Chapter 5: Complex Numbers and Quadratic Equations

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

The value of $i = \sqrt { - 1} {\text{  and }}{i^2} =  - 1$

Given expression,

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

 $\Rightarrow {(1 - i)^n}{\left( {\dfrac{{i - 1}}{i}} \right)^n}$

 $\Rightarrow {(1 - i)^n}{(i - 1)^n} \cdot {i^{ - n}} = {(1 - i)^n}{(1 - i)^n}{( - 1)^n} \cdot {i^{ - n}}$

 $\Rightarrow {\left[ {{{(1 - i)}^2}} \right]^n}{( - 1)^n} \cdot {i^{ - n}} = {\left( {1 + {i^2} - 2i} \right)^n}{( - 1)^n}{i^{ - n}}\left[ {\because {i^2} =  - 1} \right]$

 $\Rightarrow {(1 - 1 - 2i)^n}{( - 1)^n}{i^{ - n}} = {( - 2)^n} \cdot {i^n}{( - 1)^n}{i^{ - n}}$

 $\Rightarrow {( - 1)^{2n}} \cdot {2^n} = {2^n}$

 $\Rightarrow 1 \cdot {2^n} = {2^n}$ 

Ans: Given, $\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)}$ ,where $n \in N$.

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

  $\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)} ,n \in N = \sum\limits_{h = 1}^{13} {{i^n}} (1 + i)$

 $\Rightarrow (1 + i)\left[ {i + {i^2} + {i^3} + {i^4} + {i^5} + {i^6} + {i^7} + {i^8} + {i^9} + {i^{10}} + {i^{11}} + {i^{12}} + {i^{13}}} \right]$

 $\Rightarrow (1 + i)\left[ {{i^{13}}} \right]\left( {\because {i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}} = 0,{\text{ where }}n \in N{\text{ i}}{\text{.e}}{\text{., }}\sum\limits_{n = 1}^{12} {{i^n}}  = 0} \right)$

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

 $\because i = \sqrt { - 1}$

$\therefore {i^2} =  - 1$ 

 $\Rightarrow i - 1$ 

Ans: Given, ${\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^3} = x + iy$

If two complex numbers ${{\text{z}}_1} = {x_1} + i{y_1}$ and ${{\text{z}}_2} = {x_2} + i{y_2}$ are equal

  i.e.,  $ {z_1} = {z_2} \Rightarrow {x_1} + i{y_1} = {x_2} + i{y_2}$ then ${x_1} = {x_2}$ and ${y_1} = {y_2}$.

Given that, ${\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^3} = x + \dot y.....\left( i \right)$

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

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

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

$\Rightarrow{\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} =  - i.....\left( {ii} \right)$

Similarly,    ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^3} = \dfrac{{ - 1}}{i} = \dfrac{{{i^2}}}{i} = i.....\left( {iii} \right)$

Using Eqs. (ii) and (iii) in Eq. (i), we get

   $- i - i = x + iy$

 $\Rightarrow  - 2i = x + iy$ 

On comparing real and imaginary part of complex number, we get

$x = 0$ and $y =  - 2$

So,

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

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

If two complex numbers ${{\text{z}}_1} = {x_1} + i{y_1}$ and ${{\text{z}}_2} = {x_2} + i{y_2}$ are equal

 i.e.,  $ {z_1} = {z_2} \Rightarrow {x_1} + i{y_1} = {x_2} + i{y_2}$ then ${x_1} = {x_2}$ and ${y_1} = {y_2}$.

