Number Theory Test 48

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Number Theory Test 48

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Question 1
1 / -0

The number of complex numbers $$z$$ such that $$\left|z\right| = 1$$ and $$\left|z/\overline{z} + \overline{z}/z\right| = 1$$ is $$(arg(z) \in [0, 2\pi))$$

A
4

B
6

C
8

D
more than 8

Solution
Question 2
1 / -0

Consider a complex number $$z$$ which satisfies the equation $$\left |z-\left (\displaystyle\frac{4}{z}\right )\right |=2$$, then the absolute difference between the least and the greatest moduli of complex numbers is,

A
$$2$$

B
$$4$$

C
$$1$$

D
$$3$$

Solution

We have
$$\left ||z|-\left |\dfrac {4}{z}\right |\right | \leq \left |z-\dfrac {4}{z}\right |=2$$

$$\Rightarrow -2\leq |z|-\dfrac {4}{|z|}\leq 2$$

$$\Rightarrow |z|^2+2|z|-4\geq 0$$ and $$|z|^2-2|z|-4\leq 0$$

$$\Rightarrow (|z|+1)^2-5\geq 0$$ and $$(|z|-1)^2 \leq 5$$

$$\Rightarrow (|z|+1+\sqrt 5)(|z|+1-\sqrt 5) \geq 0$$

and

$$(|z|-1+\sqrt 5)\times (|z|-1-\sqrt 5) \leq 0$$

$$\Rightarrow |z|\leq -\sqrt 5-1$$ or $$|z| \geq \sqrt 5-1$$

and

$$\sqrt 5-1 \leq |z|\leq \sqrt 5+1$$

$$\Rightarrow \sqrt 5-1 \leq |z|\leq \sqrt 5+1$$

Hence, the least modulus is $$\sqrt 5-1$$ and the greatest modulus is $$\sqrt 5+1$$

Hence required difference $$=2$$
Question 3
1 / -0

Find the value of $$x$$ such that $$\displaystyle \frac{(x + \alpha)^2 - (x + \beta)^2}{ \alpha + \beta} = \frac{sin 2 \theta}{sin^2 \theta}$$. when $$\alpha$$ and $$\beta $$ are the roots of $$t^2 - 2t + 2 = 0$$

A
$$x = icot \, \, \, \theta - 1$$

B
$$x = -(icot \, \, \, \theta + 1$$)

C
$$x = icot \, \, \, \theta $$

D
$$x = itan \, \, \, \theta - 1$$

Solution

$$t^{2}-2t+2=0$$
$$(t-1)^{2}-1+2=0$$
$$(t-1)^{2}=-1$$
$$t=1\pm i$$
Hence
$$\alpha+\beta=2$$
$$\alpha-\beta=2i$$
Now let $$n=2$$
We get
$$\dfrac{2x(\alpha-\beta)+\alpha^{2}-\beta^{2}}{2}=\dfrac{sin2\theta}{sin^{2}\theta}$$
Hence
$$\dfrac{4ix+2i-(-2i)}{2}=\dfrac{2sin\theta.cos\theta}{sin^{2}\theta}$$

$$2ix+2i=2cot\theta$$
$$ix=cot\theta-i$$
$$-x=icot\theta+1$$
$$x=-(1+icot\theta)$$
Question 4
1 / -0

If $$z_1, z_2$$ be two non zero complex numbers satisfying the equation $$\displaystyle \left | \frac{z_1 + z_2}{z_1 - z_2} \right | = 1$$ then $$\displaystyle \frac{z_1}{z_2} + \left ( \frac{z_1}{z_2} \right )$$ is

A
zero

B
1

C
purely imaginary

D
2

Solution

let $$\frac{z_{1}}{z_{2}} =x$$
given $$|\frac{x+1}{x-1}| = 1$$
$$\Rightarrow |x+1|=|x-1|$$
Let $$x=a+ib$$
We get $$|a+1+ib|=|a-1+ib|$$
$$\Rightarrow a=0$$
Therefore $$x=ib$$ and $$\overline { x } =-ib$$
Therefore $$x+\overline { x } =ib-ib=0$$
So the correct option is $$A$$
Question 5
1 / -0

Find the range of real number $$\alpha$$ for which the equation $$z + \alpha |z - 1| + 2i = 0; z= x + iy$$ has a solution. Find the solution.

