Number Theory Test 30

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Number Theory Test 30

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Weekly Quiz Competition

Question 1
1 / -0

Let Z and w be complex numbers. If $$Re(z)=|z-2|, Re(w) = |w-z|$$ and $$arg(z-w)=\dfrac{\pi}{3}$$, then the value of $$Im(z+w)$$, is

A
$$\dfrac{1}{\sqrt{3}}$$

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

C
$$\sqrt{3}$$

D
$$\dfrac{4}{\sqrt{3}}$$

Solution

$$Re\left( z \right) =\left| z-2 \right| $$
$$Re\left( w \right) =\left| w-z \right| $$
$$arg\left( z-w \right) =\cfrac { \pi }{ 3 } $$
Now,
$$Re\left( z+w \right) =\left| \left| z-2 \right| +\left| w-z \right| \right| $$
$$=\left| w-2 \right| =2$$
$$arg\left( z-w \right) =\cfrac { y }{ x } $$
$$\tan { \cfrac { \pi }{ 3 } } =\cfrac { Im\left( z+w \right) }{ Re\left( z+w \right) } $$
$$\Rightarrow Im\left( z+w \right) =\cfrac { 2 }{ \sqrt { 3 } } $$
Hence (B) is the correct answer
Question 2
1 / -0

If $$ z = \dfrac {-1}{2} + i \dfrac {\sqrt3}{2} $$, then $$ 8 + 10z + 7z^2 $$ is equal to :

A
$$ - \dfrac {1} {2} - i \dfrac {\sqrt3}{2} $$

B
$$ \dfrac {1} {2} + i \dfrac {\sqrt3}{2} $$

C
$$ - \dfrac {1} {2} + i \dfrac {3\sqrt3}{2} $$

D
$$ \dfrac {\sqrt3}{2} i $$

E
$$ - \dfrac {\sqrt3}{2} i $$

Solution

Given, $$ z = - \dfrac {1} + i \dfrac {\sqrt3}{2} $$
Therefore, $$z^2 = \dfrac {1}{4} - \dfrac {3}{4} - 2i \times \dfrac {1}{2} \dfrac {\sqrt3}{2} = - \dfrac {1}{2} - i \dfrac {\sqrt3}{2} $$
$$ \Rightarrow 8 + 10z + 7z^2 = 8 + 10 \left( - \dfrac {1}{2} + i \dfrac { \sqrt3}{2} \right) + 7 \left( - \dfrac {1}{2} + \dfrac { 3 \sqrt3}{2}i \right) $$
$$ = 8 - 5 + i 5 \sqrt3 - \dfrac {7}{2} - i \dfrac {7 \sqrt3}{2}$$
$$ = - \dfrac {1}{2} + \dfrac { 3 \sqrt3}{2} i$$
Question 3
1 / -0

If the complex numbers $$z_1, z_2$$ and $$z_3$$ denote the vertices of an isosceles triangle, right angled at $$z_1$$, then $$(z_1 - z_2)^2 + (z_1 - z_3)^2$$ is equal to

A
$$0$$

B
$$(z_2 + z_3)^2$$

C
$$2$$

D
$$3$$

E
$$(z_2 - z_3)^2$$

Solution

Since, the triangleis isosceles and right angled, we can say that
$$(z_3-z_1) = (z_2-z_1) (\cos 90^o + i \sin 90^o)$$
$$\Rightarrow (z_3-z_1) = i(z_2-z_1)$$
Squaring bothe sides, we get
$$(z_3-z_1)^2 = -(z_2-z_1)^2$$
$$\Rightarrow (z_3-z_1)^2 + (z_2-z_1)^2 = 0$$
Question 4
1 / -0

If $$z$$ is a complex number such that $$z + |z| = 8 + 12i$$, then the value of $$|z^{2}|$$ is

