Number Theory Test 29

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Number Theory Test 29

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Weekly Quiz Competition

Question 1
1 / -0

The expression $$\dfrac {(1 + i)^{n}}{(1 - i)^{n - 2}}$$ equals.

A
$$-i^{n + 1}$$

B
$$i^{n + 1}$$

C
$$-2i^{n + 1}$$

D
$$1$$

Solution

$$(1 - i) = (1-i)\times \dfrac{(1+i)}{(1+i)} = \dfrac{1-i^{2}}{(1+i)} = \dfrac {2}{(1 + i)}$$
$$\therefore \dfrac {(1 + i)^{n}}{(1 - i)^{n - 2}} = \dfrac {(1 + i)^{n}}{2^{n - 2}} . (1 + i)^{n - 2}$$
$$= \dfrac {(1 + i)^{2(n - 1)}}{2^{n - 2}}$$

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

$$= \dfrac {(2i)^{n - 1}}{2^{n - 2}}$$
$$= 2i^{n - 1} = -2i^{n + 1}$$
Hence, $$\therefore \dfrac {(1 + i)^{n}}{(1 - i)^{n - 2}} = - 2 i^{n+1}$$
Question 2
1 / -0

Given : $$u = 1+i \sqrt{3}$$ and $$v = \sqrt{3} + i$$
Calculate $$\dfrac{u^3 }{ v^4}$$

A
$$(1/4) - i \sqrt{1/4}$$

B
$$(3/4) - i \sqrt{3}/4$$

C
$$(1/4) - i \sqrt{3}/4$$

D
none of these

Solution

$$u=1+i\sqrt 3 \implies \sqrt{1+3}e^{i(\tan^{-1}\sqrt 3)}\\ \implies 2e^{i\pi/3}$$
$$v=i+\sqrt 3 \implies \sqrt{1+3}e^{i(\tan^{-1} 1/\sqrt 3)}\\ \implies 2e^{i\pi/6}$$
$$\cfrac{u^3}{v^4}=\cfrac{2^3e^{i\pi}}{2^4e^{2i\pi/3}}=(1/2)e^{-i\pi/3}$$
$$\implies \cfrac{1}{2} [\cos(\pi/3)-i \sin(\pi/3)] \implies \cfrac{1}{2}(\cfrac{1}{2}-\cfrac{i\sqrt 3}{2})$$
$$ \implies (\cfrac{1}{4}-\cfrac{i\sqrt 3}{4})$$
Question 3
1 / -0

If $$z_1$$ and $$z_2$$ are complex numbers with $$|z_1|=|z_2|$$, then which of the following is/are correct?
1. $$z_1=z_2$$
2. Real part of $$z_1 =$$ Real part of $$z_2$$
3. Imaginary part of $$z_1 =$$ Imaginary part of $$z_2$$
Select the correct answer using the statements given below :

A
1 only

B
2 only

C
3 only

D
None

Solution

Solution:
We have, $$|z_1|=|z_2|$$
Let,
$$z_1=x_1+iy_1$$ and $$z_2=x_2+iy_2$$
$$\because|z_1|=|z_2|$$
$$\therefore x_1^2+y_1^2=x_2^2+y_2^2$$
or, $$(x_1^2-x_2^2)+(y_1^2-y_2^2)=0$$
or, $$x_1^2-x_2^2=0$$ and $$y_1^2-y_2^2=0$$
or, $$x_1=\pm x_2$$ and $$y_1=\pm y_2$$
Let, $$z_1=1+i$$ then $$z_2=-1-i$$
$$|z_1|=|z_2|=\sqrt2$$
But, $$Re(z_1)\neq Re(z_2)$$ and $$Im(z_1)\neq Im(z_2)$$ and $$z_1\neq z_2$$
Hence, D is the correct option.
Question 4
1 / -0

$$p + iq = (2 - 3i) (4 + 2i)$$ then $$q$$ is

A
$$14$$

B
$$-14$$

C
$$-8$$

D
$$8$$

Solution

$$p+iq=(2-3i)(4+2i)$$
$$=(8+4i-12i+6)$$
$$=14-8i$$
Hence comparing the real parts give us $$p=14$$ and comparison of the imaginary parts give us $$q=-8$$.
Question 5
1 / -0

Two complex numbers are represented by ordered pairs $$z_1: (a,0)\ \&\ z_2: (c,d)$$, which of the following is correct simplification for $$z_1\times z_2=$$?