${\text{ Given that,}}$

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

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

 $\Rightarrow \dfrac{{2i}}{{2 - i}} = x + iy$

 $\Rightarrow \dfrac{{2i(2 + i)}}{{(2 - i)(2 + i)}} = x + iy$

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

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

 $\Rightarrow \dfrac{{ - 2}}{5} + \dfrac{{4i}}{5} = x + iy$ 

On comparing both sides, we get $x =  - \dfrac{2}{5}{\text{and }}y = \dfrac{4}{5}$

$\Rightarrow  x + y = \dfrac{{ - 2}}{5} + \dfrac{4}{5} = \dfrac{2}{5}$

Ans: Given, ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib,$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

 If two complex numbers ${{\text{z}}_1} = {x_1} + i{y_1}$ and ${{\text{z}}_2} = {x_2} + i{y_2}$ are equal

 i.e.,  $ {z_1} = {z_2} \Rightarrow {x_1} + i{y_1} = {x_2} + i{y_2}$ then ${x_1} = {x_2}$ and ${y_1} = {y_2}$.

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

 $\Rightarrow  {\left[ {\dfrac{{(1 - i)}}{{(1 + i)}} \cdot \dfrac{{(1 - i)}}{{(1 - i)}}} \right]^{100}} = a + ib$

 $\Rightarrow {\left( {\dfrac{{1 + {i^2} - 2i}}{{1 - {i^2}}}} \right)^{100}} = a + ib$ 

$\Rightarrow  {\left( {\dfrac{{ - 2i}}{2}} \right)^{100}} = a + ib \left[ {\because {i^2} =  - 1} \right]$

 $\Rightarrow {\left( {{i^4}} \right)^{25}} = a + ib$

 $\Rightarrow 1 = a + ib$ 

Then, $a = 1$ and $b = 0 \left[ {\because {i^4} = 1} \right]$

$\therefore (a,b) = (1,0)$

Ans: Given, $a = \cos \theta  + i\sin \theta ,$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

To solve the above problem use the trigonometric formula, $\cos \theta  = 2{\cos ^2}\dfrac{\theta }{2}- 1 = 1 - 2{\sin ^2}\dfrac{\theta }{2}$ and $\sin \theta  = 2\sin \dfrac{\theta }{2} \cdot \cos \dfrac{\theta }{2}$

Given that,$a = \cos \theta  + i\sin \theta$

$\therefore \dfrac{{1 + a}}{{1 - a}} = \dfrac{{1 + \cos \theta  + i\sin \theta }}{{1 - \cos \theta  - i\sin \theta }}$

 $\Rightarrow \dfrac{{1 + 2{{\cos }^2}\dfrac{\theta }{2} - 1 + 2i\sin \dfrac{\theta }{2} \cdot \cos \dfrac{\theta }{2}}}{{1 - 1 + 2{{\sin }^2}\dfrac{\theta }{2} - 2i\sin \dfrac{\theta }{2} \cdot \cos \dfrac{\theta }{2}}}$

 $\Rightarrow \dfrac{{2\cos \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}\left( {\sin \dfrac{\theta }{2} - i\cos \dfrac{\theta }{2}} \right)}}$ 

 $\Rightarrow  - \dfrac{{2\cos \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}} \right)}}{{2i\sin \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}} \right)}} =  - \dfrac{1}{i}\cot \dfrac{\theta }{2}$

 $\Rightarrow \dfrac{{ + {i^2}}}{i}\cot \dfrac{\theta }{2} = i\cot \dfrac{\theta }{2} \left[ {\because \dfrac{{ - 1}}{i} = \dfrac{{{i^2}}}{i}} \right]$ 

Ans: Given, $\left( {1 + i} \right)z = \left( {1 - i} \right)\bar z,$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

${\text{We have,}}$

  $(1 + i)z = (1 - i)\bar z$

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

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

 $\Rightarrow \dfrac{z}{{\bar z}} = \dfrac{{1 + {i^2} - 2i}}{{1 - {i^2}}} \left[ {\because {i^2} =  - 1} \right]$

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

 $\Rightarrow \dfrac{z}{{\bar z}} =  - i$

$\therefore z = i\bar z$ 

Hence proved.

Ans: Given, $z = x + iy$

If $z = x + iy$ it’s conjugate is $\bar z = x - iy$.

