Chapter 5: Complex Numbers and Quadratic Equations

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

If $z = x + iy$ then it’s polar form is $z = r(\cos \theta  + i\sin \theta )$ where $\arg \left( z \right) = \theta$.

Let, 

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

${\text{also   | }}z| = r = 4{\text{ and }}\theta  = \dfrac{{5\pi }}{6}\left[ {\because \arg \left( z \right) = \theta } \right]$

$\therefore z = 4\left[ {\cos \dfrac{{5\pi }}{6} + i\sin \dfrac{{5\pi }}{6}} \right]\left[ {\because z = r\left( {\cos \theta  + i\sin \theta } \right)} \right]$ 

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

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

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

Ans: Given, $\left| {(1 + i)\dfrac{{(2 + i)}}{{(3 + i)}}} \right|$

First, convert the given expression in the formula $a + ib$, then use $\left| {a + ib} \right| = \sqrt {{a^2} + {b^2}}$.

Given that, 

$\left| {(1 + i)\dfrac{{(2 + i)}}{{(3 + i)}}} \right|$

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

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

 $\Rightarrow \left| {\dfrac{{3 + 9i - i - 3{i^2}}}{{9 - {i^2}}}} \right| = \left| {\dfrac{{3 + 8i + 3}}{{9 + 1}}} \right|$

 $\Rightarrow \left| {\dfrac{{6 + 8i}}{{10}}} \right| = \sqrt {\dfrac{{{6^2}}}{{100}} + \dfrac{{{8^2}}}{{100}}}$

 $\Rightarrow \sqrt {\dfrac{{36 + 64}}{{100}}}  = \sqrt {\dfrac{{100}}{{100}}}  = 1$ 

Ans: Given, ${\left( {1 + i\sqrt 3 } \right)^2}$

Let $z = a + ib$, then the polar form of $z$ is $r(\cos \theta  + i\sin \theta )$, where $r = |z| = \sqrt {{a^2} + {b^2}}$ and $\tan \theta  = \dfrac{b}{a} \cdot $ Here, $\theta$ is argument or amplitude of $z$ i.e., $\arg (z) = \theta$. The principal argument is a unique value of $\theta$ such that $- \pi  \leqslant \theta  \leqslant \pi$.

${\text{Given that,}}$

  $z = {\left( {1 + i\sqrt 3 } \right)^2}$

 $\Rightarrow z = 1 - 3 + 2i\sqrt 3  =  - 2 + i2\sqrt 3$

 $\Rightarrow \tan \alpha= \left| {\dfrac{{2\sqrt 3 }}{{ - 2}}} \right| = \left| { - \sqrt 3 } \right| = \sqrt 3 \left[ {\because \tan \alpha= \left| {\dfrac{{\operatorname{Im} \left( z \right)}}{{\operatorname{Re} \left( z \right)}}} \right|} \right]$

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

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

  $\because \operatorname{Re} \left( z \right) < 0{\text{and Im}}\left( z \right) > 0$

 $\Rightarrow \arg \left( z \right) = \pi  - \dfrac{\pi }{3} = \dfrac{{2\pi }}{3}$ 

Ans: Given, $\left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1$

If${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ then $\left| {{z_1}} \right| = \sqrt {x_1^2 + y_1^2}$ and $\left| {{z_2}} \right| = \sqrt {x_2^2 + y_2^2}$.

Also, use the modulus property i.e., $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}$.

Let $ z = x + iy$

Given that, $ \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \left| {\dfrac{{x + iy - 5i}}{{x + iy + 5i}}} \right|$

$\Rightarrow  \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \dfrac{{|x + i(y - 5)|}}{{|x + i(y + 5)|}}\left[ {\because \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1} \right]$

$\Rightarrow  \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \dfrac{{\sqrt {{x^2} + {{(y - 5)}^2}} }}{{\sqrt {{x^2} + {{(y + 5)}^2}} }}$

On squaring both sides, we get 

${x^2} + {(y - 5)^2} = {x^2} + {(y + 5)^2}$

$\Rightarrow   - 10y =  + 10y$

$\Rightarrow  20y = 0$

$\therefore  y = 0$

So, z lies on real axis.

Choose the correct answer from the given four options indicated against each of the Exercises from 35 to 50 (M.C.Q) 

Ans: Option C is correct answer

$z = \sin x + i\cos 2x$and $\cos x - i\sin 2x$ are conjugate to each other.

If${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ then $\left| {{z_1}} \right| = \sqrt {x_1^2 + y_1^2}$ and $\left| {{z_2}} \right| = \sqrt {x_2^2 + y_2^2}$.

Also, use the modulus property i.e., $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}$.

Let $ z = \sin x + i\cos 2x$                   …..(i)

and $\bar z = \sin x - i\cos 2x$                      …..(ii)

Given that,

$\bar z = \cos x - i\sin 2x$

$\therefore$$\sin x - i\cos 2x = \cos x - i\sin 2x$

$\Rightarrow  \sin x = \cos x$ and $\cos 2x = \sin 2x$

$\Rightarrow\tan x = 1$ and $\tan 2x = 1$

$\Rightarrow  \tan x = \tan \dfrac{\pi }{4}$ and $\tan 2x = \tan \dfrac{\pi }{4}$

$\Rightarrow  x = n\pi  + \dfrac{\pi }{4}$ and $2x = n\pi  + \dfrac{\pi }{4}$

 $\Rightarrow  2x - x = 0$

 $\Rightarrow x = 0$ 

Ans: Option C is correct answer

$\dfrac{{1 - i\sin \alpha }}{{1 + 2i\sin \alpha }}$

First, convert the given expansion into $a + ib$ form and then check whether the complex number $a + ib$ is purely real.

