Number Theory Test 35

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Number Theory Test 35

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Weekly Quiz Competition

Question 1
1 / -0

If $$i^2= -1$$, then $$1+ i^2+ i^4 +i^6+i^8 +.............to ( 2n +1)$$ terms is equal to

A
$$0$$

B
$$1$$

C
$$3i$$

D
$$4i$$

Solution

$$ 1+ i^2 + i^4 + ......(2n+1)terms$$
First term $$ a = i^2$$
Common ratio $$ r = i^2$$
$$ 1 + \cfrac{a(r^n-1)}{r-1}$$
$$ = 1 - \cfrac{i^2(i^{2(2n)} - 1)}{i-1}$$
$$ = 1 - \cfrac{i^2(i^4-1)}{i-1}$$ Since $$ i^4 = 1$$
$$ =1$$
Question 2
1 / -0

$$i \, \log \left(\dfrac{x - i}{x + i}\right)$$ is equal to

A
$$2i\log (x-i)-i\log (x^2+1)$$

B
$$2i\log (x-i)+i\log (x^2+1)$$

C
$$2i\log (x+i)-3i\log (x^2+1)$$

D
$$2i\log (x-i)-i\log (x^2+i)$$

Solution

$$i\log \left ( \cfrac{x-i}{x+i} \right )$$

$$=i\log \left ( \cfrac{x-i}{x+i}\times \cfrac{x-i}{x-i}\right )$$

$$=i\log \left ( \cfrac{(x-i)^{2}}{x^{2}-i^{2}} \right )$$

$$=i\log \left ( \cfrac{(x-i)^{2}}{x^{2}+1} \right )$$

$$=i\log (x-i)^{2}-i\log \left ( x^{2}+1 \right )$$

$$=2i\log (x-i)-i\log \left ( x^{2}+1 \right )$$
Question 3
1 / -0

If $$z = -3- i,$$ find $$|z|$$.

A
$$\sqrt{10}$$

B
$$\sqrt{9}$$

C
$$\sqrt{8}$$

D
$$\sqrt{7}$$

Solution

$$Re(z)=-3\\Im(z)=-1\\ \bar{z}=-3+i\\|z|=3^2+1^2=10$$
Question 4
1 / -0

If $$a+ ib= \sum_{k=1}^{101} i^k $$, then $$(a, b)$$ equals

A
$$(0, 1)$$

B
$$(1, 0)$$

C
$$(0,- 1)$$

D
$$(1, 1)$$

Solution

Given:-
$$a + ib = \Sigma^{101}_{k = 1}{{i}^{k}} \longrightarrow \left( 1 \right)$$

Solution:-

$$\Sigma^{101}_{k = 1}{{i}^{k}} = i + {i}^{2} + {i}^{3} + ............... + {i}^{101}$$

As te above series is in the form of G.P. with common ration $$(r)=i$$, first term $$\left( a \right) = i$$ and no. of terms $$\left( n \right) = 101$$

$$\therefore \; {S}_{101} = \cfrac{ i \left( 1 - {i}^{101} \right)}{\left( 1 - i \right)} \; \left\{ \because {S}_{n} = \cfrac{a \left( 1 - {r}^{n} \right)}{\left( 1 - r \right)}\; \text{ for } r < 1 \right\}$$

$$\Rightarrow \; {S}_{n} = \cfrac{i \left( 1 - {i}^{4 \times 25}.i \right)}{\left( 1 - i \right)}$$

$$\Rightarrow \; {S}_{n} = \cfrac{ i \left( 1 - i \right)}{\left( 1 - i \right)} \; \left\{ \because {i}^{4n} = 1 \right\}$$

$$\Rightarrow \; {S}_{n} = i$$

$$\therefore \; \Sigma^{101}_{k = 1}{{i}^{k}} = i \longrightarrow \left( 2 \right)$$

Now, from $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have,

$$a + ib = i$$

$$\Rightarrow \; a + ib = 0 + i.1$$

On comparing real and imaginary parts, we have

$$a = 0 \; \& \; b = 1$$

Hence, $$\left( a, b \right) = \left( 0, 1 \right)$$.
Question 5
1 / -0

If $$\left| {z - \dfrac{4}{z}} \right| = 2$$ , then the maximum value of$$\left| z \right|$$ is

A
$$\sqrt 5 $$

B
$$\sqrt 5 + 1$$

C
$$\sqrt 5 - 1$$

D
$$1 - \sqrt 5 $$

Solution
Question 6
1 / -0

$$\dfrac{{{{\left( {1 + i} \right)}^3}}}{{2 + i}}$$ is equal to

A
$$\dfrac{2}{5} - \dfrac{6}{5}i$$

B
$$0$$

C
$$ - \dfrac{1}{5} + \dfrac{6}{5}i$$

D
$$ - \dfrac{2}{5} + \dfrac{6}{5}i$$

Solution

$$\Rightarrow \cfrac { { \left( 1+i \right) }^{ 3 } }{ 2+i } =\cfrac { { \left( 1+i \right) }^{ 2 }\left( 1+i \right) }{ 2+i } =\cfrac { \left( 1+{ i }^{ 2 }+2i \right) \left( 1+i \right) }{ \left( 2+i \right) } =\cfrac { 2i\left( 1+i \right) }{ 2+i } $$
$$\Rightarrow \cfrac { 2\left( i-1 \right) \left( 2-i \right) }{ \left( 2+i \right) \left( 2-i \right) } =\cfrac { 2\left( 2i-{ i }^{ 2 }-2+i \right) }{ 4-{ i }^{ 2 } } =\cfrac { -2 }{ 5 } +\cfrac { 6 }{ 5 } i$$
Question 7
1 / -0