A
$$\displaystyle x = 5/2, y = - 2$$

B
$$\displaystyle x = -2, y = 5/2$$

C
$$\displaystyle x = -5/2, y = 2$$

D
$$\displaystyle x = 2, y = -5/2$$

Solution

$$x+iy+\alpha(\sqrt{(x-1)^{2}+y^{2}})+2i=0$$
$$x+i(y+2)=\alpha(\sqrt{(x-1)^{2}+y^{2}}$$
$$x^{2}-(y+2)^{2}+i2x(y+2)=\alpha^{2}((x-1)^{2}+y^{2})$$
Real part
$$x^{2}-\alpha(x-1)^{2}-\alpha(y^{2})=0$$ and
Imaginary part
$$2x(y+2)=0$$
$$x(y+2)=0$$
Either $$x=0 $$or $$y=-2$$ ...if the above equations have a solution.
Considering y=-2,
Hence
$$\alpha\epsilon[-\dfrac{\sqrt{5}}{2},\dfrac{\sqrt{5}}{2}]$$
Hence
$$x=\dfrac{5}{2}$$ if $$|\alpha|\neq1$$.
Question 6
1 / -0

Find the regions of the z-plane for which $$\displaystyle \left | \frac{z - a}{z + \overline a} \right | < 1, = 1$$ or $$> 1$$. when the real part of a is positive.

A
The required regions are the right half of the z-pane, the imaginary axis and the left half of the z-plane respectively.

B
The required regions are the left half of the z-pane, the imaginary axis and the right half of the z-plane respectively.

C
The required regions are the right half of the z-pane the real axis and the left half of the z-plane respectively.

D
The required regions are the left half of the z-pane the real axis and the right half of the z-plane respectively.

Solution

Let
$$z=x+iy$$ and $$a$$ be a fixed complex number $$(a+ib)$$.
Hence
$$|\dfrac{z-a-ib}{z+a-ib}|=1$$
$$|(x-a)+i(y-b)|=|(x+a)+i(y-b)|$$
$$(x-a)^{2}=(x+a)^{2}$$
$$2x(a)=0$$
$$x=0$$ ... imaginary axis.
If
$$|\dfrac{z-a-ib}{z+a-ib}|<1$$
Then
$$(x-a)^{2}<(x+a)^{2}$$
Or
$$x>0$$ ... positive real axis.
If
$$|\dfrac{z-a-ib}{z+a-ib}|>1$$
$$(x-a)^{2}>(x+a)^{2}$$
$$x<0$$ ... negative real axis.
Question 7
1 / -0

Find all complex numbers satisfying the equation $$2|z|^2 + z^2 - 5 + i \sqrt{3} = 0$$

A
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} + \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} + \frac{3}{2} i \right )$$

B
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{2}{\sqrt{6}} - \frac{3}{2} i \right )$$

C
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{3}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{2} i \right )$$

D
$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{\sqrt{2}} i \right )$$

Solution

$$Let\quad z=a+ib\\ Thus:\quad 2({ a }^{ 2 }+{ b }^{ 2 })+({ a }^{ 2 }-{ b }^{ 2 }+2abi)-5+i\sqrt { 3 } =0\\ =>3{ a }^{ 2 }+{ b }^{ 2 }-5=0\quad and\quad 2ab+\sqrt { 3 } =0=>a=-\frac { \sqrt { 3 } }{ b } \\ Substituting\quad in\quad the\quad first:\quad \frac { 9 }{ 4{ b }^{ 2 } } +{ b }^{ 2 }=5=>{ b }^{ 2 }=\frac { 9 }{ 2 } \quad or\quad \frac { 1 }{ 2 } \\ =>b=\pm \frac { 3 }{ \sqrt { 2 } } \quad or\quad \pm \frac { 1 }{ \sqrt { 2 } } =>a=\mp \frac { 1 }{ \sqrt { 6 } } \quad or\quad \mp \frac { \sqrt { 3 } }{ \sqrt { 2 } } =\mp \frac { \sqrt { 6 } }{ 2 } \\ =>z=\pm \left( \frac { \sqrt { 6 } }{ 2 } -\frac { i }{ \sqrt { 2 } } \right) \quad or\quad \pm \left( \frac { 1 }{ \sqrt { 6 } } -\frac { 3i }{ \sqrt { 2 } } \right) \\ $$
Hence, (d) is correct.
Question 8
1 / -0

Find the real values of the parameter $$a$$ for which at least one complex number $$z = x + iy$$ satisfies both the equality $$|z + \sqrt{2}| = a^2 - 3a + 2$$ and the inequality $$|z + i \sqrt{2}| < a^2$$.

A
$$ a \in (2, \infty)$$

B
$$ a \in (-\infty, 2]$$

C
$$ a \in [2, \infty)$$

D
$$ a \in (-\infty, 2)$$

Solution

Let $$z = x + iy$$
$$\Rightarrow |z+\sqrt{2}|=\sqrt{x^2 + 2\sqrt{2}x + 2 + y^2} = a^2 - 3a + 2$$ .................(1)

Also,
$$|z+i\sqrt{2}|=\sqrt{y^2 + 2\sqrt{2}y + 2 + x^2} < a^2$$..........................(2)

Since, modulus of a complex number is greater than zero,
Hence, $$a^2 - 3a + 2 > 0$$ $$[\text {from (1)}]$$
$$\Rightarrow (a-2)(a-1)>0$$
Solving the inequality, we get,
$$\Rightarrow a \in (-\infty, 1) \cup (2,\infty)$$ ... (3)

and because of (2), $$a$$ cannot be zero
So, $$a≠0$$ .... (4)

