Number Theory Test 38

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Number Theory Test 38

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Weekly Quiz Competition

Question 1
1 / -0

If $$z_{1},\ z_{2}$$ are two complex numbers such that $$arg\left( { z }_{ 1 }+{ z }_{ 2 } \right) =0$$ and $$Im\left( { z }_{ 1 }{ z }_{ 2 } \right) =0$$, then

A
$$z_{1}=-z_{2}$$

B
$$z_{1}=z_{2}$$

C
$$z_{1}=\bar { { z }_{ 2 } } $$

D
$$none\ of\ these$$

Solution

Let the complex numbers be,

$$z_1=a+ib,z_2=x+iy$$

$$arg(z_1+z_2)=arg(a+x+i(b+y))=0$$

$$\tan^{-1}\left ( \dfrac{b+y}{a+x} \right )=0$$

$$ \dfrac{b+y}{a+x}=0$$

$$b+y=0$$.....(1)

$$Im(z_1z_2)=Im(ax-by+i(ay+bx))=0$$

$$(ay+bx)=0$$

from (1),

$$ay=xy\Rightarrow a=x$$

$$\Rightarrow b=-y,a=x$$

$$\therefore z_1=x-iy,z_2=x+iy$$

Hence proved, $$z_1=\bar{z}_2$$
Question 2
1 / -0

$$\sqrt{-2}\sqrt{-3}=$$

A
-$$\sqrt{6}$$

B
$$\sqrt{6}$$

C
$$i\sqrt{6}$$

D
$$-i\sqrt{6}$$

Solution

$${\textbf{Step - 1: Writing the product in terms of i (iota)}}$$

$${\text{We know that, }}\sqrt {{\text{ - 1}}} {\text{ = i and }}\sqrt {{\text{ab}}} {\text{ = }}\sqrt {\text{a}} \sqrt {\text{b}} $$

  $$\therefore {\text{ }}\sqrt {{\text{ - 2}}} {\text{ = }}\sqrt {{\text{ - 1}}} \sqrt {\text{2}} {\text{ = i}}\sqrt 2 $$ $$\quad \quad \quad \text{.....eqn(i)}$$

  $${\text{and }}\sqrt {{\text{ - 3}}} {\text{ = }}\sqrt {{\text{ - 1}}} \sqrt {\text{3}} {\text{ = i}}\sqrt {\text{3}} $$ $$\quad \quad \quad \text{.....eqn(ii)}$$

$${\textbf{Step - 2: Multiplying the terms}}$$

  $$\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ = }}\left( {{\text{i}}\sqrt {\text{2}} } \right)\left( {{\text{i}}\sqrt {\text{3}} } \right)$$ $$\quad \quad \quad \textbf{[From eqn(i) and eqn(ii)]}$$

  $$ \Rightarrow {\text{ }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ = }}{{\text{i}}^{\text{2}}}\sqrt {{\text{3}} \times 2} $$

  $$ \Rightarrow {\text{ }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ =   - }}\sqrt {\text{6}} {\quad \quad \quad \quad \quad \quad \textbf{ [}}\because {\text{ }}{{\textbf{i}}^{\textbf{2}}}{\textbf{ =   - 1]}}$$

$$\mathbf{{\text{Thus, the value of the product }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ is -}}\sqrt {\text{6}} }$$
Question 3
1 / -0

If $$\left| z \right| = 1$$, then $$\left| z - 1 \right|$$ is

A
< $$\left| arg (z) \right|$$

B
> $$\left| arg (z) \right|$$

C
= $$\left| arg (z) \right|$$

D
None of these

Solution

Given $$|z|=1\ \Rightarrow \ z=1\ e^{i\theta} $$ where $$\theta =\arg (z)$$

$$|z-1|=|e^{i\theta}-1|=|(\cos \theta -1)+i\sin \theta|$$

$$=\sqrt {(i\cos \theta -1)^2+\sin^2 \theta}$$

$$=\sqrt {2(1-\cos \theta)}$$

$$=2\sin \dfrac {\theta}{2}$$

For $$\theta \ge 0,\ \sin \theta \le \theta \ \Rightarrow \ |z-1| =2\sin \dfrac {\theta}{2}\le 2\dfrac {\theta }{2}$$

