Number Theory Test 1

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Number Theory Test 1

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\alpha,\ \beta$$ be real and $$\mathrm{z}$$ be a complex number. If $$\mathrm{z}^{2}+\alpha \mathrm{z}+\beta=0$$ has two distinct roots on the line $$Re(z) =1$$, then it is necessary that:

A
$$\beta\in(0,1)$$

B
$$\beta\in(-1,0)$$

C
$$|\beta|=1$$

D
$$\beta\in(1, \infty)$$
Question 2
1 / -0

Let z be a complex number such that $$\left|\dfrac{z-i}{z+2i}\right|=1$$ and $$|z|=\dfrac{5}{2}$$. Then the value of $$|z+3i|$$ is?

A
$$\dfrac{7}{2}$$

B
$$\dfrac{15}{4}$$

C
$$2\sqrt{3}$$

D
$$\sqrt{10}$$

Solution

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

$$\Rightarrow |z - i| = |z + 2i|$$

Let $$z = x + iy$$

So, $$x^2 + (y- 1)^2 = x^2 + (y + 2)^2$$

$$-2y + 1 = 4y + 4$$

$$6y = -3 \Rightarrow y = \dfrac{-1}{2}$$

$$x^2 + y^2 = \dfrac{25}{4} \Rightarrow x^2 = \dfrac{24}{4} = 6$$

$$z = \pm \sqrt{6} - \dfrac{i}{2}$$

$$|z + 3i| = \sqrt{6 + \dfrac{25}{4}} = \sqrt{\dfrac{49}{4}} = \dfrac{7}{2}$$

$$\therefore$$ $$\boxed{|2 + 3i| = \dfrac{7}{2}}.....Answer$$

Hence option $$'A'$$ is the answer.
Question 3
1 / -0

If $$ z$$ is a complex number such that $$ |z|\geq 2$$, then the minimum value of $$ |z+\displaystyle \frac{1}{2}|$$

A
is equal to $$ \displaystyle \frac{5}{2}$$

B
lies in the interval $$ \left ( 1,2 \right )$$

C
is strictly greater than $$ \displaystyle \frac{5}{2}$$

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

Solution

Given $$ |z|\geq 2$$

$$\therefore |z+\displaystyle \frac{1}{2}|\geq \left | |z|-|\frac{1}{2}|\right |$$ $$[\because | {z_1+z_2}|\geq| z_1|-| z_2|]$$

$$\therefore |z+\displaystyle \frac{1}{2}|\geq |2-\displaystyle \frac{1}{2}|\geq \frac{3}{2}$$

Hence,minimum value of $$ |z+\displaystyle \frac{1}{2}|$$ is $$ \displaystyle \frac{3}{2}$$ which indicates that it lies in the interval $$(1,2)$$
Question 4
1 / -0

If $$z = x + iy$$ and $$\omega = \dfrac{(1 -iz)}{(z-i)}$$, then $$\left|\omega\right| = 1$$ implies that in the complex plane

A
z lies on the imaginary axis

B
z lies on the real axis

C
z lies on the unit circle

D
none of these

Solution

Given $$w=\dfrac { 1-iz }{ z-i } $$ and

$$\left| w \right| =1$$

$$\Rightarrow \left| \dfrac { 1-iz }{ z-i } \right| =1$$ ...(1)

Substitute $$z=x+iy$$ in equation (1)

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

$$\Rightarrow \left| 1+y-ix \right| =\left| x+i\left( y-1 \right) \right| \\ \Rightarrow { \left( 1+y \right) }^{ 2 }+{ x }^{ 2 }={ x }^{ 2 }+{ \left( y-1 \right) }^{ 2 }\\ \Rightarrow y=0$$

Therefore z lies on the real axis.

Ans: B
Question 5
1 / -0

Let z be a complex number such that the imaginary part of z is nonzero and a = $$z^2 + z + 1$$ is real. Then a cannot take the value