A
$$(ac,-ad)$$

B
$$(ad,ac)$$

C
$$(ac,ad)$$

D
None of these

Solution

$$z_1$$ : (a,0) = a + 0i = a
$$z_2$$ : (c,d) = c + id

$$\therefore$$ $$z_1\times z_2$$ = (a)(c + id)

= ac + i(ad) = (ac , ad)

Hence, option C is correct.
Question 6
1 / -0

If $${ x }^{ 2 }+{ y }^{ 2 }=1$$ then value of $$\dfrac { 1+x+iy }{ 1+x-iy } $$ is

A
$$x-iy$$

B
$$2x$$

C
$$-2iy$$

D
$$x+iy$$

Solution

Since $$x^2+y^2=1$$
So let $$x=\cos\theta$$ and $$y=\sin\theta$$
$$\therefore \dfrac{1+x+iy}{1+x-iy}=\dfrac{(1+\cos\theta)+i\sin\theta}{(1+\cos\theta)-i\sin\theta}$$

$$=\dfrac{2\cos^2\cfrac{\theta}{2}+2i\sin\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{2\cos^2\cfrac{\theta}{2}-2i\sin\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}$$

$$=\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}-i\sin\cfrac{\theta}{2}}$$

$$=\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}-i\sin\cfrac{\theta}{2}}\times$$ $$\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}$$

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

$$=\dfrac{\cos^2\cfrac{\theta}{2}+i^2\sin^2\cfrac{\theta}{2}+2i\cos\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{\cos^2\cfrac{\theta}{2}+\sin^2\cfrac{\theta}{2}}$$

$$=\dfrac{\cos^2\cfrac{\theta}{2}-\sin^2\cfrac{\theta}{2}+2i\cos\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{1}$$

$$=\cos\theta+i\sin\theta=x+iy$$
Question 7
1 / -0

The value of $$ \sum _{ k=0 }^{ n }{ (i^k + i^{k+1} ) } , $$ where $$ i^2 = -1 ,$$ is equal to :

A
$$ i - i^n $$

B
$$ - i + i^{n+1} $$

C
$$ i - i^{n+1} $$

D
$$ i - i^{n+2} $$

E
$$ - i - i^{n} $$

Solution

The value of $$ \sum _{ k=0 }^{ n }{ (i^k + i^{k+1} ) } = \sum _{ k=0 }^{ n }{ i^k ( {i +1} ) } $$ is,
$$ = (i+1) \left[ \dfrac {1(1-i^{n+1})} {(1-i)} \right] \times \dfrac {1+i}{1+i} $$
$$ = \dfrac {(1+i)^2 }{1+1} = \dfrac {(1-1+2i)(1-i^{n+1} )}{2} $$
$$ = \dfrac {2i}{2} ( 1 - i^{n+1} ) $$
$$= i - i^{n+2} $$
Question 8
1 / -0

If $$f\left( z \right) =\dfrac { 1-{ z }^{ 3 } }{ 1-z } $$, where $$z=x+iy$$ with $$z\neq 1$$, then $$Re\overline { \left\{ f\left( z \right) \right\} } =0$$ reduces to

A
$${ x }^{ 2 }+{ y }^{ 2 }+x+1=0$$

B
$${ x }^{ 2 }-{ y }^{ 2 }+x-1=0$$

C
$${ x }^{ 2 }-{ y }^{ 2 }-x+1=0$$

D
$${ x }^{ 2 }-{ y }^{ 2 }+x+1=0$$

E
$${ x }^{ 2 }-{ y }^{ 2 }+x+2=0$$

Solution

$$f\left( z \right) =\dfrac { 1-{ z }^{ 3 } }{ 1-z } =\dfrac { \left( 1-z \right) \left( 1+z+{ z }^{ 2 } \right) }{ \left( 1-z \right) } $$
$$=1+z+{ z }^{ 2 }$$
Put $$z=x+iy$$, we get
$$f\left( z \right) =1+x+iy+{ \left( x+iy \right) }^{ 2 }$$
$$=1+x+iy+{ x }^{ 2 }+2xyi+{ i }^{ 2 }{ y }^{ 2 }$$
$$=\left( 1+x+{ x }^{ 2 }-{ y }^{ 2 } \right) +i\left( y+2xy \right) $$
$$\Rightarrow f\left( \overline { z } \right) =\left( 1+x+{ x }^{ 2 }-{ y }^{ 2 } \right) -i\left( y+2xy \right) $$
Now, $$Re\left\{ \overline { f\left( z \right) } \right\} =0$$
$$\Rightarrow 1+x+{ x }^{ 2 }-{ y }^{ 2 }=0$$
$$\Rightarrow { x }^{ 2 }-{ y }^{ 2 }+x+1=0$$
Question 9
1 / -0

If $$z_{1}$$ and $$z_{2}$$ be complex numbers such that $$z_{1} + i(\overline {z_{2}}) = 0$$ and $$arg (\overline {z_{1}}z_{2}) = \dfrac {\pi}{3}$$. Then, $$arg (\overline {z_{1}})$$ is equal to