Hence, $$|z-1| < |\arg (z)| $$

$$\therefore \ $$ Option $$A$$ is correct
Question 4
1 / -0

If z is a complex number such that $$|z|\ge 2$$, then the minimumm value of $$\left|z+\dfrac{1}{2}\right|$$:

A
is equal to $$\dfrac{5}{2}$$

B
lies in the interval $$(1,2)$$

C
is strictly greater then $$\dfrac{5}{2}$$

D
is strictly greater than $$\dfrac{3}{2}$$ but less than $$\dfrac{5}{2}$$

Solution
Question 5
1 / -0

If $$|z|=1$$ and $$|\omega -1| =1$$ where $$z, \omega \in C$$, then the largest set of values of $$|2z - 1|^2 + | 2\omega -1|^2$$ equals

A
$$[1, 9]$$

B
$$[2, 6]$$

C
$$[2, 12]$$

D
$$[2, 18]$$

Solution

$$|z|=1$$ implies that Z in Q circle with radius with a center (0,0) in the complex plane.
$$|w-1|=1$$ implies that it in a circle with the radius unit, center $$(1,0)$$ in the complex plane.
$$|2z-1|^2+|2w-1|^2=4\left(\left|z-\dfrac{1}{2}\right|^2+\left|w-\dfrac{1}{2}\right|^2\right)$$
It means it is the range of $$4 $$ times the sum of the square of the distance of points from $$(0.5,)$$ from the respective circles of low of the z,w
The point $$(0.5,0)$$ in symmetric with the both circles on the center of a radius on the X-axis so both of them get equal ranges so it simply becomes $$8$$ times of annual circle range.
The min or max possible values of the distance from the point become the distance b/w the point and the vertices of the diameter on choice it lies that is
$$\rightarrow \dfrac{1}{2},\dfrac{3}{2}$$
$$=8\times \left[\left(\dfrac{1}{2}\right)^2,\left(\dfrac{3}{2}\right)^2\right]=[2,18]$$
Question 6
1 / -0

If $$Z$$ is a complex number such that $$|z| \ge 2$$,
then the minimum value of $$\left|z + \dfrac{1}{2}\right|$$

A
Is equal to $$\dfrac{5}{2}$$

B
Lies in the interval $$(1, 2)$$

C
Is strictly grater than $$\dfrac{5}{2}$$

D
Is strictly greater than $$\dfrac{3}{2}$$ but less than $$\dfrac{5}{2}$$

Solution

R.E.F image

graph of $$|z|\geq 2 $$

everything outside the

circle and on the circle is $$|z|\geq 2 $$

Minimum distance bet$$^{n}$$ $$ |z+1/2| $$

$$ \Rightarrow $$ distance $$ AB $$

By using distance formula

$$ \Rightarrow \dfrac 32 $$

$$ \therefore $$ Option B is correct
Question 7
1 / -0

If $$\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = 1$$ and $$\arg \left( {{z_1}{z_2}} \right) = 0$$ , then

