Chapter 5: Complex Numbers and Quadratic Equations

Ans: Option A is correct answer

complex numbers ${z_1}{\text{ and }}{z_2}$.

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.

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

 $\Rightarrow  {\left| {{z_1}} \right| = {r_1}.....\left( i \right)}$ 

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

$\Rightarrow  \left| {{z_2}} \right| = {r_2}.....\left( {ii} \right)$

Now, 

  ${z_1}{z_2} = {r_1}{r_2}\left[ {\cos {\theta _1}\cos {\theta _2} + i\sin {\theta _1}\cos {\theta _2} + i\cos {\theta _1}\sin {\theta _2} + {i^2}\sin {\theta _1}\sin {\theta _2}} \right]$

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

 $\Rightarrow \left| {{z_1}{z_2}} \right| = {r_1}{r_2}$

$\therefore \left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|\left[ {{\text{Using equation }}\left( i \right){\text{ and }}\left( {ii} \right)} \right]$ 

Ans: Option B is correct answer

complex number $(2 - i)$

Here, $z < i\alpha $ is a complex number, where modulus is $r$ and argument $(\theta  + \alpha )$ If $P(z)$ rotates in clockwise sense through an angle $\alpha $, then its new position will be $z(\theta  - i\alpha )$

Given that, $z = 2 - i$

It is rotated about origin through an angle $\dfrac{\pi }{2}$ in the clockwise direction.

$\therefore  {\text{ New position}} = z{e^{ - i\pi /2}}$

 $\Rightarrow (2 - i){e^{ - i\pi /2}} = (2 - i)\left[ {\cos \left( {\dfrac{{ - \pi }}{2}} \right) + i\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right]$

 $\Rightarrow (2 - i)[0 - i] =  - 2i - 1 =  - 1 - 2i$ 

Ans: Option D is correct answer

Given, $x,y \in R$

$x + iy$ is non-real complex number if and only if $y \ne 0$.

Given that,

$x,y \in R$

Then, $x + iy$ is non-real complex number if and only if $y \ne 0$.

Ans: Option D is correct answer

$a + ib = c + id$

If two complex numbers ${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ are equal, then

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

 $\Rightarrow \sqrt {x_1^2 + y_1^2}  = \sqrt {x_2^2 + y_2^2}$ 

Given that, $a + ib = c + id$

$\Rightarrow  |a + ib| = |c + id|$

$\Rightarrow  \sqrt {{a^2} + {b^2}}  = \sqrt {{c^2} + {d^2}}$

On squaring both sides, we get 

${a^2} + {b^2} = {c^2} + {d^2}$

Ans: Option B is correct answer

$\left| {\dfrac{{i + z}}{{i - z}}} \right| = 1$

Let $z$ be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

Given that, $\left| {\dfrac{{i + z}}{{i - z}}} \right| = 1$

Let $ z = x + iy$

$\therefore  \left| {\dfrac{{x + i(y + 1)}}{{ - x - i(y - 1)}}} \right| = 1$

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

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

 $\Rightarrow  4y = 0$

 $\Rightarrow y = 0$ 

So, $z$ lies on $x$-axis (real axis).

Ans: Option B is correct answer

$z$ is a complex number.

If $z$ is a complex number, then $z = x + iy$ $|z| = |x + |y\mid$ and $|z{|^2} = |x + iy{|^2}$

If $z$ is a complex number, then $z = x + iy$ $|z| = |x + |y\mid$ and $|z{|^2} = |x + iy{|^2}$

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

and $ {z^2} = {(x + iy)^2} = {x^2} + {i^2}{y^2} + i2xy$

${z^2} = {x^2} - {y^2} + i2xy$

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

$\Rightarrow  \left| {{z^2}} \right| = \sqrt {{x^4} + {y^4} - 2{x^2}{y^2} + 4{x^2}{y^2}}$

$\Rightarrow  \left| {{z^2}} \right| = \sqrt {{x^4} + {y^4} + 2{x^2}{y^2}}  = \sqrt {{{\left( {{x^2} + {y^2}} \right)}^2}}$