A
$$-1$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{1}{2}$$

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

Solution

Let $$z=\alpha+i\beta$$. Then

$$Im(z^2+z+1)=2\alpha\beta+\beta=0 $$

$$\Rightarrow \alpha=-\dfrac{1}{2}$$ since $$\beta\neq0$$. Then

$$a=\alpha^2-\beta^2+\alpha+1$$

$$a=\dfrac{3}{4}+\beta$$

$$a\neq\dfrac{3}{4}$$. $$[\because \beta\neq0]$$
Question 6
1 / -0

$$\displaystyle \left|\dfrac{\sqrt{3}+i}{(1+i)(1+\sqrt{3}i)}\right|=$$

A
$$1$$

B
$$\sqrt{2}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{1}{\sqrt{2}}$$

Solution

The value of $$\displaystyle \left |\frac{\sqrt{3} + i}{(1 + i)(1 + \sqrt{3}i)} \right |$$
$$ = \dfrac{|\sqrt{3} + i|}{|1 + i||1 + \sqrt{3}i|} = \dfrac{\sqrt{3 + 1}}{\sqrt{1 + 1} \times \sqrt{1 + 3}} $$
$$= \dfrac{2}{2\sqrt{2}} $$
$$= \dfrac{1}{\sqrt{2}}$$
Question 7
1 / -0

The modulus of $$\sqrt{2}i-\sqrt{-2}i$$ is:

A
2

B
$$\sqrt{2}$$

C
0

D
$$2\sqrt{2}$$

Solution

Let $$z=\sqrt{2}i-\sqrt{-2}i$$
$$\Rightarrow z=\sqrt{2}i-\sqrt{2}i^2$$
$$\Rightarrow z=\sqrt{2}+\sqrt{2}i$$
Now, $$|z|=\sqrt{2+2}=2$$
Question 8
1 / -0

If $$z =3+5i$$, then $$z^3+z+198=$$

A
$$3 - 15i$$

B
$$-3 - 15i$$

C
$$-3 + 15i$$

D
$$3 + 15i$$

Solution

$$z=3+5i$$

$$z^{3}=(3+5i)^{3}$$

$$=3^{3}+3.3^{2}(5i)+3.3(5i)^{2}+(5i)^{3}$$

$$=27-125i+135i-225$$

$$=-225+27+(135-125)i$$

$$=-198+10i$$

$$\therefore z^{3}+z+198$$

$$=-198+10i+3+5i+198$$

$$=3+15i$$
Question 9
1 / -0

If $$z=2-3i$$ then $$z^2-4z+13=$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$z=2-3i$$
$$z^{2}=2^{2}-3^{2}-12i$$
$$=-5-12i$$
$$\therefore z^{2}-4z+13$$
$$=(-5-12i)-4(2-3i)+13$$
$$=-5-12i-8+12i +13$$
$$=-13+13$$
$$=0$$
Question 10
1 / -0

The complex number $$\displaystyle \frac{1+2i}{1-i}$$ lies in the quadrant :

A
I

B
II

C
III

D
IV

Solution

Let $$z =\dfrac{1+2i}{1-i}$$

$$\Rightarrow z =\dfrac{(1+2i)}{1-i}\times \dfrac{1+i}{1+i}$$

$$= \dfrac{1+2i+i+2i^2}{1-i^2}$$

$$=\dfrac{1+3i-2}{1+1}$$ ............ $$[\because i^2 = -1]$$

$$\Rightarrow z =\dfrac{-1+3i}{2}$$

$$ =-\dfrac{1}{2}+\dfrac{3}{2}i $$
$$=x+iy$$
Clearly $$x<0$$ and $$y>0$$
Hence $$z$$ lies in $$\text{II}$$ quadrant

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Number Theory Test 1

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Number Theory Test 1

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SHARING IS CARING

Question 1

$$\beta\in(0,1)$$

$$\beta\in(-1,0)$$

$$|\beta|=1$$

$$\beta\in(1, \infty)$$

Question 2

$$\dfrac{7}{2}$$

$$\dfrac{15}{4}$$

$$2\sqrt{3}$$

$$\sqrt{10}$$

Solution

Question 3

is equal to $$ \displaystyle \frac{5}{2}$$

lies in the interval $$ \left ( 1,2 \right )$$

is strictly greater than $$ \displaystyle \frac{5}{2}$$

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

Solution

Question 4

z lies on the imaginary axis

z lies on the real axis

z lies on the unit circle

none of these

Solution

Question 5

$$-1$$

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

$$\dfrac{1}{2}$$

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

Solution

Question 6

$$1$$

$$\sqrt{2}$$

$$\dfrac{1}{2}$$

$$\dfrac{1}{\sqrt{2}}$$

Solution

Question 7

2

$$\sqrt{2}$$

0

$$2\sqrt{2}$$

Solution

Question 8

$$3 - 15i$$

$$-3 - 15i$$

$$-3 + 15i$$

$$3 + 15i$$

Solution

Question 9

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 10

I

II

III

IV

Solution

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