A
$$\dfrac {\pi}{3}$$

B
$$\pi$$

C
$$\dfrac {\pi}{2}$$

D
$$\dfrac {5\pi}{12}$$

E
$$\dfrac {5\pi}{6}$$

Solution

Given, $$z_{1} + i \,\overline {(z_{2})} = 0$$
$$\Rightarrow z_{1} = (-i) \overline {z_{2}}$$
Taking argument on both sides, we get
$$arg (z_{1}) = arg \left \{(-i) \cdot \overline {z_{2}}\right \}$$
$$\Rightarrow arg (z_{1}) = arg (-i) + arg (\overline {z_{2}})$$ (by property)
$$\Rightarrow arg(z_{1}) - arg (\overline {z_{2}}) = \tan^{-1} \left (\dfrac {-1}{0}\right ) = -\dfrac {\pi}{2}$$
$$\Rightarrow arg (z_{1}) + arg (z_{2}) = \dfrac {-\pi}{2} ..... (i)$$
$$[\because arg (\overline {z}) = -arg (z)]$$
and $$arg (\overline {z_{1}}z_{2}) = \dfrac {\pi}{3}$$
$$\Rightarrow arg (\overline {z_{1}}) + arg (z_{2}) = \dfrac {\pi}{3}$$
$$\Rightarrow -arg (z_{1}) + arg (z_{2}) = \dfrac {\pi}{3}... (ii)$$
On adding Eqs. (i) and (ii), we get
$$2 arg (z_{2}) = \dfrac {-\pi}{6}$$
$$\Rightarrow arg (z_{2}) = \dfrac {-\pi}{12}$$
Therefore, from Eq. (i),
$$arg (z_{1}) = \dfrac {-\pi}{2} + \dfrac {\pi}{12} = \dfrac {-5\pi}{12}$$
$$\Rightarrow -arg (z_{1}) = \dfrac {5\pi}{12}$$
$$\Rightarrow arg (z_{1}) = \dfrac {5\pi}{12}$$.
Question 10
1 / -0

If $$\left( \dfrac{1 + i}{1 - i} \right)^m = 1$$, then the least positive integral value of m is

A
1

B
4

C
2

D
3

Solution

Consider $$(\dfrac{1+i}{1-i})$$

Rationalize the denominator we get,

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

Here we see that denominator is in the form of $$(a+b)(a-b)=a^2-b^2$$

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

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

we know that $$(a+b)^2=a^2+2ab+b^2$$

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

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

We know that $$i^2=-1$$ , substituting this we get,

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

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

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

$$\Rightarrow \dfrac{1+i}{1-i}=i$$

Given that $$\Rightarrow (\dfrac{1+i}{1-i})^m=1$$

$$\Rightarrow (\dfrac{1+i}{1-i})^m=i^m$$

So, $$i^m=1$$

We know that $$i^2=-1$$

Squaring on both sides we get,

$$i^4=(-1)^2$$

$$i^4=1$$

Therefore the smallest value of $$m$$ for which $$ (\dfrac{1+i}{1-i})^m=1$$ is $$m=4$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 29

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Number Theory Test 29

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SHARING IS CARING

Question 1

$$-i^{n + 1}$$

$$i^{n + 1}$$

$$-2i^{n + 1}$$

$$1$$

Solution

Question 2

$$(1/4) - i \sqrt{1/4}$$

$$(3/4) - i \sqrt{3}/4$$

$$(1/4) - i \sqrt{3}/4$$

none of these

Solution

Question 3

1 only

2 only

3 only

None

Solution

Question 4

$$14$$

$$-14$$

$$-8$$

$$8$$

Solution

Question 5

$$(ac,-ad)$$

$$(ad,ac)$$

$$(ac,ad)$$

None of these

Solution

Question 6

$$x-iy$$

$$2x$$

$$-2iy$$

$$x+iy$$

Solution

Question 7

$$ i - i^n $$

$$ - i + i^{n+1} $$

$$ i - i^{n+1} $$

$$ i - i^{n+2} $$

$$ - i - i^{n} $$

Solution

Question 8

$${ x }^{ 2 }+{ y }^{ 2 }+x+1=0$$

$${ x }^{ 2 }-{ y }^{ 2 }+x-1=0$$

$${ x }^{ 2 }-{ y }^{ 2 }-x+1=0$$

$${ x }^{ 2 }-{ y }^{ 2 }+x+1=0$$

$${ x }^{ 2 }-{ y }^{ 2 }+x+2=0$$

Solution

Question 9

$$\dfrac {\pi}{3}$$

$$\pi$$

$$\dfrac {\pi}{2}$$

$$\dfrac {5\pi}{12}$$

$$\dfrac {5\pi}{6}$$

Solution

Question 10

1

4

2

3

Solution

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