Number Theory Test 43

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Number Theory Test 43

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Weekly Quiz Competition

Question 1
1 / -0

The principle amplitude of $$(\sin 40^{o}+i \cos 40^{o})^{5}$$ is

A
$$70^{o}$$

B
$$-1100^{o}$$

C
$$70^{110}$$

D
$$70^{-70}$$

Solution

$$40°=2\cfrac { \pi }{ 9 } ,50°=\cfrac { 5\pi }{ 18 } $$

So, $$\left( \sin { 40° } +i\cos { 40° } \right) =\cos { 50° } +i\sin { 50° } $$

Let $$T={ \left( \sin { 40° } +i\cos { 40° } \right) }^{ 5 }={ \left( \cos { 50° } +i\sin { 50° } \right) }^{ 5 }$$

$$T={ \left( { e }^{ i50° } \right) }^{ 5 }=\left( { e }^{ i\cfrac { 5\pi }{ 18 } \times 5 } \right) ={ e }^{ i\cfrac { 25\pi }{ 18 } }$$

So, $$25\cfrac { \pi }{ 18 } =250°$$

Thus, principal argument$$=-180°+250°=70°$$
Question 2
1 / -0

The value of the sum $$\sum _{ n=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n+1 }) } $$ , where $$i=\sqrt { -1 } $$ , equals

A
$$i$$

B
$$i-1$$

C
$$-i$$

D
$$0$$

Solution

As we know that $$i+i^2+i^3+i^4=0$$
$$\displaystyle\sum ^{13}_{n=1} (i^n+i^{n+1})=\sum ^{13}_{n=1}(1+i)(i^n)=(1+i)\sum ^{13}_{n=1} i^{n}=(1+i)(0+0+i^{13})=(1+i)i=i+i^2=i-1$$
Question 3
1 / -0

$$z_1$$ and $$z_2$$ are two non-zero complex numbers such that $$z_1=2+4i\\z_2=5-6i$$, then $$z_2-z_1$$ equals

A
$$3-10i$$

B
$$3+10i$$

C
$$7-2i$$

D
$$10-24i$$

Solution

$$z_1=2+4i\\\\z_2=5-6i\\\\z_2-z_1=5-6i-2-4i=3-10i$$
Question 4
1 / -0

The imaginary part of $$t ; t \in R$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-1$$

Solution
Question 5
1 / -0

IF $$z_1=1+i,z_2=1-i$$ find $$z_1z_2$$

A
$$z_1+z_2$$

B
$$z_1-z_2$$

C
$$z_1/z_2$$

D
None.

Solution

$$z_1=1+i$$

$$z_2=1-i$$

$$z_1z_2=1-(-1)=2$$

$$=1+1$$

$$=1-i+1+i$$

$$z_1z_2=z_1+z_2$$
Question 6
1 / -0

The real value of $$'\theta '$$, for which the expression $$\frac { 1+i\cos { \theta } }{ 1-2i\cos { \theta } } $$ is a real number is

A
$$2n\pi +\frac { 3\pi }{ 2 } ,n\in I$$

B
$$2n\pi -\frac { 3\pi }{ 2 } ,n\in I$$

C
$$2n\pi \pm \frac { \pi }{ 2 } ,n\in I$$

D
$$2n\pi +\frac { \pi }{ 4 } ,n\in I$$

Solution
Question 7
1 / -0

The greatest and least value of $$\left | z \right |$$ if $$z$$ satisfies $$\left | z - 5 + 5i \right |$$ $$\leq 5$$ are

A
$$10$$ , $$5\sqrt{2}$$

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

C
$$10$$ , $$0$$

D
$$5 + 5\sqrt{2}$$ , $$5\sqrt{2} - 5$$

Solution

$$\left | z - 5 + 5i \right |$$ $$\leq 5$$ represents equation of circle, where centre of circle is (5,-5)

Maximum value of $$z=c+\sqrt{a^2+b^2}$$

Maximum value of $$z=c-\sqrt{a^2-b^2}$$

Here, $$a=5, b=-5$$ and $$c=5$$

Maximum value of $$|z|=|5+\sqrt{25+25}|$$

$$=|5+\sqrt{50}|$$

$$=5+5\sqrt{2}$$

Minimum value of $$|z|=|5-\sqrt{50}|$$

$$=|5-5\sqrt{2}|$$

$$=5\sqrt{2}-5$$
Question 8
1 / -0

let $$z=\left| \begin{matrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{matrix} \right|$$ then