A
$$228$$

B
$$144$$

C
$$121$$

D
$$169$$

E
$$189$$

Solution

Let $$z = x + iy$$
Then, we have $$z + |z| = 8 + 12i$$
$$\Rightarrow(x + iy) + |x + iy| = 8 + 12l$$
$$\Rightarrow (x + \sqrt {x^{2} + y^{2}}) + iy = 8 + 12i$$
On comparing the real and imaginary part, we get
$$y = 12$$
and $$x + \sqrt {x^{2} + y^{2}} = 8$$
$$\Rightarrow \sqrt {x^{2} + 144} = 8 - x$$
On squaring both sides, we get
$$x^{2} + 144 = 64 + x^{2} - 16x$$
$$\Rightarrow 16x = -80$$
$$\Rightarrow x = -5$$
$$\therefore z = x + iy = -5 + i\cdot 12$$
Then, $$|z| = \sqrt {25 + 144} = \sqrt {169} = 13$$
$$\Rightarrow |z|^{2} = 169$$
$$\Rightarrow |z^{2}| = 169 (\because |z^{n}| = |z|^{n})$$.
Question 5
1 / -0

If $$iz^{3} + z^{2} - z + i = 0$$, then $$|z|$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
None of these

Solution

Given : $$iz^{3} + z^{2} - z + i = 0$$
Dividing both sides by $$i$$ and using $$\dfrac {1}{i} = -i$$.
We have,
$$z^{3} - iz^{2} + iz + 1 = 0$$
$$\Rightarrow z^{2}(z - i) + i(z - i) = 0$$ ...... $$(\because i^{2} = -1)$$
$$\Rightarrow (z - i)(z^{2} + i) = 0$$
$$\therefore z = i$$ or $$z^{2} = -i$$
$$\therefore |z| = |i| = 1$$
and $$|z|^{2}=|z^{2}| = |-i| = 1$$
$$\therefore |z| =1$$.
Question 6
1 / -0

The principal argument of the complex number $$z=\cfrac { 1+\sin { \cfrac { \pi }{ 3 } } +i\cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } -i\cos { \cfrac { \pi }{ 3 } } } $$ is?

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

B
$$\cfrac { \pi }{ 6 } $$

C
$$\cfrac { 2\pi }{ 3 } $$

D
$$\cfrac { \pi }{ 2 } $$

E
$$\cfrac { \pi }{ 4 } $$

Solution

$$arg(z)=arg(Numerator)-arg(Denominator)$$
$$=\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| } +\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| } $$
$$=2\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| } $$
$$=2\tan ^{ -1 }{ (2-\sqrt { 3 } } )=2\times {15}^{o}=\cfrac { \pi }{ 6 } $$
Question 7
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If '$$\omega$$' is a complex cube root of unity,then $$\omega ^{ \begin{pmatrix} \frac { 1 }{ 3 } & +\frac { 2 }{ 9 } +\frac { 4 }{ 27 } ...\infty \end{pmatrix} }+\omega^{ \begin{pmatrix} \frac { 1 }{ 2 } & +\frac { 3 }{ 8 } +\frac { 9 }{ 32 } ...\infty \end{pmatrix} }=$$

A
$$1$$

B
$$-1$$

C
$$\omega$$

D
$$i$$

Solution

$$\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{4}{27}+......+\infty$$ (infinte G.P.)

$$S_\omega = \dfrac{a}{1-r} =1$$

$$\dfrac{1}{2}+\dfrac{3}{8}+\dfrac{9}{32}+......+\infty$$ (infinte G.P.)

$$S_\omega = \dfrac{a}{1-r} =2$$

$$\therefore\, \omega^1+\omega^2=-1$$
Question 8
1 / -0

The inequality $$\left| z-4 \right| <\left| z-2 \right| $$ represents the region given by:

A
$$Re(z)\ge 0$$

B
$$Re(z)<0$$

C
$$Re(z)>0$$

D
None of these

Solution

We know that, $$\left| z-{ z }_{ 1 } \right| >\left| z-{ z }_{ 2 } \right| $$ represents the region on right side of perpendicular bisector of $${z}^{1}$$ and $${z}^{2}$$.
Thus, the given inequality $$\left| z-2 \right| >\left| z-4 \right| $$ represents the region on the right side of perpendicular bisector of $$2$$ and $$4$$.
$$\Rightarrow Re(z)>3$$ and $$\text{Im}(z)\in R$$
Question 9
1 / -0

Let$$ z$$ = $$\cos\theta + i \sin\theta$$. Then the value of $$\sum\limits_{m=1}^15Im( z^{2m-1})$$ at $$\theta = 2^0$$ is

