Number Theory Test 13

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Number Theory Test 13

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Weekly Quiz Competition

Question 1
1 / -0

How many two-digit prime numbers are there having the digit $$3$$ in their units place?

A
$$10$$

B
$$8$$

C
$$6$$

D
$$5$$

Solution

$$2$$ digit prime nos. having $$3$$ in their units place are-

$$13,23,43,53,73$$

$$\therefore$$ There are $$5$$ such nos.
Question 2
1 / -0

Argument and modulus of $$\left[\dfrac {1+i}{1-i}\right]^{2013}$$ are respectively ____

A
$$\dfrac {-\pi}{2}$$ and $$1$$

B
$$\dfrac {\pi}{2}$$ and $$\sqrt {2}$$

C
$$0$$ and $$\sqrt {2}$$

D
$$\dfrac {\pi}{2}$$ and $$1 $$

Solution

Let $$P={ [\cfrac { 1+i }{ 1-i } ] }^{ 2013 }={ [\cfrac { \sqrt { 2 } { e }^{ \cfrac { i\pi }{ 4 } } }{ \sqrt { 2 } { e }^{ \cfrac { -i\pi }{ 4 } } } ] }^{ 2013 }=({ e }^{ \cfrac { i2013\pi }{ 2 } })$$
$$P=\cos { (2013\cfrac { \pi }{ 2 } ) } +i\sin { (2013\cfrac { \pi }{ 2 } ) } =i\sin { (-\cfrac { \pi }{ 2 } ) } $$
So, argument $$=-\cfrac { \pi }{ 2 } $$
and modulus$$=1$$
Question 3
1 / -0

If $$z_1=\sqrt { 3 } -i,z_2=1+i\sqrt { 3 } ,$$ then amp$$(z_1+z_2)=$$

A
$$\dfrac { \pi }{ 12 } $$

B
$$\dfrac { \pi }{ 15 } $$

C
$$\dfrac { \pi }{ 6 } $$

D
$$\dfrac { \pi }{ 4 } $$

Solution

$$z_1=\sqrt { 3 } -i,z_2=1+i\sqrt { 3 } $$
$$ z_1+z_2=(\sqrt { 3 }+1)+i(\sqrt{3}-1) $$
$$ z_1+z_2=(\sqrt { 3 }+1)+i(\sqrt{3}-1) $$
$$\left| z_1+z_2 \right|= \sqrt {(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}=2\sqrt{2}$$
Question 4
1 / -0

If $$z=\dfrac {1+2i}{1-2i}$$ and.consider $$z=x+iy$$ then which of the following relationship is correct:

A
$$x^{2}+y^{2}=1$$

B
$$x^{2}+y^{2}=4$$

C
$$x^{2}+y^{2}=2$$

D
$$x^{2}+y^{2}=3$$

Solution
Question 5
1 / -0

Find the least positive value of n, if $$(\dfrac{1+i}{1-i})^n=1$$

A
1

B
2

C
3

D
4

Solution

We have,
$$\dfrac{1+i}{1-i} =\dfrac{1+i}{1-i} \times \dfrac{1+i}{1+i}=\dfrac{(1+i)^2}{1-i^2}=\dfrac{1+2i+i^2}{1-i^2}=\dfrac{1+2i-1}{1+1}=i$$

Therefore,
$$(\dfrac{1+i}{1-i})^n=1$$

$$i^n=1$$
$$ \implies$$ n is a multiple of 4.
The smallest positive value of n is 4.
Question 6
1 / -0

Inequality $$a + i b > c + i d$$ can be explained only when :

A
$$b=0,c=0$$

B
$$b=0,d=0$$

C
$$a=0,c=0$$

D
$$a=0,d=0$$

Solution

$$\begin{array}{l} a+ib>c+id \\ is\, \, valid\, \, only\, \, when\, \, b=0\, \, \& \, \, d=0\, \, as\, \, inequality\, \, \, in \\ complex\, no.\, \, is\, \, not\, \, defined. \end{array}$$
Question 7
1 / -0

If $$z_1=3+4i\\z_2=4-5i$$ Then find $$z_1+z_2$$

A
7-i

B
7+i

C
7+9i

D
None of these

Solution

$$z_1=3+4i\\z_2=4-5i\\z_1+z_2=3+4i+4-5i\\7-i$$
Question 8
1 / -0

If $$z_1=3+4i,z_2=2-i$$ find $$z_2-z_1$$

A
-1-5i

B
2-5i

C
1+5i

D
1-5i

Solution

Given $$z_1=3+4i\\z_2=2-i\\z_2-z_1=2-i-3-4i=-1-5i$$
Question 9
1 / -0

The complex numbers $$z_1=8+9i, z_2=4-6i$$ then $$z_1-z_2$$

A
$$4+15i$$

B
$$4-3i$$

C
$$12+3i$$

D
$$12-15i$$

Solution

$$z_1=8+9i\\z_2=4-6i\\z_1-z_2=8+9i-4+6i=4+15i$$
Question 10
1 / -0

If $$z$$ is a complex number such that $$|z|=1$$, then $$\left|\dfrac 1{\bar z}\right|$$ is

A
$$0$$

B
$$-1$$

C
$$\sqrt{2}$$

D
$$1$$

Solution

Let $$z=x+iy\\|z|=\sqrt{x^2+y^2}=1\\x^2+y^2=1\\\dfrac 1z=\dfrac1{x+iy}\\\dfrac{x-iy}{\sqrt{x^2+y^2}}=x-iy\\\left|\dfrac 1z\right|=\sqrt{x^2+y^2}=1$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 13

Result

Number Theory Test 13

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SHARING IS CARING

Question 1

$$10$$

$$8$$

$$6$$

$$5$$

Solution

Question 2

$$\dfrac {-\pi}{2}$$ and $$1$$

$$\dfrac {\pi}{2}$$ and $$\sqrt {2}$$

$$0$$ and $$\sqrt {2}$$

$$\dfrac {\pi}{2}$$ and $$1 $$

Solution

Question 3

$$\dfrac { \pi }{ 12 } $$

$$\dfrac { \pi }{ 15 } $$

$$\dfrac { \pi }{ 6 } $$

$$\dfrac { \pi }{ 4 } $$

Solution

Question 4

$$x^{2}+y^{2}=1$$

$$x^{2}+y^{2}=4$$

$$x^{2}+y^{2}=2$$

$$x^{2}+y^{2}=3$$

Solution

Question 5

1

2

3

4

Solution

Question 6

$$b=0,c=0$$

$$b=0,d=0$$

$$a=0,c=0$$

$$a=0,d=0$$

Solution

Question 7

7-i

7+i

7+9i

None of these

Solution

Question 8

-1-5i

2-5i

1+5i

1-5i

Solution

Question 9

$$4+15i$$

$$4-3i$$

$$12+3i$$

$$12-15i$$

Solution

Question 10

$$0$$

$$-1$$

$$\sqrt{2}$$

$$1$$

Solution

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