Number Theory Test 12

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Number Theory Test 12

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Weekly Quiz Competition

Question 1
1 / -0

The number $$111111111$$ is a

A
Prime number

B
Composite number

C
Divisible by $$\frac{10^7-1}{9}$$

D
None of these

Solution

A whole number that can be divided exactly by numbers other than 1 or itself is known as composite numbers.
111111111 is an odd composite number. It is composed of three distinct prime numbers multiplied together.
Prime factorisation of $$111111111 = {3}^{2} \times 37 \times 333667$$
Question 2
1 / -0

If $$(1+i)^{-20}=a+ib$$, then the values of $$a$$ and $$b$$ are

A
$$a=2^{-10}, b=-2^{-10}$$

B
$$a=-2^{-10}, b=0$$

C
$$a=2^{-10}, b=0$$

D
$$none\ of\ these$$

Solution

$$(1+i)^{-20}\\((1+i)^2){-10}\\(1+i^2+2i)^{-10}\\=(2i)^{-10}\\=(\frac{2^{-10}}{i^10})\\=-2^{-10}\\\therefore \>by\>comparison\>a\>=\>-2^{-10},\>b=0$$
Question 3
1 / -0

Find the which of the complex number has greatest modulus.

A
$$7-5i$$

B
$$\sqrt{3}+\sqrt{2}i$$

C
$$-8+15i$$

D
$$-3(1-i)$$

Solution

$$|7-5i| = \sqrt {54 } $$

$$|\sqrt {3} +i \sqrt 2| = \sqrt 5$$

$$ | -8+15i| = \sqrt {289}$$

$$ |-3+3i| = 3\sqrt 2$$

Thus option C has the greatest modulus.
Question 4
1 / -0

If $$\dfrac {z-1}{z+1}$$ is purely imaginary then

A
$$|z|=1$$

B
$$|z|>1$$

C
$$|z|<1$$

D
$$|z|<2$$

Solution

$$\cfrac{z-1}{z+1}$$
Let $$z=x+iy$$
$$\cfrac{x+iy-1}{x+1+iy}=\cfrac{x-1+iy}{x+1+iy}\times \cfrac{x+1-iy}{x+1-iy}\\k=\cfrac{x^2+x-ixy-x+iy-1+ixy+iy+y^2}{(x+1)^2+y^2}\\k=\cfrac{x^2+y^2+2iy-1}{(x+1)^2+y^2}$$
If k is purely imaginary then real part $$=0$$
$$\cfrac{x^2+y^2-1}{(x+1)^2+y^2}=0\\x^2+y^2-1=0\\x^2+y^2=1\\ \therefore |z|=1$$
Question 5
1 / -0

For any complex number $$z$$ the minimum value of $$|z|+|z-2013i|$$ is...

A
$$2010$$

B
$$2011$$

C
$$2013$$

D
$$2012$$

Solution

Q:- minimum value of $$\left | z \right |+\left | z-2013i \right |$$
Solution For ZEG
$$ \left | 2013i \right | = \left | z+(2013i-z) \right | \leqslant \left | z \right |+\left | 2013i-z \right |$$
$$ \Rightarrow 2013 \leqslant \left | z \right |+\left | z-2013i \right |$$
$$ \Rightarrow $$ minimum value of $$ \left | z \right |+\left | z-2013i \right | $$ is
2013
which is attained when $$ z = i$$
so, c is correct option.
$$ x^{2}+y^{2} \leqslant 1 $$ & $$ y^2 \leqslant 1-x . $$
Question 6
1 / -0

If $$z=x+iy(x,y\epsilon R,x\neq -1/2),$$ the number of values of z satisfying $$\left | z \right |^{n}=z^{2}\left | z \right |^{n-2}+1.(n\epsilon N,n> 1)$$is

A
0

B
1

C
2

D
3

Solution

$$\begin{array}{l} we\, \, write\, , \\ { \left| z \right| ^{ n } }={ \left| z \right| ^{ n-2 } }\left( { { z^{ 2 } }+z } \right) +1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left\{ { we\, know,\, \, all\, \, are\, real\, no. } \right. \\ so,\, \, \, \, { z^{ 2 } }+z={ \overline { z } ^{ 2 } }+\overline { z } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { where\, ,\, \, { z^{ 2 } }+z\, \, \, \, is\, real\, no.\, } \right. \\ \Rightarrow { z^{ 2 } }\, -{ \overline { z } ^{ 2 } }+z-\overline { z } =0 \\ \Rightarrow (z-\overline { z } )\, \, (z+\overline { z } +1)=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| \begin{array}{l} \, z=x+i\, y \\ \, \overline { z } =x-i\, y \end{array} \right. \\ \, \, \therefore \, \, \, \, 2\, i\, y\, (2x+1)=0 \\ \, \, \, y=0\, \, \, \, \, \, \, \, \, \, \, or\, \, \, \, \, \, x=-\frac { 1 }{ 2 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (Here,\, \, \, z\, \, is\, a\, \, \, real\, \, \, number. \\ Let\, see, \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| z \right| =x \\ \left| \begin{array}{l} \, \, { x^{ n } }={ x^{ n-2 } }.{ x^{ 2 } }+1 \\ \, \, \, { \left| x \right| ^{ n-2 } }.\, x=-1 \end{array} \right. \\ \, \, \, \therefore \, \, \, \, x=-1\, \\ so,\, number\, of\, z\, =1\, \, and\, \, the\, correct\, option\, is\, B.\, \, \, \, \, \end{array}$$
Question 7
1 / -0

$$z_{1}$$ and $$z_{2}$$ are two non-zero complex numbers such that $$|z_{1}|=|z_{2}|$$ and $$argz_{1}+argz_{2}=\pi$$, then $$z_{2}$$ equals

