CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 8 - MCQExams.com

The question must be which of the following are relatively primes?

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

We have,

$${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{8n}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{8n}}$$

$$ {{\left[ {{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}}+{{\left[ {{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{2}} \right]}^{4n}} $$

$$ ={{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{{{1}^{2}}+{{i}^{2}}-2i}{2} \right]}^{4n}} $$

$$ ={{\left[ \dfrac{1-1+2i}{2} \right]}^{4n}}+{{\left[ \dfrac{1-1-2i}{2} \right]}^{4n}} $$

$$ ={{\left( {{i}^{4}} \right)}^{n}}+{{\left( {{\left( -i \right)}^{4}} \right)}^{n}} $$

$$ ={{1}^{n}}+{{1}^{n}} $$

$$ =1+1 $$

$$ =2 $$

Hence, this is the answer.

$$Assume,z=a+ib,where \ i=\sqrt { -1 } $$

$$let\quad { z }_{ 1 }=a+ib\quad \quad and\quad \quad { z }_{ 2 }=c+id\\ \Rightarrow \quad \left| \quad { z }_{ 1 }+\quad { z }_{ 2 } \right| \quad \quad \quad \quad =\quad \quad \quad \left| \quad { z }_{ 1 } \right| +\left| \quad { z }_{ 2 } \right| \\ \Rightarrow \quad \quad \left| a+ib+c+id \right| \quad =\quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \left| a+c+i\left( b+d \right)  \right| \quad \quad \quad =\quad \quad \quad \left| a+ib \right| +\left| c+id \right| \\ \Rightarrow \sqrt { \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 } } =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad \quad squaring\quad both\quad sides\\ \Rightarrow \left( a+c \right) ^{ 2 }+\left( b+d \right) ^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+d^{ 2 }+2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow 2ac+2bd\quad \quad =\quad \quad 2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \Rightarrow ac+bd\quad \quad =\quad \quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } .\sqrt { { c }^{ 2 }+d^{ 2 } } \\ \quad \quad \quad again\quad squaring\quad both\quad sides\\ \Rightarrow { \left( ac \right)  }^{ 2 }+{ \left( bd \right)  }^{ 2 }+2abcd\quad =\quad \left( { a }^{ 2 }+{ b }^{ 2 } \right) \left( { c }^{ 2 }+d^{ 2 } \right) \\ \Rightarrow { \left( ac \right)  }^{ 2 }+{ \left( bd \right)  }^{ 2 }+2abcd\quad =\quad { \left( ac \right)  }^{ 2 }+{ \left( ad \right)  }^{ 2 }+{ \left( bc \right)  }^{ 2 }+{ \left( bd \right)  }^{ 2 }\\ \Rightarrow { \left( ad \right)  }^{ 2 }+{ \left( bc \right)  }^{ 2 }-2abcd\quad =\quad 0\\ \Rightarrow \left( ad-bc \right) ^{ 2 }\quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \Rightarrow ad-bc\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ad\quad \quad =\quad \quad \quad bc\quad \quad \quad \quad \longrightarrow \left( 1 \right) \\ Now\quad \quad \dfrac { { z }_{ 1 } }{ { z }_{ 2 } } \quad =\quad \dfrac { a+ib }{ c+id } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { \left( a+ib \right) \left( c-id \right)  }{ \left( c+id \right) \left( c-id \right)  } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac-iad+ibc-{ i }^{ 2 }bd }{ { c }^{ 2 }+{ d }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd+i\left( bc-ad \right)  }{ { c }^{ 2 }+{ d }^{ 2 } } \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \dfrac { ac+bd }{ { c }^{ 2 }+{ d }^{ 2 } } \quad +\quad 0\quad \quad \quad using\quad equation\quad \left( 1 \right) \\ Imaginary\quad part\quad is\quad zero\quad as\quad there\quad is\quad only\quad real\quad part\\ \therefore \quad Im\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } }  \right) =\quad 0$$

The prime numbers between $$1$$ to$$ 10$$ is, 

Let $$z_2=a+ib$$

$$\cfrac { \cos { \theta  } -i\sin { \theta  }  }{ \sin { \theta  } -i\cos { \theta  }  } $$

$$3+2i\sin \theta$$ will be real if imaginary term is zero

$$|z_1| = \sqrt 2, \ |z_2| = \sqrt 3, \ |z_3| = \sqrt 4, \ |z_4| = \sqrt 5$$

$$ | z _1 + z _2 | ^2 + |z _1-z_2|^2 =  (z _1+z_2 ) ( \bar{z_1} + \bar {z_2}) + (z_1-z_2)(\bar{z_1} - \bar {z_2})$$

$${\textbf{Step  - 1: Writing the product in terms of i (iota)}}$$

                  $${\text{We know that, }}\sqrt {{\text{ - 1}}} {\text{  =  i and }}\sqrt {{\text{ab}}} {\text{  =  }}\sqrt {\text{a}} \sqrt {\text{b}} $$

                  $$\therefore {\text{ }}\sqrt {{\text{ - 2}}} {\text{  =  }}\sqrt {{\text{ - 1}}} \sqrt {\text{2}} {\text{  =  i}}\sqrt 2 $$     $$\quad \quad \quad \text{.....eqn(i)}$$

                  $${\text{and }}\sqrt {{\text{ - 3}}} {\text{  =  }}\sqrt {{\text{ - 1}}} \sqrt {\text{3}} {\text{  =  i}}\sqrt {\text{3}} $$   $$\quad \quad \quad \text{.....eqn(ii)}$$

$${\textbf{Step  - 2: Multiplying the terms}}$$

                  $$\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{  =  }}\left( {{\text{i}}\sqrt {\text{2}} } \right)\left( {{\text{i}}\sqrt {\text{3}} } \right)$$   $$\quad \quad \quad \textbf{[From eqn(i) and eqn(ii)]}$$

                  $$ \Rightarrow {\text{ }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{  =  }}{{\text{i}}^{\text{2}}}\sqrt {{\text{3}} \times 2} $$

                  $$ \Rightarrow {\text{ }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{  =   - }}\sqrt {\text{6}} {\quad \quad \quad \quad \quad \quad \textbf{  [}}\because {\text{ }}{{\textbf{i}}^{\textbf{2}}}{\textbf{  =   - 1]}}$$

$$\mathbf{{\text{Thus, the value of the product }}\sqrt {{\text{ - 2}}} \sqrt {{\text{ - 3}}} {\text{ is  -}}\sqrt {\text{6}} }$$

$$|z|=1$$ implies that Z in Q circle with radius with a center (0,0)  in the complex plane.

R.E.F image 

Given $$z_1=3+7i$$

$$87, 45, 34, 33, 27$$ are composite numbers