A
$$\dfrac{1}{sin2^0}$$

B
$$\dfrac{1}{3sin2^0}$$

C
$$\dfrac{1}{2sin2^0}$$

D
$$\dfrac{1}{4sin2^0}$$

Solution

using the sum of series of sine
$$\sum_{m=1}^{n}sin(a+k.d)=\dfrac{sin\dfrac{n\times d}{2}}{sin\dfrac{d}{2}}\times sin(\dfrac{2a+(n-1).d}{2})$$
Given,
$$\sum_{m=1}^{15}Im(z^{2m-1})\rightarrow$$

$$sin\theta+sin3\theta+sin5\theta+.......sin29\theta=$$

$$\rightarrow\dfrac{sin\dfrac{15\times 2\theta}{2}}{sin\dfrac{2\theta}{2}}\times sin(\dfrac{2\theta+(14\times2\theta)}{2})$$

$$\rightarrow\dfrac{sin^215\theta}{sin\theta}$$

put $$\theta = 2^{\circ}$$

$$\rightarrow\dfrac{sin^230^{\circ}}{sin2^{\circ}}$$

$$\rightarrow\dfrac{1}{4.sin2^{\circ}}$$
Question 10
1 / -0

If $$z_1, z_2$$ are two complex numbers such that $$arg(z_1+z_2)=0$$ and $$Im(z_1z_2)=0$$, then.

A
$$z_1=-z_2$$

B
$$z_1=z_2$$

C
$$z_1=\vec{z_2}$$

D
none of these

Solution

Let $$ z_1=a+ib$$
and $$z_2=c+id$$
Now $$z_1+z_2=(a+c)+i(b+d)$$
Given $$arg(z_1+z_2)=0$$
$$\Rightarrow tan^{-1}[\dfrac{b+d}{a+c}]=0$$
$$\Rightarrow b+d=0$$.........$$(1)$$
Now,
$$z_1z_2=(ac-bd)+i(ad+bc)$$
Given $$Im(z_1z_2)=0$$
$$\Rightarrow ad+bc=0$$.........$$(2)$$
Putting $$b=-d$$ from $$(1)$$ in $$(2)$$ we get $$a=c$$
Hence $$z_1=a+ib=c-id= $$conjugate of $$z_2=z_2^{\rightarrow}$$
Hence option C is correct.

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Number Theory Test 30

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Number Theory Test 30

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Question 1

$$\dfrac{1}{\sqrt{3}}$$

$$\dfrac{2}{\sqrt{3}}$$

$$\sqrt{3}$$

$$\dfrac{4}{\sqrt{3}}$$

Solution

Question 2

$$ - \dfrac {1} {2} - i \dfrac {\sqrt3}{2} $$

$$ \dfrac {1} {2} + i \dfrac {\sqrt3}{2} $$

$$ - \dfrac {1} {2} + i \dfrac {3\sqrt3}{2} $$

$$ \dfrac {\sqrt3}{2} i $$

$$ - \dfrac {\sqrt3}{2} i $$

Solution

Question 3

$$0$$

$$(z_2 + z_3)^2$$

$$2$$

$$3$$

$$(z_2 - z_3)^2$$

Solution

Question 4

$$228$$

$$144$$

$$121$$

$$169$$

$$189$$

Solution

Question 5

$$0$$

$$1$$

$$2$$

None of these

Solution

Question 6

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

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

$$\cfrac { 2\pi }{ 3 } $$

$$\cfrac { \pi }{ 2 } $$

$$\cfrac { \pi }{ 4 } $$

Solution

Question 7

$$1$$

$$-1$$

$$\omega$$

$$i$$

Solution

Question 8

$$Re(z)\ge 0$$

$$Re(z)<0$$

$$Re(z)>0$$

None of these

Solution

Question 9

$$\dfrac{1}{sin2^0}$$

$$\dfrac{1}{3sin2^0}$$

$$\dfrac{1}{2sin2^0}$$

$$\dfrac{1}{4sin2^0}$$

Solution

Question 10

$$z_1=-z_2$$

$$z_1=z_2$$

$$z_1=\vec{z_2}$$

none of these

Solution

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