A
$${z_1} = {z_2}$$

B
$${\left| {{z_2}} \right|^2} = {z_1}{z_2}$$

C
$${z_1}{z_2} = 1$$

D
$${ z_{ 1 } }=-{ z_{ 2 } }$$

Solution

If $$\left|\dfrac{z_1}{z_2}\right|=1$$ and $$arg(z_1z_2)=0$$

$$z_1=x_1+iy, z_2=x_2+iy_2$$

then if $$\left|\dfrac{z_1}{z_2}\right|=1$$

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

$$x_1^2+y_1^2=x_2^2+y_2^2$$

$$arg(z_1z_2)=arg(x_1x_2-y_1y_2+i(x_1y_2+y_1x_2))=0$$

$$\dfrac{x_1y_2+y_1x_2}{x_1x_2-y_1y_2}=0$$

$$x_1y_2=-y_1x_2$$

$$\dfrac{x_1}{x_2}=\dfrac{-y_1}{y_2}$$

$$\dfrac{x_1}{y_2}=\dfrac{-x_2}{y_2}$$

Both $$z_1z_2$$ make $$180^o$$ angle

So $$z_1+z_2=0$$

if both have an equal modulus

$$z_1+z_2=0$$

$$z_1=-z_2$$

$$D$$ is correct
Question 8
1 / -0

$$I_m$$ $$\left( {\sqrt {a + i\sqrt {{a^4} + {a^2} + 1} } } \right) = $$

A
$$\dfrac{1}{2}\sqrt {{a^2} - a + 1} $$

B
$$\sqrt {\dfrac{{{a^2} - a + a}}{2}} $$

C
$$\dfrac{1}{2}\sqrt {{a^2} + a + 1} $$

D
$$\sqrt {\dfrac{{{a^2} - a + 1}}{2}} $$

Solution

Let $$z=\sqrt{a+i\sqrt{a^4+a^2+1}}$$
$$x+iy=\sqrt{a+i\sqrt{a^4+a^2+1}}$$
Find y which $$I_m(z)$$
Squaring both sides
$$x^2-y^2+2ixy=a+i\sqrt{a^4+a^2+1}$$
$$x^2-y^2=a$$ ……..$$(1)$$
$$2xy=\sqrt{a^4+a^2+1}$$
$$2xy=\sqrt{(a^2+a+1)(a^2+1-a)}$$
$$xy=\sqrt{\left(\dfrac{a^2+a+1}{2}\right)\left(\dfrac{a^2-a+1}{2}\right)}$$ ……$$(2)$$
From $$(1)$$ & $$(2)$$ we get
$$x=\sqrt{\dfrac{a^2+a+1}{2}}, y=\sqrt{\dfrac{a^2-a+1}{2}}$$
$$I_m(z)=\sqrt{\dfrac{a^2-a+1}{2}}$$
D is correct.
Question 9
1 / -0

If for complex number $$z_{1}and z_{2}arg(z_{1})-arg(z_{2})=0then \mid z_{ 1}-z_{2}\mid $$ is equal to:

A
$$ \mid z_{1}+z_{2}\mid $$

B
$$ \mid z_{1}\mid +\mid z_{2}\mid$$

C
$$ \parallel z_{1}\mid -\mid z_{2}\parallel$$

D
0
Question 10
1 / -0

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.

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Re-Attempt Weekly Quiz Competition

Number Theory Test 38

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Number Theory Test 38

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SHARING IS CARING

Question 1

$$z_{1}=-z_{2}$$

$$z_{1}=z_{2}$$

$$z_{1}=\bar { { z }_{ 2 } } $$

$$none\ of\ these$$

Solution

Question 2

-$$\sqrt{6}$$

$$\sqrt{6}$$

$$i\sqrt{6}$$

$$-i\sqrt{6}$$

Solution

Question 3

< $$\left| arg (z) \right|$$

> $$\left| arg (z) \right|$$

= $$\left| arg (z) \right|$$

None of these

Solution

Question 4

is equal to $$\dfrac{5}{2}$$

lies in the interval $$(1,2)$$

is strictly greater then $$\dfrac{5}{2}$$

is strictly greater than $$\dfrac{3}{2}$$ but less than $$\dfrac{5}{2}$$

Solution

Question 5

$$[1, 9]$$

$$[2, 6]$$

$$[2, 12]$$

$$[2, 18]$$

Solution

Question 6

Is equal to $$\dfrac{5}{2}$$

Lies in the interval $$(1, 2)$$

Is strictly grater than $$\dfrac{5}{2}$$

Is strictly greater than $$\dfrac{3}{2}$$ but less than $$\dfrac{5}{2}$$

Solution

Question 7

$${z_1} = {z_2}$$

$${\left| {{z_2}} \right|^2} = {z_1}{z_2}$$

$${z_1}{z_2} = 1$$

$${ z_{ 1 } }=-{ z_{ 2 } }$$

Solution

Question 8

$$\dfrac{1}{2}\sqrt {{a^2} - a + 1} $$

$$\sqrt {\dfrac{{{a^2} - a + a}}{2}} $$

$$\dfrac{1}{2}\sqrt {{a^2} + a + 1} $$

$$\sqrt {\dfrac{{{a^2} - a + 1}}{2}} $$

Solution

Question 9

$$ \mid z_{1}+z_{2}\mid $$

$$ \mid z_{1}\mid +\mid z_{2}\mid$$

$$ \parallel z_{1}\mid -\mid z_{2}\parallel$$

0

Question 10

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

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

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

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

Solution

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