Number Theory Test 36

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Number Theory Test 36

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Weekly Quiz Competition

Question 1
1 / -0

If $$|Z|=2,|z_{2}|=3,|z_{3}=4|$$ and $$|z_{1}+z_{2}+z_{3}|=5$$ then $$|4z_{2}z_{3}+9z_{3}z_{1}+16z_{1}z_{2}|=$$

A
$$20$$

B
$$24$$

C
$$48$$

D
$$120$$

Solution

Solution:- (D) 120
$$\left| {z}_{1} \right| = 2, \left| {z}_{2} \right| = 3, \left| {z}_{3} \right| = 4, \left| {z}_{1} + {z}_{2} + {z}_{3} \right| = 5$$
$$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = \cfrac{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|}{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|} \left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right|$$
$$= \left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right| \left| \cfrac{4}{{z}_{1}} + \cfrac{9}{{z}_{2}} + \cfrac{16}{{z}_{3}} \right|$$
$$= \left( 2 \times 3 \times 4 \right) \left| \cfrac{4 \bar{{z}_{1}}}{{\left| {z}_{1} \right|}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{\left| {z}_{2} \right|}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{\left| {z}_{3} \right|}^{2}}\right|$$
$$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{{2}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{3}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{4}^{2}} \right|$$
$$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{4} + \cfrac{9 \bar{{z}_{2}}}{9} + \cfrac{16 \bar{{z}_{3}}}{16} \right|$$
$$= 24 \left| \bar{{z}_{1}} + \bar{{z}_{2}} + \bar{{z}_{3}} \right|$$
$$= 24 \left| \overline{{z}_{1} + {z}_{2} + {z}_{3}} \right|$$
$$= 24 \times 5$$
$$= 120$$
Hence, $$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = 120$$.
Question 2
1 / -0

If $$\dfrac{x - 3}{3 + i} + \dfrac{y - 3}{3 - i} = i $$ where $$x , y \in R$$ then

A
$$x = 2$$ & $$y = -8$$

B
$$x = -2$$ & $$y = 8$$

C
$$x = -2$$ & $$y = -6$$

D
$$x = 2$$ & $$y = 8$$

Solution

$$\dfrac{x-3}{3+i}+\dfrac{y-3}{3-i}=i$$
$$\dfrac{(x-3)(3-i)}{10}+\dfrac{(y-3)(3+i)}{10}=i$$
$$3(x-3)+3(y-3)-i((x-3+(3-y))=10i$$
$$3(x+y-6)-i(x-y)=10i$$
compare real and imaginary point
$$x+y=6$$
$$y-x=10$$
$$\Rightarrow y=8, x=-2$$
Question 3
1 / -0

The locus of $$z$$ such that $$\left| {\dfrac{{z + i}}{{z - 1}}} \right| = 2$$

A
straight line

B
circle with radius $$2$$

C
circle with radius $$\frac{{2\sqrt 2 }}{3}$$

D
none of these

Solution

$$\left|\dfrac{z+i}{z-1}\right|=2$$

Put, $$z = x + iy$$

$$\therefore\,\left|\dfrac{x + iy+i}{x + iy-1}\right|=2$$

$$\Rightarrow\,\left|\dfrac{x + i\left(y+1\right)}{\left(x-1\right)+iy}\right|=2$$

$$\Rightarrow\,\left|x + i\left(y+1\right)\right|=2\left|\left(x-1\right)+iy\right|$$

$$\Rightarrow\,\sqrt{{x}^{2}+{\left(y+1\right)}^{2}}=2\sqrt{{\left(x-1\right)}^{2}+{y}^{2}}$$

Squaring both sides, we get
$$\Rightarrow\,{x}^{2}+{\left(y+1\right)}^{2}=4{\left(x-1\right)}^{2}+4{y}^{2}$$

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

$$\Rightarrow\,\left(x-2x+2\right)\left(x+2x-2\right)=\left(2y-y-1\right)\left(2y+y+1\right)$$

$$\Rightarrow\,\left(-x+2\right)\left(3x-2\right)=\left(y-1\right)\left(3y+1\right)$$
$$\Rightarrow\,-3{x}^{2}+2x+6x-4=3{y}^{2}+y-3y-1$$

$$\Rightarrow\,-3{x}^{2}+8x-4=3{y}^{2}-2y-1$$

$$\Rightarrow\,-3{x}^{2}+8x-4-3{y}^{2}+2y+1=0$$

$$\Rightarrow\,3{x}^{2}+3{y}^{2}-8x-2y+3=0$$

$$\Rightarrow\,{x}^{2}+{y}^{2}-\dfrac{8}{3}x-\dfrac{2}{3}y+1=0$$ is of the form $${x}^{2}+{y}^{2}+2gx+2fy+c=0$$