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

From Eqs. (i) and (ii), ${\left| {z} \right|^2} = \left| {{z^2}} \right|$

Ans: Option C is correct answer

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

Let $z$ be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

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

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

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

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

$\Rightarrow  \sqrt {r_1^2{{\cos }^2}{\theta _1} + r_2^2{{\cos }^2}{\theta _2} + 2{r_1}{r_2}\cos {\theta _1}\cos {\theta _2} + r_1^2{{\sin }^2}{\theta _1} + r_2^2{{\sin }^2}{\theta _2}}$

$\sqrt { + 2{r_1}{r_2}\sin {\theta _1}\sin {\theta _2}}  = {r_1} + {r_2}$

$\Rightarrow \sqrt {r_1^2 + r_2^2 + 2{r_1}{r_2}\left[ {\cos \left( {{\theta _1} - {\theta _2}} \right)} \right]}  = {r_1} + {r_2}$

On squaring both sides, we get 

$r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {{\theta _1} - {\theta _2}} \right) = r_1^2 + r_2^2 + 2{r_1}{r_2}$

$\Rightarrow  2{r_1}{r_2}\left[ {1 - \cos \left( {{\theta _1} - {\theta _2}} \right)} \right] = 0$

$\Rightarrow1 - \cos \left( {{\theta _1} - {\theta _2}} \right) = 0$

$\Rightarrow\cos \left( {{\theta _1} - {\theta _2}} \right) = 1$

$\Rightarrow \cos \left( {{\theta _1} - {\theta _2}} \right) = \cos {0^\circ }$

$\Rightarrow  {\theta _1} - {\theta _2} = {0^\circ }$

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

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

Ans: Option C is correct answer

$\dfrac{{1 + i\cos \theta }}{{1 - 2i\cos \theta }}$

Simplify the given equation and use the formula $\cos \dfrac{\pi }{2} = 0$.

Given expression,

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

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

 $\Rightarrow \dfrac{{1 + 3i\cos \theta  - 2{{\cos }^2}\theta }}{{1 + 4{{\cos }^2}\theta }}$ 

For real value of $\theta ,$

$\dfrac{{3\cos \theta }}{{1 + 4{{\cos }^2}\theta }} = 0$

 $\Rightarrow 3\cos \theta  = 0$

 $\Rightarrow \cos \theta  = \cos \dfrac{\pi }{2}$

 $\Rightarrow \theta  = 2n\pi  \pm \dfrac{\pi }{2}$ 

Ans: Option C is correct answer

$x < 0$

Let $z$be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

Let

  $z = x + 0i{\text{ and }}x < 0$

  $|z| = \sqrt {{{( - 1)}^2} + \left( {{0^2}} \right)}  = 1$ 

Since, the point $(x, 0)$ represent $z = x + 0i$ lies on the negative side of real axis. 

$\therefore$ Principal $\arg (z) = \pi$

Ans: Option A is correct answer

$f(z) = \dfrac{{7 - z}}{{1 - {z^2}}}$, where $z = 1 + 2i$

Let $z$ be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

${{\text{ Let }}} {z = 1 + 2i}$ 

 $\Rightarrow {|z| = \sqrt {1 + 4}  = \sqrt 5 } $

Now, 

  $f(z) = \dfrac{{7 - z}}{{1 - {z^2}}} = \dfrac{{7 - 1 - 2i}}{{1 - {{(1 + 2i)}^2}}}$

 $\Rightarrow \dfrac{{6 - 2i}}{{1 - 1 - 4{i^2} - 4i}} = \dfrac{{6 - 2i}}{{4 - 4i}}$ 

$\Rightarrow \dfrac{{(3 - i)(2 + 2i)}}{{(2 - 2i)(2 + 2i)}} = \dfrac{{6 - 2i + 6i - 2{i^2}}}{{4 - 4{i^2}}}$

 $\Rightarrow \dfrac{{6 + 4i + 2}}{{4 + 4}} = \dfrac{{8 + 4i}}{8}$

 $\Rightarrow 1 + \dfrac{1}{2}i$ 

 $f(z) = 1 + \dfrac{1}{2}i$

$\therefore |f(z)| = \sqrt {1 + \dfrac{1}{4}}$ 

$\Rightarrow \sqrt {\dfrac{{4 + 1}}{4}}  = \dfrac{{\sqrt 5 }}{2} = \dfrac{{|z|}}{2}$

An overview of NCERT Exemplar for Class 11 Mathematics - Complex Numbers and Quadratic Equations

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