Chapter 5: Complex Numbers and Quadratic Equations

Ans: Given, $\left| z \right| = z + 1 + 2i$

Simplify $\left| z \right| = z + 1 + 2i$ by using the formula $|z| + \sqrt {{x^2} = {y^2}}$.

The given equation is $|z| = z + 1 + 2i.....\left( i \right)$

Let $ z = x + iy$

From Eq. (i), $ |x + iy| = x + iy + 1 + 2i$

 $\Rightarrow  \sqrt {{x^2} + {y^2}}  = x + iy + 1 + 2i \left[ {\because |z| + \sqrt {{x^2} = {y^2}} } \right]$

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

On squaring both sides, we get

${x^2} + {y^2} = {(x + 1)^2} + {i^2}{(y + 2)^2} + 2i(x + 1)(y + 2)$

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

On comparing real and imaginary parts,

${x^2} + {y^2} = {x^2} + 2x + 1 - {y^2} - 4y - 4$

${\text{i}}{\text{.e}}{\text{., }}2{y^2} = 2x - 4y - 3.....\left( {ii} \right)$ 

and

  $2(x + 1)(y + 2) = 0$

  $(x + 1) = 0{\text{ or }}(y + 2) = 0$

 $\Rightarrow x =  - 1{\text{ or }}y =  - 2$ 

For $x =  - 1$,

$2{y^2} =  - 2 - 4y - 3$[using Eq. (ii)]

$2{y^2} + 4y + 5 = 0$

 $\Rightarrow y = \dfrac{{ - 4 \pm \sqrt {16 - 2 \times 4 \times 5} }}{4}$

 $\Rightarrow y = \dfrac{{ - 4 \pm \sqrt { - 24} }}{4} \notin R$ 

${\text{For }}y =  - 2,$

  $2{( - 2)^2} = 2x - 4( - 2) - 3[{\text{using}}{\text{Eq}}.({\text{ii}})]$ 

 $\Rightarrow 8 = 2x + 8 - 3$

 $\Rightarrow 2x = 3$

 $\Rightarrow x = \dfrac{3}{2}$ 

$\therefore  z = x + iy = \dfrac{3}{2} - 2i$

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

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

Given that, $ |z + 1| = z + 2(1 + i).....\left( i \right)$

Let  $z = x + iy$

Then,

 $\Rightarrow |x + iy + 1| = x + iy + 2(1 + i)$

 $\Rightarrow |x + 1 + \mathop {iy}\limits^. | = (x + 2) + i(y + 2)$

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

On squaring both sides, we get

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

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

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

On comparing real and imaginary parts, we get

 $\Rightarrow {x^2} + {y^2} + 2x + 1 = {x^2} - {y^2} + 4x - 4y$

 $\Rightarrow 2{y^2} - 2x + 4y + 1 = 0.....\left( {ii} \right)$

${\text{and }}2(x + 2)(y + 2) = 0$

 $\Rightarrow x + 2 = 0{\text{ or }}y + 2 = 0$

 $\Rightarrow x =  - 2{\text{ or }}y =  - 2.....\left( {iii} \right)$ 

${\text{For  }}x =  - 2,$

  $2{y^2} + 4 + 4y + 1 = 0 \left[ {{\text{Using Eq}}{\text{. }}\left( {ii} \right)} \right]$

 $\Rightarrow 2{y^2} + 4y + 5 = 0$

 $\Rightarrow 16 - 4 \times 2 \times 5 < 0$ 

$\therefore$ Discriminant, $D = {b^2} - 4ac < 0$

$\Rightarrow  2{y^2} + 4y + 5$ has no real roots.

${\text{For  }}y =  - 2,$

 $\Rightarrow 2{( - 2)^2} - 2x + 4( - 2) + 1 = 0\left[ {{\text{Using Eq}}{\text{. }}\left( {ii} \right)} \right]$

 $\Rightarrow 8 - 2x - 8 + 1 = 0$

 $\Rightarrow x = \dfrac{1}{2}$

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

Ans: Given, arg $(z - 1) = \arg (z + 3i)$, where $z = x + iy$.

Apply $\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$, where $z = x + iy$.