The distance between $$-\sqrt2$$ and $$-i\sqrt2$$ is $$2$$. So, for the first circle to overlap with the interior of the second circle the following two inequalities need to hold:

$$r_1 + r_2 > 2$$
$$\Rightarrow 2a^2 - 3a + 2 > 2$$
$$\Rightarrow \left(a-\dfrac34\right)^2 > \dfrac9{16}$$

$$\Rightarrow a\in(-\infty, 0) \cup \left(\dfrac32,\infty\right)$$ ... (5)

and $$r_1 < r_2 + 2$$
$$a^2 - 3a + 2 < a^2 + 2$$
$$a>0$$ ... (6)

From (3), (4), (5), (6),
We can say that
$$a>2$$
Question 9
1 / -0

If z be a complex number satisfying$$\displaystyle\ z^{4}+z^{3}+2z^{2}+z+1=0$$ then $$\displaystyle\ |z|$$ is

A
$$\displaystyle\ \frac{1}{2}$$

B
$$\displaystyle\ \frac{3}{4}$$

C
$$\displaystyle\ 1$$

D
None of these

Solution

$${ z }^{ 4 }+{ z }^{ 3 }+2{ z }^{ 2 }+z+1=0$$

$$\left[ { z }^{ 4 }+2{ z }^{ 2 }+1 \right] +\left[ { z }^{ 3 }+z \right] =0$$

$$\left[ { z }^{ 2 }+1 \right] \left[ { z }^{ 2 }+1+z \right] =0$$

$${ z }^{ 2 }+1=0\Longrightarrow z=\pm i$$

$$\left| z \right| =1$$
or
$${ z }^{ 2 }+z+1=0$$

$$z=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } =\cfrac { -1\pm \sqrt { 3i } }{ 2 } $$

$$\left| z \right| =\sqrt { \cfrac { 1 }{ 4 } +\cfrac { 3 }{ 4 } } =1$$
$$\therefore \boxed { \left| z \right| =1 } $$
Question 10
1 / -0

$$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }=$$

A
$$-2$$

B
$$2$$

C
$$-1$$

D
$$1$$

Solution

We have $$\displaystyle \frac { \sqrt { 3 } +i }{ 2 } =\frac { i\sqrt { 3 } +{ i }^{ 2 } }{ 2i } $$ (Here we multiplied the numerator and denominator by $$(-i)$$ on the LHS)
We get: $$-i\left( \frac { -1+\sqrt { 3 } i }{ 2 } \right) =i\omega $$
Simliarly applying the same idea we get the steps below:
and $$\displaystyle \frac { i-\sqrt { 3 } }{ 2 } =\frac { { i }^{ 2 }-i\sqrt { 3 } }{ 2i } =-i\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) =-i{ \omega }^{ 2 }$$
Hence $$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }={ \left( -i\omega \right) }^{ 6 }+{ \left( -i{ \omega }^{ 2 } \right) }^{ 6 }\\ \\ ={ i }^{ 6 }\left( { \omega }^{ 6 }+{ \omega }^{ 12 } \right) =-1\left( 1+1 \right) =-2$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 48

Result

Number Theory Test 48

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SHARING IS CARING

Question 1

4

6

8

more than 8

Solution

Question 2

$$2$$

$$4$$

$$1$$

$$3$$

Solution

Question 3

$$x = icot \, \, \, \theta - 1$$

$$x = -(icot \, \, \, \theta + 1$$)

$$x = icot \, \, \, \theta $$

$$x = itan \, \, \, \theta - 1$$

Solution

Question 4

zero

1

purely imaginary

2

Solution

Question 5

$$\displaystyle x = 5/2, y = - 2$$

$$\displaystyle x = -2, y = 5/2$$

$$\displaystyle x = -5/2, y = 2$$

$$\displaystyle x = 2, y = -5/2$$

Solution

Question 6

The required regions are the right half of the z-pane, the imaginary axis and the left half of the z-plane respectively.

The required regions are the left half of the z-pane, the imaginary axis and the right half of the z-plane respectively.

The required regions are the right half of the z-pane the real axis and the left half of the z-plane respectively.

The required regions are the left half of the z-pane the real axis and the right half of the z-plane respectively.

Solution

Question 7

$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} + \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} + \frac{3}{2} i \right )$$

$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{2}{\sqrt{6}} - \frac{3}{2} i \right )$$

$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{3}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{2} i \right )$$

$$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{\sqrt{2}} i \right )$$

Solution

Question 8

$$ a \in (2, \infty)$$

$$ a \in (-\infty, 2]$$

$$ a \in [2, \infty)$$

$$ a \in (-\infty, 2)$$

Solution

Question 9

$$\displaystyle\ \frac{1}{2}$$

$$\displaystyle\ \frac{3}{4}$$

$$\displaystyle\ 1$$

None of these

Solution

Question 10

$$-2$$

$$2$$

$$-1$$

$$1$$

Solution

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