A
z is purely real

B
z is purely imaginary

C
$$\left( z-\bar { z } \right) $$ i is real and imaginary both

D
$$\left( z+\bar { z } \right) =0$$

Solution

From given, we have,

$$z=\begin{vmatrix}1&1+2i&-5i\\ 1-2i&-3&5+3i\\ 5i&5-3i&7\end{vmatrix}$$

$$=1 \times \begin{vmatrix}-3&5+3i\\ 5-3i&7\end{vmatrix}-(1+2i) \times \begin{vmatrix}1-2i&5+3i\\ 5i&7\end{vmatrix}-5i \times \begin{vmatrix}1-2i&-3\\ 5i&5-3i\end{vmatrix}$$

$$=1\cdot \left(-55\right)-\left(1+2i\right)\left(22-39i\right)-5i\left(-1+2i\right)$$

$$=-145$$

Hence $$z$$ is purely real
Question 9
1 / -0

How many prime numbers are there in the following series.
$$1,2,7,9,13,15,21,23,27,29$$

A
7

B
6

C
5

D
4

Solution
Question 10
1 / -0

The imaginary roots of the equation $${ ({ x }^{ 2 }+2) }^{ 2 }+8{ x }^{ 2 }=6x({ x }^{ 2 }+2)$$ are ____________.

A
$$1+i$$

B
$$2\pm i$$

C
$$-1\pm i$$

D
$$none of these$$

Solution

$${\left({x}^{2}+2\right)}^{2}+8{x}^{2}=6x\left({x}^{2}+2\right)$$

$${\left({x}^{2}+2\right)}^{2}-6x\left({x}^{2}+2\right)+8{x}^{2}=0$$

Let $$t={x}^{2}+2$$

$$\Rightarrow\,{t}^{2}-6xt+8{x}^{2}=0$$

$$\Rightarrow\,{t}^{2}-2t\times 3x+9{x}^{2}-{x}^{2}=0$$

$$\Rightarrow\,{\left(t-3x\right)}^{2}-{x}^{2}=0$$

$$\Rightarrow\,{\left({x}^{2}+2-3x\right)}^{2}-{x}^{2}=0$$ where $$t={x}^{2}+2$$

$$\Rightarrow\,{\left({x}^{2}-3x+2\right)}^{2}-{x}^{2}=0$$

$$\Rightarrow\,\left({x}^{2}-3x+2+x\right)\left({x}^{2}-3x+2-x\right)=0$$

$$\Rightarrow\,\left({x}^{2}-2x+2\right)\left({x}^{2}-4x+2\right)=0$$

$$\Rightarrow\,{x}^{2}-2x+2=0,\,or\,{x}^{2}-4x+2=0$$

$$\Rightarrow\,x=\dfrac{2\pm\sqrt{4-4\times\,1\times 2}}{2},\,or\,x=\dfrac{-4\pm\sqrt{16-4\times 1\times 2}}{2}$$

$$\Rightarrow\,x=\dfrac{2\pm\sqrt{4-8}}{2},\,or\,x=\dfrac{-4\pm\sqrt{16-8}}{2}$$

$$\Rightarrow\,x=\dfrac{2\pm\sqrt{-4}}{2},\,or\,x=\dfrac{-4\pm\sqrt{8}}{2}$$

$$\Rightarrow\,x=\dfrac{2\pm 2i}{2},\,or\,x=\dfrac{-4\pm\,2\sqrt{2}}{2}$$

$$\Rightarrow\,x=1\pm i,\,or\,x=-2\pm\,\sqrt{2}$$

Hence the roots are $$\left\{1+i,\,1-i,\,-2+\sqrt{2},\,-2-\sqrt{2}\right\}$$

Imaginary roots are $$\left\{1+i,\,1-i\right\}$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 43

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Number Theory Test 43

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SHARING IS CARING

Question 1

$$70^{o}$$

$$-1100^{o}$$

$$70^{110}$$

$$70^{-70}$$

Solution

Question 2

$$i$$

$$i-1$$

$$-i$$

$$0$$

Solution

Question 3

$$3-10i$$

$$3+10i$$

$$7-2i$$

$$10-24i$$

Solution

Question 4

$$0$$

$$1$$

$$2$$

$$-1$$

Solution

Question 5

$$z_1+z_2$$

$$z_1-z_2$$

$$z_1/z_2$$

None.

Solution

Question 6

$$2n\pi +\frac { 3\pi }{ 2 } ,n\in I$$

$$2n\pi -\frac { 3\pi }{ 2 } ,n\in I$$

$$2n\pi \pm \frac { \pi }{ 2 } ,n\in I$$

$$2n\pi +\frac { \pi }{ 4 } ,n\in I$$

Solution

Question 7

$$10$$ , $$5\sqrt{2}$$

$$5\sqrt{2}$$ , $$5$$

$$10$$ , $$0$$

$$5 + 5\sqrt{2}$$ , $$5\sqrt{2} - 5$$

Solution

Question 8

z is purely real

z is purely imaginary

$$\left( z-\bar { z } \right) $$ i is real and imaginary both

$$\left( z+\bar { z } \right) =0$$

Solution

Question 9

7

6

5

4

Solution

Question 10

$$1+i$$

$$2\pm i$$

$$-1\pm i$$

$$none of these$$

Solution

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