Given that,

$\arg (z - 1) = \arg (z + 3i)$

Let  $z = x + iy$

${\text{Now,}}$

 $\Rightarrow \arg (x + \mathop {iy}\limits^.  - 1) = \arg (x + \mathop {iy}\limits^.  + 3i)$

 $\Rightarrow \arg (z - 1) = \arg (z + 3i)$

 $\Rightarrow \arg (x - 1 + \dot y) = \arg [x + i(y + 3)]$

 $\Rightarrow  {\tan ^{ - 1}}\dfrac{y}{{x - 1}} = {\tan ^{ - 1}}\dfrac{{y + 3}}{x}$

 $\Rightarrow  \dfrac{y}{{x - 1}} = \dfrac{{y + 3}}{x}$

 $\Rightarrow xy = (x - 1)(y + 3)$

 $\Rightarrow xy = xy - y + 3x - 3$

 $\Rightarrow 3x - 3 = y$

 $\Rightarrow \dfrac{{3\left( {x - 1} \right)}}{y} = 1$

 $\Rightarrow \dfrac{{x - 1}}{y} = \dfrac{1}{3}$

$\therefore \left( {x - 1} \right):y = 1:3$ 

Ans: Given, $\left| {\dfrac{{z - 2}}{{z - 3}}} \right| = 2$

If ${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ are two complex numbers, then $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}},\left( {{z_2} \ne 0} \right)$, use this concept to solve the above problem. Also, the general equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$, with centre $( - g, - f)$ and radius $= \sqrt {{g^2} + {f^2} - c}$.

Let $z = x + iy$

Given,

Equation is $\left| {\dfrac{{z - 2}}{{z - 3}}} \right| = 2$

$\Rightarrow \left| {\dfrac{{z - 2}}{{z - 3}}} \right| = 2$

$\Rightarrow \left| {\dfrac{{x + iy - 2}}{{x + iy - 3}}} \right| = 2$

$\Rightarrow |x - 2 + iy| = 2|x - 3 + iy|$

$\Rightarrow \sqrt {{{(x - 2)}^2} + {y^2}}  = 2\sqrt {{{(x - 3)}^2} + {y^2}}  \left[ {\because |x + iy| = \sqrt {{x^2} + {y^2}} } \right]$

On squaring both sides, we get

 $\Rightarrow {x^2} - 4x + 4 + {y^2} = 4\left( {{x^2} - 6x + 9 + {y^2}} \right)$

 $\Rightarrow 3{x^2} + 3{y^2} - 20x + 32 = 0$

 $\Rightarrow {x^2} + {y^2} - \dfrac{{20}}{3}x + \dfrac{{32}}{3} = 0.....\left( i \right)$ 

On comparing the above equation with ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get

 $\Rightarrow  2g = \dfrac{{ - 20}}{3}$

 $\Rightarrow g = \dfrac{{ - 10}}{3}$ 

and

 $\Rightarrow 2f = 0$

 $\Rightarrow f = 0{\text{ and }}c = \dfrac{{32}}{3}$

$\therefore {\text{ Centre }} = \left( { - g, - f} \right) = \left( {\dfrac{{10}}{3},0} \right)$ 

Also,

radius \[(r) = \sqrt {{{\left( {\dfrac{{10}}{3}} \right)}^2} + 0 - \dfrac{{32}}{3}}  \left[ {\because r = \sqrt {{g^2} + {f^2} - c} } \right]\]

$= \dfrac{1}{3}\sqrt {(100 - 96)}  = \dfrac{2}{3}$

Ans: Let,

$z = x + iy$

 $\Rightarrow \dfrac{{z - 1}}{{z + 1}} = \dfrac{{x + iy - 1}}{{x + iy + 1}}z \ne  - 1$

 $\Rightarrow \dfrac{{x - 1 + iy}}{{x + 1 + \mathop {iy}\limits^. }} = \dfrac{{(x - 1 + iy)(x + 1 - iy)}}{{(x + 1 + iy)(x + 1 - iy)}}$ 

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

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

Given that, $\dfrac{{z - 1}}{{z + 1}}$ is a purely imaginary number.

Then,

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

 $\Rightarrow  {x^2} - 1 + {y^2} = 0$

 $\Rightarrow {x^2} + {y^2} = 1$

 $\Rightarrow \sqrt {{x^2} + {y^2}}  = \sqrt 1 \left[ {\because |z| = \sqrt {{x^2} + {y^2}} } \right]$

$\therefore |z| = 1$ 

Ans: Given, $\left| {{z_1}} \right| = \left| {{z_2}} \right|$ and $\arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right) = \pi$

$\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$, where $z = x + iy$.

