\left( {\dfrac{{ - 1}}{2},\dfrac{{ - 3}}{2}} \right)$ lies in the third quadrant.

(viii) Given that,

  $z = 1 - i$

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

 $\Rightarrow \dfrac{{1 + i}}{{\left( {1 - i} \right)\left( {1 + i} \right)}} = \dfrac{{1 + i}}{{1 - {i^2}}}$

 $\Rightarrow \dfrac{1}{2}\left( {1 + i} \right)$ 

Hence, $\left( {\dfrac{1}{2},\dfrac{1}{2}} \right)$ lies in first quadrant.

Ans: Given, $\dfrac{{2 - i}}{{{{\left( {1 - 2i} \right)}^2}}}$

If $z = x + iy{\text{ and }}\bar z = x - iy$.

Given that,

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

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

 $\Rightarrow \dfrac{{2 - i}}{{ - 3 - 4i}} = \dfrac{{(2 - i)}}{{ - (3 + 4i)}}$

 $\Rightarrow  - \left[ {\dfrac{{(2 - i)(3 - 4i)}}{{(3 + 4i)(3 - 4i)}}} \right] =  - \left( {\dfrac{{6 - 8i - 3i + 4{i^2}}}{{9 + 16}}} \right)$

 $\Rightarrow  - \dfrac{{( - 11i + 2)}}{{25}} = \dfrac{{ - 1}}{{25}}(2 - 11i)$

 $\Rightarrow z = \dfrac{1}{{25}}( - 2 + 11i)$

 $\Rightarrow \bar z = \dfrac{1}{{25}}( - 2 - 11i) = \dfrac{{ - 2}}{{25}} - \dfrac{{11}}{{25}}i$ 

Ans: Given, $\left| {{z_1}} \right| = \left| {{z_2}} \right|$

If $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

Given that,

$\left| {{z_1}} \right| = \left| {{z_2}} \right|$

${\text{Let,}}$

${z_1} = {x_1} + i{y_1}{\text{ and }}{z_2} = {x_2} + i{y_2}$

 $\Rightarrow \left| {{x_1} + i{y_1}} \right| = \left| {{x_2} + i{y_2}} \right|$

 $\Rightarrow x_1^2 + y_1^2 = x_2^2 + y_2^2$

 $\Rightarrow x_1^2 = x_2^2,y_1^2 = y_2^2$

 $\Rightarrow {x_1} =  \pm {x_2},{y_1} =  \pm {y_2}$

 $\Rightarrow {z_1} = {x_1} + i{y_1}{\text{ or }}{z_1} =  \pm {x_2} \pm i{y_2}$ 

Hence, it is not necessary that ${z_1} = {z_2}$.

Ans: Given, $\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{2a - i}} = x + iy$

Rationalize the given expression and compare the real and imaginary parts.

Given that,

$\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{2a - i}} = x + iy$

 $\Rightarrow \dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{2a - i}} \times \dfrac{{2a + i}}{{2a + i}} = x + iy$

 $\Rightarrow \dfrac{{{{\left( {{a^2} + 1} \right)}^2}\left( {2a + i} \right)}}{{4{a^2} + 1}} = x + iy$

 $\Rightarrow \dfrac{{2a{{\left( {{a^2} + 1} \right)}^2} + i{{\left( {a + 1} \right)}^2}}}{{4{a^2} + 1}} = x + iy$ 

On comparing the real and imaginary parts, we get

 $\Rightarrow x = \dfrac{{2a{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}{\text{  and  }}y = \dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}$

$\therefore {x^2} + {y^2} = 4{a^2}{\left[ {\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}} \right]^2} + {\left[ {\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}} \right]^2}$

$= \dfrac{{\left( {4{a^2} + 1} \right){{\left( {{a^2} + 1} \right)}^4}}}{{{{\left( {4{a^2} + 1} \right)}^2}}} = \dfrac{{{{\left( {{a^2} + 1} \right)}^4}}}{{\left( {4{a^2} + 1} \right)}}$