where $$g=\dfrac{4}{3},\,f=\dfrac{1}{3}$$ and $$c=1$$

Radius$$=r=\sqrt{{\left(\dfrac{4}{3}\right)}^{2}+{\left(\dfrac{1}{3}\right)}^{2}+1}=\sqrt{\dfrac{16}{9}+\dfrac{1}{9}-1}=\sqrt{\dfrac{17-9}{9}}=\dfrac{\sqrt{8}}{3}=\dfrac{2\sqrt{2}}{3}$$

The locus of $$z$$ is a circle with radius $$\dfrac{2\sqrt{2}}{3}$$
Question 4
1 / -0

If $$\left| {z - 1} \right| = 2$$, then the value of $$z\overline z - z - \overline z $$ is equal to:

A
$$-3$$

B
$$4$$

C
$$3$$

D
$$-4$$

Solution

Given $$|z-1|=2$$
we know that $$|z|^{2}=z.\bar{z}$$
so taking square root on both side.
$$|z-1|^{2}=4$$
$$(z-1)(\bar{z-1})=4$$
$$(z-1)(\bar{z}-1)=4$$
$$z.\bar{z}-\bar{z}-z+1=4$$
$$z.\bar{z}-z-\bar{z}=3$$
so answer is 3.
Question 5
1 / -0

If z is a complex number such that $$\left|\dfrac{z-3i}{z+3i}\right|=1$$ then z lies on?

A
The real axis

B
The line Im(z)$$=3$$

C
A circle

D
None of these

Solution

Given $$\left| \cfrac { z-3i }{ z+3i } \right| =1$$
$$ \Rightarrow \left| \cfrac { z-3i }{ z+3i } \right| =1$$
$$\Rightarrow z$$ is equidistant from $$(0,3)$$ & $$(0,-3)$$
It lies on real axis.
Question 6
1 / -0

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

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

$$\left( \dfrac{1+i}{1-i} \right)^{4}+ \left( \dfrac{1-i}{1+i} \right)^{4}$$
$$= \left( \dfrac{(1+i)^{2}}{1-i^{2}} \right)^{4}+ \left( \dfrac{(1-i)^{2}}{1-i^{2}} \right)^{4}$$
$$=\dfrac{1}{2^{4}} (1+i^{2} +2 i)^{4}+ \dfrac{1}{2^{4}} (1+i^{2}-2i)^{4}$$
$$=\dfrac{1}{2^{4}} (1-1+ 2i)^{4}+ \dfrac{1}{2^{4}} (1-1-2i)^{4}$$
$$=\dfrac{2^{4}}{2^{4}} (i^{4})+ \dfrac{2^{4}}{2^{4}} (i)^{4}$$
$$= 1+1 =2$$
Question 7
1 / -0

If $$|z_1+z_2|=|z_1|+|z_2|$$ where $$z_1$$ and $$z_2$$ are different non - zero complex number, then ?