Let ${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and ${z_2} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$ are two complex numbers.

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

and $ \arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right) = \pi$

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

${r_1} = {r_2}$

and if $ \arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right) = \pi$

${\theta _1} + {\theta _2} = \pi$

${\theta _1} = \pi  - {\theta _2}$

Now, $ {z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$

$\Rightarrow  {z_1} = {r_2}\left[ {\cos \left( {\pi  - {\theta _2}} \right) + i\sin \left( {\pi  - {\theta _2}} \right)} \right] \left[ {\because {r_1} = {r_2}} \right.$ and $\left. {{\theta _1} = \left( {\pi  - {\theta _2}} \right)} \right]$

$\Rightarrow  {z_1} = {r_2}\left( { - \cos {\theta _2} + i\sin {\theta _2}} \right)$

$\Rightarrow  {z_1} =  - {r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)$

$\Rightarrow  {z_1} =  - \left[ {{r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)} \right]$

$\Rightarrow  {z_1} =  - {\bar z_2}\left[ {\because {{\bar z}_2} = {r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)} \right]$

Ans: Given, $\left| {{z_1}} \right| = 1\left( {{z_1} \ne  - 1} \right)$ and ${z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + {1^\prime }}}$.

Take ${z_1} = x + iy$ and use the formula $\left| {z} \right| = \sqrt {{x^2} + {y^2}}$.

Let,

${z_1} = x + iy$

 $\Rightarrow \left| {{z_1}} \right| = \sqrt {{x^2} + {y^2}}  = 1 \left[ {\because \left| {{z_1}} \right| = 1,{\text{ given }}} \right].....\left( i \right)$ 

Now,

${z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + 1}} = \dfrac{{x + iy - 1}}{{x + iy + 1}}$

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

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

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

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

$= 0 + \dfrac{{2yi}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$ 

Hence, the real part of \[{z_2}\] is zero.

Ans: Given, ${z_1},{z_2}$ and ${z_3},{z_4}$ are two pairs of conjugate complex numbers.

First let, $z = r(\cos \theta  + i\sin \theta )$ then conjugate of $z$ i.e., $\bar z = r(\cos \theta  - i\sin \theta )$. Use the

property arg $\left( {\dfrac{{{z_1}}}{{{z_2}}}} \right) = \arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right)$.

Let ${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$

Then, 

  $ {z_2} = {{\bar z}_1}$

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

${\text{Also,}}$

${\text{Let }}{{\text{z}}_3} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right),$

${\text{Then,}}$

${{\text{z}}_4} = {{\bar z}_3} = {r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)$ 

$\arg \left( {\dfrac{{{z_1}}}{{{z_4}}}} \right) + \arg \left( {\dfrac{{{z_2}}}{{{z_3}}}} \right) = \arg \left( {{z_1}} \right) - \arg \left( {{z_4}} \right) + \arg \left( {{z_2}} \right) - \arg \left( {{z_3}} \right)$

${ = {\theta _1} - \left( { - {\theta _2}} \right) + \left( { - {\theta _1}} \right) - {\theta _2}}{[\because \arg (z) = \theta ]}$ 

${ = {\theta _1} + {\theta _2} - {\theta _1} - {\theta _2} = 0}$ 

Ans: Given, $\left| {{z_1}} \right| = \left| {{z_2}} \right| =  \cdots= \left| {{z_n}} \right| = 1$

Using formula ${\left| {z} \right|^2} = z.\bar z{\text{and }}\left| {{z_1} + {z_2} + {z_3}....{z_n}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right| + \left| {{z_3}} \right| + ....\left| {{z_n}} \right|$ 

${\text{ Given that,}}$

${\left| {{z_1}} \right| = \left| {{z_2}} \right| =  \cdots= \left| {{z_n}} \right| = 1}$ 

${ \Rightarrow } {{{\left| {{z_1}} \right|}^2} = {{\left| {{z_2}} \right|}^2} =  \cdots  = {{\left| {{z_n}} \right|}^2} = 1}$ 