The modulus of the complex number $$z=\dfrac{1-i}{3-4i}$$ is

A
$$\dfrac{5}{\sqrt{2}}$$

B
$$\dfrac{\sqrt{2}}{5}$$

C
$$\sqrt{\dfrac{2}{5}}$$

D
none of these

Solution

Now,
$$z=\dfrac{1-i}{3-4i}$$
or, $$z=\dfrac{(1-i)(3+4i)}{3^2+4^2}$$
or, $$z=\dfrac{7+i}{25}$$.
Then $$|z|=\sqrt{\left(\dfrac{7}{25}\right)^2+\left(\dfrac{1}{25}\right)^2}=\dfrac{\sqrt{50}}{25}=\dfrac{\sqrt{2}}{5}$$.
Question 8
1 / -0

The real part of $$(1 - \cos\theta + 2i \sin\theta)^{-1}$$ is:

A
$$\dfrac{2\sin \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

B
$$\dfrac{1+\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

C
$$\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$$

D
$$\dfrac{1-\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

Solution

$$\Rightarrow \cfrac { 1 }{ \left( 1-\cos { \theta } \right) +2i\sin { \theta } } =\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }-4{ i }^{ 2 }\sin ^{ 2 }{ \theta } } $$
$$=\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$$
Question 9
1 / -0

Which of the following pair of numbers are $$\text{relatively prime}$$ :-

A
36 and 54

B
52 and 78

C
54 and 114

D
59 and 61

Solution

The question must be which of the following are relatively primes?
$$(a)$$HCF$$\left(36,54\right)$$
Factors of $$36=2\times 3\times 2\times 3$$
Factors of $$54=2\times 3\times 3\times 3$$
Common factor$$=2\times 3\times 3=18$$
Hence HCF$$=18$$
$$\therefore\,36$$ and $$54$$ are not relatively primes
$$(b)$$HCF$$\left(52,78\right)$$
Factors of $$52=2\times 2\times 13$$
Factors of $$78=2\times 3\times 13$$
Common factor$$=2\times 13=26$$
Hence HCF$$=26$$
$$\therefore\,52$$ and $$78$$ are not relatively primes
$$(c)$$HCF$$\left(54,114\right)$$
Factors of $$54=2\times 3\times 3\times 3$$
Factors of $$114=2\times 3\times 19$$
Common factor$$=2\times 3=6$$
Hence HCF$$=6$$
$$\therefore\,54$$ and $$114$$ are not relatively primes
$$(d)$$HCF$$\left(59,61\right)$$
Factors of $$59=1\times 59$$
Factors of $$61=1\times 61$$
Common factor$$=1$$
Hence HCF$$=1$$
$$\therefore\,59$$ and $$61$$ are relatively primes
As $$61-59=2$$ they are also twin primes.
Question 10
1 / -0

If $$z=\dfrac{1+i}{\sqrt{2}}$$, then the value of $$z^{1929}$$ is

A
$$1+i$$

B
$$-1$$

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

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

Solution

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

$$\Rightarrow |z|=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}}=1$$

$$\Rightarrow Arg=tan^{-1}(1)=\dfrac{\pi}{4}$$

$$\therefore z^{1929}=(1\cdot e^{\dfrac{\pi i}{4}})^{1929}$$

$$=e^{1929 \times \dfrac{\pi i}{4}}$$

$$=e^{i\left(482\pi+\dfrac{\pi}{4}\right)}$$

$$=e^{i\dfrac{\pi}{4}}$$

$$=\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}$$

$$\therefore z^{1929}=\dfrac{1+i}{\sqrt{2}}$$

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Number Theory Test 35

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Number Theory Test 35

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SHARING IS CARING

Question 1

$$0$$

$$1$$

$$3i$$

$$4i$$

Solution

Question 2

$$2i\log (x-i)-i\log (x^2+1)$$

$$2i\log (x-i)+i\log (x^2+1)$$

$$2i\log (x+i)-3i\log (x^2+1)$$

$$2i\log (x-i)-i\log (x^2+i)$$

Solution

Question 3

$$\sqrt{10}$$

$$\sqrt{9}$$

$$\sqrt{8}$$

$$\sqrt{7}$$

Solution

Question 4

$$(0, 1)$$

$$(1, 0)$$

$$(0,- 1)$$

$$(1, 1)$$

Solution

Question 5

$$\sqrt 5 $$

$$\sqrt 5 + 1$$

$$\sqrt 5 - 1$$

$$1 - \sqrt 5 $$

Solution

Question 6

$$\dfrac{2}{5} - \dfrac{6}{5}i$$

$$0$$

$$ - \dfrac{1}{5} + \dfrac{6}{5}i$$

$$ - \dfrac{2}{5} + \dfrac{6}{5}i$$

Solution

Question 7

$$\dfrac{5}{\sqrt{2}}$$

$$\dfrac{\sqrt{2}}{5}$$

$$\sqrt{\dfrac{2}{5}}$$

none of these

Solution

Question 8

$$\dfrac{2\sin \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

$$\dfrac{1+\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

$$\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$$

$$\dfrac{1-\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

Solution

Question 9

36 and 54

52 and 78

54 and 114

59 and 61

Solution

Question 10

$$1+i$$

$$-1$$

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

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

Solution

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