Complex Numbers and Quadratic Equations Test 32

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Complex Numbers and Quadratic Equations Test 32

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Weekly Quiz Competition

Question 1
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Argument and modules of $$[\dfrac{1+i}{1-i}]^{2\pi i}$$ are respectively.................

A
$$\dfrac{-\pi}{2}$$ and $$1$$

B
$$\dfrac{\pi}{2}$$ and $$\sqrt{2}$$

C
$$0$$ and $$\sqrt{2}$$

D
$$\dfrac{\pi}{2}$$ and $$1$$

Solution

$$ (\dfrac{1+i}{1-i})^{2\pi i}$$
let $$ 2 = (\dfrac{1+i}{1-i})^{2\pi i}$$
Rationalizing the above, we get
$$ = \dfrac{(1+i)^2}{(1)^2-(i)^2} = \dfrac{1+(i)^2+2i}{1-(-1)} = \dfrac{1-1+2i}{1+1} = \dfrac{2i}{2} = i $$
$$ z = 0+i = 1$$
here x =0 ,y = 1
modules of $$ 2 = \sqrt{x^2 + y^2 } = \sqrt{0+1^2} = 1 $$
modules of z = 1
$$ z =i , $$ As per the question, $$(i)^{2\pi i }$$
& $$ z = r(cos\theta + sin\theta) $$
$$ \Rightarrow (r(cos\theta + 1sin \theta))^{2\pi i}$$
$$ = [r(cos(2\pi i))+\theta +i sin (2\pi i )\theta ]$$
$$ (i)^{2\pi i} = (r)^{2\pi i}$$
$$ \Rightarrow r = i $$
$$ r^2 cos^2 \theta = 0 ; r^2 sin^2 \theta = 1$$
$$ tan \theta = \dfrac{\pi}{2}$$
$$ 0+1 = r^2 cos^2\theta +r sin^2 \theta $$
$$ 1 = r^2(cos^2 \theta + sin^2 \theta)$$
$$ 1 = r^2 \Rightarrow r = 1 $$
argument of$$ z = \dfrac{\pi}{2}$$
Option (D) is correct.
Question 2
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If $$z_{1}$$ and $$z_{2}$$ are two non zero complex numbers such that $$|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$$ then $$arg\ z_{1} $$-$$arg\ z_{2}$$ is equal to

A
$$-\pi$$

B
$$\dfrac {\pi}{2}$$

C
$$-\dfrac {\pi}{2}$$

D
$$0$$

Solution

A] Let $$ z_{1} = cos\theta _{1}+i\,sin\theta _{1} $$
$$ z_{2} = cos\theta _{2}+i\,sin\theta _{2} $$
$$ z_{1}+z_{2} = (cos\theta _{1}+cos\theta _{2})+i(sin\theta _{1}+sin\theta _{2}) $$
As $$ |z_{1}+z_{2}| = |z_{1}|+|z_{2}| $$
$$ = \sqrt{(cos\theta _{1}+cos\theta _{2})^{2}+(sin\theta _{1}+sin\theta _{2})^{2}} = 1+1 $$
$$ = 2(1+cos(\theta _{1}-\theta _{2})) = 4 $$ [squaring]
$$ = cos(\theta _{1}-\theta _{2}) = 1 $$
$$ = \theta _{1}-\theta _{2} = 0 $$ [as $$ cos0^{\circ} = 1$$]
$$ = Arg z_{1}-Argz_{2} = 0 $$
Question 3
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If $$\theta$$ and $$\phi$$ are the roots of the equation $$8x^{2}+22x+5=0$$, then

A
both $$\sin^{-1}{\theta}$$ and $$\sin^{-1}{\phi}$$ are real

B
both $$\sec^{-1}{\theta}$$ and $$\sec^{-1}{\phi}$$ are real

C
both $$\tan^{-1}{\theta}$$ and $$\tan^{-1}{\phi}$$ are real

D
None of these

Solution

Solving given quadratic eq., we get

$$8x^2+22x+5=0$$

$$\implies (4x+1)(2x+5)=0$$

$$\implies x=\dfrac{-1}{4}, \dfrac{-5}{2}$$

Hence these values are not in the range of $$sec^{-1}x$$ and $$sin^{-1}x$$

So, $$tan^{-1}\theta$$ and $$tan^{-1}\phi$$ are real
Question 4
1 / -0

The figure formed by four points $$1+0i;-1+0i,3+4i$$ and $$\cfrac{25}{-3-4i}$$ on the argand plane is