 $\Rightarrow  {{z_1}{{\bar z}_1} = {z_2}{{\bar z}_2} = {z_3}{{\bar z}_3} =  \cdots  = {z_n}{{\bar z}_n} = 1}$ 

 $\Rightarrow { {z_1} = \dfrac{1}{{{{\bar z}_1}}},{z_2} = \dfrac{1}{{{{\bar z}_2}}} =  \cdots  = {z_n} = \dfrac{1}{{{{\bar z}_n}}}}$ 

Now, 

$\left| {{z_1} + {z_2} + {z_3} + {z_4} +  \cdots  + {z_n}} \right|$

 $= \left| {\dfrac{{{z_1}{{\bar z}_1}}}{{{{\bar z}_1}}} + \dfrac{{{z_2}{{\bar z}_2}}}{{{{\bar z}_2}}} + \dfrac{{{z_3}{{\bar z}_3}}}{{{{\bar z}_3}}} +  \cdots  + \dfrac{{{z_n}{{\bar z}_n}}}{{{{\bar z}_n}}}} \right| \left[ {\because {z_1}\bar z = 1,{\text{ where }}{z_1} = \dfrac{1}{z},{z_1} = \dfrac{{\bar z}}{{\bar z - \bar z}},{z_1} = \bar z} \right]$ 

$= \left| {\dfrac{{{{\left| {{z_1}} \right|}^2}}}{{{{\bar z}_1}}} + \dfrac{{{{\left| {{z_2}} \right|}^2}}}{{{{\bar z}_2}}} + \dfrac{{{{\left| {{z_3}} \right|}^2}}}{{{z_3}}} +  \cdots  + \dfrac{{{{\left| {{z_n}} \right|}^2}}}{{{{\bar z}_n}}}} \right|$

$= \left| {\dfrac{1}{{{{\bar z}_1}}} + \dfrac{1}{{{{\bar z}_2}}} + \dfrac{1}{{{{\bar z}_3}}} +  \cdots  + \dfrac{1}{{{{\bar z}_n}}}} \right| = \sqrt {\dfrac{1}{{{{\bar z}_1}}} + \dfrac{1}{{{{\bar z}_2}}} + \dfrac{1}{{{{\bar z}_3}}}}$

$= \left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} +  \cdots  + \dfrac{1}{{{z_n}}}} \right|$

Hence proved.

Ans: Given, arg $\left( {{z_1}} \right) - \arg \left( {{z_2}} \right) = 0$.

Apply $\arg (z) = \theta$, where $z = x + iy$ and $z = r(\cos \theta  + i\sin \theta )$.

Let,

${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$

${z_2} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

 $\Rightarrow \arg \left( {{z_1}} \right) = {\theta _1}{\text{ and }}\arg \left( {{z_2}} \right) = {\theta _2}$ 

Given that, 

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

${\theta _1} - {\theta _2} = 0$

 $\Rightarrow {\theta _1} = {\theta _2}$ 

${z_2} = {r_2}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)\left[ {\because {\theta _1} = {\theta _2}} \right]$

${z_1} - {z_2} = \left( {{r_1}\cos {\theta _1} - {r_2}\cos {\theta _1}} \right) + i\left( {{r_1}\sin {\theta _1} - {r_2}\sin {\theta _1}} \right)$

$\left| {{z_1} - {z_2}} \right| = \sqrt {{{\left( {{r_1}\cos {\theta _1} - {r_2}\cos {\theta _1}} \right)}^2} + {{\left( {{r_1}\sin {\theta _1} - {r_2}\sin {\theta _1}} \right)}^2}}$

$= \sqrt {r_1^2 + r_2^2 - 2{r_1}{r_2}{{\cos }^2}{\theta _1} - 2{r_1}{r_2}{{\sin }^2}{\theta _1}}$

$= \sqrt {r_1^2 + r_2^2 - 2{q_1}{r_2}\left( {{{\sin }^2}{\theta _1} + {{\cos }^2}{\theta _1}} \right)}$

$= \sqrt {r_1^2 + r_2^2 - 2{r_1}{r_2}}  = \sqrt {{{\left( {{r_1} - {r_2}} \right)}^2}}$

$\left| {{z_1} - {z_2}} \right| = {r_1} - {r_2}  [\because r = |z|]$

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

Hence proved.