Given expression, 

$z = \dfrac{{1 - i\sin \alpha }}{{1 + 2i\sin \alpha }}$

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

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

 $\Rightarrow \dfrac{{1 - 3i\sin \alpha  - 2{{\sin }^2}\alpha }}{{1 + 4{{\sin }^2}\alpha }}$

 $\Rightarrow \dfrac{{1 - 2{{\sin }^2}\alpha }}{{1 + 4{{\sin }^2}\alpha }} - \dfrac{{3i\sin \alpha }}{{1 + 4{{\sin }^2}\alpha }}$ 

It is given that $z$ is a purely real.

  ${\therefore  \dfrac{{ - 3\sin \alpha }}{{1 + 4{{\sin }^2}\alpha }} = 0}$ 

 $\Rightarrow   - 3\sin \alpha  = 0$

 $\Rightarrow \sin \alpha  = 0$ 

  ${\alpha  = n\pi }$ 

Ans: Option B is correct answer

Given, $z = x + iy$ lies in third quadrant. $x < 0$ and $y < 0$.

If $z = x + iy$ then $\bar z = x - iy$.

Given that, $z = x + iy$ lies in third quadrant. $x < 0$ and $y < 0$.

Now,

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

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

${\text{ and }}$

$\dfrac{{\bar z}}{z} = \dfrac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} - \dfrac{{2ixy}}{{{x^2} + {y^2}}}$ 

Since, $\dfrac{{\bar z}}{z}$ also lies in the third quadrant.

$\therefore \dfrac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} < 0{\text{ and }}\dfrac{{ - 2xy}}{{{x^2} + {y^2}}} < 0$

${x^2} - {y^2} < 0{\text{ and }} - 2xy < 0$

${x^2} < {y^2}{\text{ and }}xy > 0$ 

So,

$x < y < 0$

Ans: Option A is correct answer

Given, $(z + 3)(\bar z + 3)$

If $z = x + iy$ then $\bar z = x - iy$.

Given that, $(z + 3)(\bar z + 3)$

Let 

  $z = x + iy$

 $\Rightarrow (z + 3)(\bar z + 3) = (x + iy + 3)(x + 3 - iy)$

 $\Rightarrow {(x + 3)^2} - {(iy)^2} = {(x + 3)^2} + {y^2}$

 $\Rightarrow |x + 3 + iy{|^2} = |z + 3{|^2}$ 

Ans: Option B is correct answer

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

${i^{4n}} = 1,n \in N$

Given that, 

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

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

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

 $\Rightarrow  {\left[ {\dfrac{{2i}}{{1 + 1}}} \right]^x} = 1$

 $\Rightarrow {\left[ {\dfrac{{2i}}{2}} \right]^x} = 1$ 

 $\Rightarrow  {i^x} = 1$

 $\Rightarrow {i^x} = \left( {{i^{4n}}} \right)\left[ {\because {i^{4n}} = 1,n \in N} \right]$ 

$\Rightarrow  x = 4n$

Ans: Option A is correct answer

$\left( {\dfrac{{3 - 4ix}}{{3 + 4ix}}} \right) = \alpha  - i\beta (\alpha ,\beta  \in R)$

If $z = x + iy$ then $\bar z = x - iy$.

$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

Given equation, 

$ \left( {\dfrac{{3 - 4ix}}{{3 + 4ix}}} \right) = \alpha  - i\beta (\alpha ,\beta  \in R)$

$\Rightarrow  \left[ {\dfrac{{3 - 4ix}}{{3 + 4ix}}} \right] = \alpha  - i\beta$

Now, $ (\alpha  - i\beta ) = \dfrac{{(3 - 4ix)(3 - 4ix)}}{{(3 + 4ix)(3 - 4ix)}} = \dfrac{{9 + 16{i^2}{x^2} - 24ix}}{{9 - 16{i^2}{x^2}}}$

$\Rightarrow  \alpha  - i\beta  = \dfrac{{9 - 16{x^2} - 24ix}}{{9 + 16{x^2}}}$

$\Rightarrow  \alpha  - i\beta  = \dfrac{{9 - 16{x^2}}}{{9 + 16{x^2}}} - \dfrac{{i24x}}{{9 + 16{x^2}}}$                               …..(i)

$\therefore$$\alpha  + i\beta  = \dfrac{{9 - 16{x^2}}}{{9 + 16{x^2}}} + \dfrac{{i24x}}{{9 + 16{x^2}}}$                                   …..(ii)

${\text{So, }} (\alpha  - i\beta )(\alpha  + i\beta ) = {\left( {\dfrac{{9 - 16{x^2}}}{{9 + 16{x^2}}}} \right)^2} - {\left( {\dfrac{{i24x}}{{9 + 16{x^2}}}} \right)^2}$

$\therefore  {\alpha ^2} + {\beta ^2} = \dfrac{{81 + 256{x^4} - 288{x^2} + 576{x^2}}}{{{{\left( {9 + 16{x^2}} \right)}^2}}}$ 

 $\Rightarrow \dfrac{{81 + 256{x^4} + 288{x^2}}}{{{{\left( {9 + 16{x^2}} \right)}^2}}}$

 $\Rightarrow \dfrac{{{{\left( {9 + 16{x^2}} \right)}^2}}}{{{{\left( {9 + 16{x^2}} \right)}^2}}} = 1$