A
parallelogram but not a rectangle

B
a trapesium which is not equilateral

C
cyclic quadrilateral

D
none of these

Solution

$$1+0i, -1+0i, 3+4i, \dfrac{25}{-3-4i}$$
$$\dfrac{25}{-3-4i}\dfrac{(-3+4i)}{(-3+4i)}$$
$$=\dfrac{25(-3+4i)}{25}=-3+4i$$.
Question 5
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The amplitude of $$\cfrac { 1+\sqrt { 3i } }{ \sqrt { 3 } +1 } $$ is

A
$$\cfrac { \pi }{ 3 } $$

B
$$-\cfrac { \pi }{ 3 } $$

C
$$\cfrac { \pi }{ 6 } $$

D
$$-\cfrac { \pi }{ 6 } $$

Solution

$$\displaystyle \frac{1}{\sqrt{3}+1}+\frac i{\sqrt{3}}{\sqrt{3}+1}$$

$$\displaystyle \tan^{-1}\left ( \frac{\sqrt{3}/\sqrt{3}+1}{1/\sqrt{3}+1} \right )$$

$$=\tan^{-1}(\sqrt{3})$$

$$=\dfrac{\pi}{3}$$
Question 6
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If $$z=(3+7i)(p+iq)$$, where $$p,q\in I-\left\{ 0 \right\} $$, is a purely imaginary, then minimum value of $${ \left| z \right| }^{ 2 }$$ is

A
$$0$$

B
$$58$$

C
$$\cfrac{3364}{3}$$

D
$$3364$$

Solution

$$z=(3+7i)(p+iq) \quad \dots (1)$$

$$z=(3p-7q)+i(3q+7p)$$

Now, $$z$$ is purely imaginary

$$3p-7q=0$$

or $$\cfrac{p}{q}=\cfrac{7}{3}$$

$$\cfrac{p}{q}+i=\cfrac{7}{3}+i$$

Therefore, $$\cfrac{(p+qi)}{q}=\cfrac{7+3i}{3}$$

$$p+iq=7+3i \quad \dots (2)$$

Substitute $$(2)$$ in $$(1)$$

$$z=21+9i+49i-21$$

Thus, $$z=58i$$

$$\left| { z }^{ 2 } \right| =3364$$
Question 7
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If $$z=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i$$ then $$z\bar{z}$$ is

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

Conjugate of $$\dfrac{\sqrt{3}+i}{2}=\dfrac{\sqrt{3}-i}{2}$$
$$\therefore\,\bar{z}=\dfrac{\sqrt{3}-i}{2}$$
Now,$$z\bar{z}=\left(\dfrac{\sqrt{3}+i}{2}\right)\left(\dfrac{\sqrt{3}-i}{2}\right)=\dfrac{3-{i}^{2}}{4}=\dfrac{3+1}{4}=\dfrac{4}{4}=1$$
Question 8
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If $$Z = \frac{{1 - \sqrt 3 i}}{{1 + \sqrt 3 i}}$$ then find $$arg(z).$$

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

B
$$ \dfrac{2\pi }{5} $$

C
$$ \dfrac{\pi }{3} $$

D
$$ \dfrac{2\pi }{3} $$

Solution

Consider the following question.

$$ Z=\dfrac{1-\sqrt{3}i}{1+\sqrt{3i}} $$

$$ =\dfrac{\left( 1-\sqrt{3}i \right)}{\left( 1+\sqrt{3i} \right)}\times \dfrac{\left( 1-\sqrt{3}i \right)}{\left( 1-\sqrt{3}i \right)} $$

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

$$ =\dfrac{1+3{{i}^{2}}-2\sqrt{3}}{4} $$

$$ =\dfrac{1-3-2\sqrt{3}}{4} $$

$$ =\dfrac{-1-i\sqrt{3}}{2} $$

$$ =-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} $$

$$ \arg \left( -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right)=-\pi +\dfrac{\pi }{3} $$

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

Hence, this is the required answer.
Question 9
1 / -0

If $$\left| {z + 2 - i} \right| = 5$$ then the maximum value of $$\left| {3z + 9 - 7i} \right|$$ is

A
18

B
19

C
20

D
8

Solution
Question 10
1 / -0

What is the modulus of following complex number:$$-2+2\sqrt { 3i } $$

A
4

B
5

C
2

D
3

Solution