If $z = x + iy$, then show that $z\bar z + 2(z + \bar z) + b = 0$, where $b \in R$, represents

a circle.

Given that, 

$z = x + iy$

Then, $\bar z = x - iy$

Now,

$z\bar z + 2(z + \bar z) + b = 0$

$\Rightarrow  (x + iy)(x - iy) + 2(x + iy + x - iy) + b = 0$

$\Rightarrow {x^2} + {y^2} + 4x + b = 0$, which is the equation of a circle.

Ans: Given, Real part of $\dfrac{{\bar z + 2}}{{\bar z - 1}}$ is 4

Evaluate the value of $\dfrac{{\bar z + 2}}{{\bar z - 1}}$ and if it satisfies the equation of the circle the locus of the point representing $z$ in the complex plane is a circle.

Let $ z = x + iy$

Now,

$\dfrac{{\bar z + 2}}{{\bar z - 1}} = \dfrac{{x - iy + 2}}{{x - iy - 1}}$

$= \dfrac{{[(x + 2) - iy][(x - 1) + iy]}}{{[(x - 1) - iy][(x - 1) + iy]}}$

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

$= \dfrac{{(x - 1)(x + 2) + {y^2} + i[(x + 2)y - (x - 1)y]}}{{{{(x - 1)}^2} + {y^2}}} \left[ {\because  - {i^2} = 1} \right]$

${\text{Let,}}$

$\dfrac{{(x - 1)(x + 2) + {y^2}}}{{{{(x - 1)}^2} + {y^2}}} + \dfrac{{i[(x + 2)y - (x - 1)y]}}{{{{(x - 1)}^2} + {y^2}}} = x + iy$ 

Taking real part,

$\dfrac{{(x - 1)(x + 2) + {y^2}}}{{{{(x - 1)}^2} + {y^2}}} = 4$              (Given, that the real part = 4)

 $\Rightarrow {x^2} - x + 2x - 2 + {y^2} = 4\left( {{x^2} - 2x + 1 + {y^2}} \right)$

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

$\Rightarrow 3{x^2} + 3{y^2} - 9x + 6 = 0$, which represents a circle.

Hence, $z$ lies on the circle.

Ans: Given, The complex number $z$First use, $\arg \left( {\dfrac{{{z_1}}}{{{z_2}}}} \right) = \arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right)$. Also apply $\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$, where $z = x + iy$ and then use the property ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$

${\text{Let}},$

$z = x + iy$

${\text{Given that,}}$

 $\Rightarrow \arg (z - 1) - \arg (z + 1) = \dfrac{\pi }{4}$

 $\Rightarrow \arg (x + iy - 1) - \arg (x + iy + 1) = \dfrac{\pi }{4}$

 $\Rightarrow \arg (x - 1 + iy) - \arg (x + 1 + iy) = \dfrac{\pi }{4}$ 

$\Rightarrow  {\tan ^{ - 1}}\dfrac{y}{{x - 1}} - {\tan ^{ - 1}}\dfrac{y}{{x + 1}} = \dfrac{\pi }{4}$

$\Rightarrow  {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{y}{{x - 1}} - \dfrac{y}{{x + 1}}}}{{1 + \left( {\dfrac{y}{{x - 1}}} \right)\left( {\dfrac{y}{{x + 1}}} \right)}}} \right] = \dfrac{\pi }{4}$

$\Rightarrow  \dfrac{{y\left[ {\dfrac{{x + 1 - x + 1}}{{{x^2} - 1}}} \right]}}{{\dfrac{{{x^2} - 1 + {y^2}}}{{{x^2} - 1}}}} = \tan \dfrac{\pi }{4}$

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

$\Rightarrow  {x^2} + {y^2} - 1 = 2y$

$\Rightarrow  {x^2} + {y^2} - 2y - 1 = 0$, which represents a circle.