A
$$\bar{z_{1}}$$

B
$$-\bar{z_{1}}$$

C
$$z_{1}$$

D
$$-z_{1}$$

Solution

$$-z_1$$

as, $$arg z_+argz_=\pi$$
Question 8
1 / -0

If $$ z=1+i\sqrt { 3, } \left| arg\left( z \right) \right| +\left| arg\left( \overline { z } \right) \right|$$

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

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

C
$$0$$

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

Solution

$$z=i+i\sqrt{3}$$ $$\bar{z}=1-i\sqrt{3}$$
$$arg z=\dfrac{\pi}{3}$$ $$arg(\bar{z})=\dfrac{\pi}{3}$$
$$|arg z|+|arg (\bar{z})|=\dfrac{2\pi}{3}$$.
Question 9
1 / -0

If arg $$\left( {{Z_1} + {Z_2}} \right) = 0\;$$ and $$|{z_1}|\; = \;|{z_2}|\; = 1.$$ then.

A
$${z_1} + {z_2} = 0$$

B
$${z_1}{\overline z _2} = 1$$

C
$${z_1} = {\overline z _2}$$

D
none

Solution

We have,
$$\arg \left( {{Z_1} + {Z_2}} \right) = 0\,\,\,\,\,\,\,\,\,\,\,\left| {{Z_1}} \right| = 1\,\,\,\,\left| {{Z_2}} \right| = 1$$
$$\begin{matrix} { Z_{ 1 } }+{ Z_{ 2 } }=k\, \, \left( { k>0 } \right) \\ { Z_{ 1 } }=\cos \theta +i\sin \theta \\ { Z_{ 2 } }=\cos \theta +i\sin \theta \\ \end{matrix}$$
$$\cos \theta + \cos \theta + \left( {i\sin \theta + i\sin \theta } \right) = k$$
$$ \Rightarrow \sin \varphi = - \sin \theta $$
$$ \Rightarrow \cos \varphi = - \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,as\left| {{Z_2}} \right| = 1$$
$${Z_1} = \cos \theta \, + i\sin \theta $$
$${Z_2} = \cos \theta \, - i\sin \theta $$
$${Z_1} = \overline {{Z_2}} $$
Then,
Option $$C$$ is correct answer.
Question 10
1 / -0

The value of $${\left( {1 + i} \right)^5} \times {\left( {1 - i} \right)^5}$$ is

A
$$-8$$

B
$$8i$$

C
$$8$$

D
$$32$$

Solution

We have,

$$ {{\left( 1+i \right)}^{5}}{{\left( 1-i \right)}^{5}} $$

$$ ={{\left( 1+i \right)}^{4}}{{\left( 1-i \right)}^{4}}\left( 1+i \right)\left( 1-i \right) $$

$$ ={{\left( 1+i \right)}^{4}}{{\left( 1-i \right)}^{4}}\left( {{1}^{2}}-{{i}^{2}} \right) $$

$$ ={{\left[ {{\left( 1+i \right)}^{2}} \right]}^{2}}{{\left[ {{\left( 1-i \right)}^{2}} \right]}^{2}}\left( {{1}^{2}}-\left( -1 \right) \right)\,\,\,\,\therefore {{i}^{2}}=-1 $$

$$ ={{\left[ \left( {{1}^{2}}+{{i}^{2}}+2i \right) \right]}^{2}}{{\left[ \left( {{1}^{2}}+{{i}^{2}}-2i \right) \right]}^{2}}\left( 2 \right) $$

$$ =2{{\left[ \left( {{1}^{2}}+\left( -1 \right)+2i \right) \right]}^{2}}{{\left[ \left( {{1}^{2}}+\left( -1 \right)-2i \right) \right]}^{2}}\,\,\,\,\,\,\therefore {{i}^{2}}=-1 $$

$$ =2{{\left[ 2i \right]}^{2}}{{\left[ -2i \right]}^{2}} $$

$$ =32 $$
Hence, this is the answer.

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Number Theory Test 12

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Number Theory Test 12

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SHARING IS CARING

Question 1

Prime number

Composite number

Divisible by $$\frac{10^7-1}{9}$$

None of these

Solution

Question 2

$$a=2^{-10}, b=-2^{-10}$$

$$a=-2^{-10}, b=0$$

$$a=2^{-10}, b=0$$

$$none\ of\ these$$

Solution

Question 3

$$7-5i$$

$$\sqrt{3}+\sqrt{2}i$$

$$-8+15i$$

$$-3(1-i)$$

Solution

Question 4

$$|z|=1$$

$$|z|>1$$

$$|z|<1$$

$$|z|<2$$

Solution

Question 5

$$2010$$

$$2011$$

$$2013$$

$$2012$$

Solution

Question 6

0

1

2

3

Solution

Question 7

$$\bar{z_{1}}$$

$$-\bar{z_{1}}$$

$$z_{1}$$

$$-z_{1}$$

Solution

Question 8

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

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

$$0$$

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

Solution

Question 9

$${z_1} + {z_2} = 0$$

$${z_1}{\overline z _2} = 1$$

$${z_1} = {\overline z _2}$$

none

Solution

Question 10

$$-8$$

$$8i$$

$$8$$

$$32$$

Solution

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