A
$$Re\left(\dfrac{z_1}{z_2}\right)=0$$

B
$$Im\left(\dfrac{z_1}{z_2}\right)=0$$

C
$$z_1+z_2=0$$

D
None

Solution

$$let\quad { z }_{ 1 }=a+ib\quad \quad and\quad \quad { z }_{ 2 }=c+id\\ \Rightarrow \quad \left| \quad { z }_{ 1 }+\quad { z }_{ 2 } \right| \quad \quad \quad \quad =\quad \quad \quad \left| \quad { z }_{ 1 } \right| +\left| \quad { z }_{ 2 } \right| \\ \Rightarrow \quad \quad \left| a+ib+c+id \right| \quad =\quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \left| a+c+i\left( b+d \right) \right| \quad \quad \quad =\quad \quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \sqrt { \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 } } =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad \quad squaring\quad both\quad sides\\ \Rightarrow \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+d^{ 2 }+2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow 2ac+2bd\quad \quad =\quad \quad 2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow ac+bd\quad \quad =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad again\quad squaring\quad both\quad sides\\ \Rightarrow { \left( ac \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }+2abcd\quad =\quad \left( { a }^{ 2 }+{ b }^{ 2 } \right) \left( { c }^{ 2 }+d^{ 2 } \right) \\ \Rightarrow { \left( ac \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }+2abcd\quad =\quad { \left( ac \right) }^{ 2 }+{ \left( ad \right) }^{ 2 }+{ \left( bc \right) }^{ 2 }+{ \left( bd \right) }^{ 2 }\\ \Rightarrow { \left( ad \right) }^{ 2 }+{ \left( bc \right) }^{ 2 }-2abcd\quad =\quad 0\\ \Rightarrow \left( ad-bc \right) ^{ 2 }\quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \Rightarrow ad-bc\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ad\quad \quad =\quad \quad \quad bc\quad \quad \quad \quad \longrightarrow \left( 1 \right) \\ Now\quad \quad \dfrac { { z }_{ 1 } }{ { z }_{ 2 } } \quad =\quad \dfrac { a+ib }{ c+id } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { \left( a+ib \right) \left( c-id \right) }{ \left( c+id \right) \left( c-id \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac-iad+ibc-{ i }^{ 2 }bd }{ { c }^{ 2 }+{ d }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd+i\left( bc-ad \right) }{ { c }^{ 2 }+{ d }^{ 2 } } \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd }{ { c }^{ 2 }+{ d }^{ 2 } } \quad +\quad 0\quad \quad \quad using\quad equation\quad \left( 1 \right) \\ Imaginary\quad part\quad is\quad zero\quad as\quad there\quad is\quad only\quad real\quad part\\ \therefore \quad Im\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =\quad 0$$
Question 8
1 / -0

The complex number $$z$$ satisfies $$z+|z|=2+8i$$. The value of $$|z|$$ is

A
$$10$$

B
$$13$$

C
$$17$$

D
$$23$$

Solution

$$Assume,z=a+ib,where \ i=\sqrt { -1 } $$

$$Then,z+|z|=a+ib+\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =2+8i$$

$$Thus,b=8.$$

$$And\quad a+\sqrt { { a }^{ 2 }+64 } =2$$

$${ a }^{ 2 }+64={ (2−a) }^{ 2 }$$

$${ a }^{ 2 }+64=4−4a+{ a }^{ 2 }$$

$$a=−15$$

$$Thus\quad z=−15+8i$$

$$Thus|z|=\sqrt { { (-15) }^{ 2 }+{ 8 }^{ 2 } } =17$$
Question 9
1 / -0

The number of prime numbers between $$1 \ and \ 10$$ is

A
$$12$$

B
$$4$$

C
$$3$$

D
$$2$$

Solution

The prime numbers between $$1$$ to$$ 10$$ is,
$$2,3,5,7,$$
There are four prime numbers in between $$1 $$ to $$10$$.
Question 10
1 / -0

if $$z_1=3+4i$$ and $$Im(z_1z_2)=0$$ Find $$z_2$$

A
$$z_2=3-4i$$

B
$$z_2=3+4i$$

C
$$z_2=3\pm 4i$$

D
None of these

Solution

Let $$z_2=a+ib$$

$$z_1z_2=(3+4i)(a+ib)$$

$$=(3a-4b)+i(4a+3b)$$

$$Im(z_1z_2)=0$$

$$\Rightarrow 4a+3b=0$$

$$\Rightarrow 4a=-3b$$

$$\dfrac ab=\dfrac 3{-4}$$

$$\implies a=3$$

$$b=-4$$

So $$z_2=3-4i$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 36

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Number Theory Test 36

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SHARING IS CARING

Question 1

$$20$$

$$24$$

$$48$$

$$120$$

Solution

Question 2

$$x = 2$$ & $$y = -8$$

$$x = -2$$ & $$y = 8$$

$$x = -2$$ & $$y = -6$$

$$x = 2$$ & $$y = 8$$

Solution

Question 3

straight line

circle with radius $$2$$

circle with radius $$\frac{{2\sqrt 2 }}{3}$$

none of these

Solution

Question 4

$$-3$$

$$4$$

$$3$$

$$-4$$

Solution

Question 5

The real axis

The line Im(z)$$=3$$

A circle

None of these

Solution

Question 6

$$0$$

$$1$$

$$2$$

$$4$$

Solution

Question 7

$$Re\left(\dfrac{z_1}{z_2}\right)=0$$

$$Im\left(\dfrac{z_1}{z_2}\right)=0$$

$$z_1+z_2=0$$

None

Solution

Question 8

$$10$$

$$13$$

$$17$$

$$23$$

Solution

Question 9

$$12$$

$$4$$

$$3$$

$$2$$

Solution

Question 10

$$z_2=3-4i$$

$$z_2=3+4i$$

$$z_2=3\pm 4i$$

None of these

Solution

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