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Number Theory Test 32

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Number Theory Test 32
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  • Question 1
    1 / -0
    If $${z}_{1}=-3+5i;{z}_{2}=-5-3i$$ and $$z$$ is a complex number lying on the line segment joining $${z}_{1}$$ and $${z}_{2}$$, then $$arg(z)$$ can be:
  • Question 2
    1 / -0
    Which of the following is not a binary number?
    Solution

  • Question 3
    1 / -0
    Find a complex number z satisfying the equation $$z+\sqrt{2}|z+1|+i=0$$
    Solution
    Let $$z=x+iy$$. Then,

    $$z+\sqrt{2}|z+1|+i=0$$

    $$x+iy+\sqrt{2}|(x+1)+iy|+i=0$$

    $$x+\sqrt{2}\sqrt{(x+1)^2+y^2}+(y+1)i=0$$

    $$x+\sqrt{2(x+1)^2+2y^2}+(y+1)i=0$$

    $$x+\sqrt{2(x+1)^2+2}=0$$ and $$y=-1$$

    $$\sqrt{2(x+1)^2+2}=-x$$ and $$y=-1$$

    $$2(x+1)^2+2=x^2$$ and $$y=-1$$

    $$x^2+4x+4=0$$ and $$y=-1$$

    $$(x+2)^2=0$$ and $$y=-1$$

    $$x=-2$$ and $$y=-1$$

    Hence, $$z=x+iy=-2-i$$
  • Question 4
    1 / -0
    If $${ a }^{ 2 }+{ b }^{ 2 }=1$$, then $$\dfrac {\left( 1+b+ia \right) }{\left( 1+b-ia \right)} $$ is
    Solution
    $$a^2+b^2=1$$

    $$\implies 1-a^2=b^2$$

    $$\dfrac{1+b+ia}{1+b-ia}$$

    $$=\dfrac{1+b+ia}{1+b-ia}\times \dfrac{1+b+ia}{1+b+ia}$$

    $$=\dfrac{(1+b)^2-a^2+2ia(1+b)}{(1+b)^2+a^2}$$

    $$=\dfrac{b^2+2b+1-a^2+2ia(1+b)}{1+2b+b^2+a^2}$$

    $$=\dfrac{2b^2+2b+2ia(1+b)}{2+2b}$$

    $$=\dfrac{(2b+2ia)(1+b)}{2(1+b)}$$

    $$=b+ia$$

    Answer-(C)
  • Question 5
    1 / -0
    Multiplication of $$111_2$$ by $$101_2$$ is?
    Solution

    Multiplication of 1112 by 1012 is 1000112

    A binary multiplier is an electronic circuit used in digital electronics, such as a computer, to multiply two binary numbers. The lattice method is an alternative to long multiplication for numbers. In this approach, a lattice is first constructed, sized to fit the numbers being multiplied. If we are multiplying an -digit number by an -digit number, the size of the lattice is .

  • Question 6
    1 / -0
    If $$\dfrac {lz_{2}}{mz_{1}}$$ is purely imaginary number, then $$\left |\dfrac {\lambda z_{1} + \mu z_{2}}{\lambda z_{1} - \mu z_{2}}\right |$$ is equal to
    Solution
    $$\cfrac { { lz }_{ 2 } }{ m{ z }_{ 1 } } =ki\\ \cfrac { { z }_{ 2 } }{ { z }_{ 1 } } =\cfrac { mki }{ l } \\ \implies |\cfrac { { \lambda z }_{ 1 }+\mu { z }_{ 2 } }{ { \lambda z }_{ 1 }-\mu { z }_{ 2 } } |\\ \implies |\cfrac { \lambda +\mu { z }_{ 2 }/{ z }_{ 1 } }{ \lambda -\mu { z }_{ 2 }/{ z }_{ 1 } } |\\ \implies |\cfrac { \lambda +\mu \cfrac { mki }{ l }  }{ \lambda -\mu \cfrac { mki }{ l }  } |\\ \implies|\cfrac { \lambda l+\mu mki }{ \lambda l-\mu mki } |$$
    $$\implies|\cfrac{re^{i\theta}}{re^{-i\theta}}|$$
    $$\implies |e^{2i\theta}|$$
    $$\implies 1$$
  • Question 7
    1 / -0
    An output device that converts data from a binary format in main storage to coded hole patterns punched into a paper tape is?
    Solution

    An output device that converts data from a binary format in main storage to coded hole patterns punched into a paper tape is Paper tape punch

    Punched tape or perforated paper tape is a form of data storage, consisting of a long strip of paper in which holes are punched to store data. Now effectively obsolete, it was widely used during much of the twentieth century for teleprinter communication, for input to computers of the 1950s and 1960s, and later as a storage medium for minicomputers and CNC machine tools.

  • Question 8
    1 / -0
    The $$2$$'s complement number of $$110010$$ is?
    Solution

    The 2's complement number of 110010 is None of the above.

    2's complement number of (110010)=001110.

  • Question 9
    1 / -0
    Instructions and memory addresses are represented by.
    Solution

    Instructions and memory addresses are represented by binary codes.

    In computing, a memory address is a reference to a specific memory location used at various levels by software and hardware. Memory addresses are fixed-length sequences of digits conventionally displayed and manipulated as unsigned integers.

    ASCII code. The American Standard Code for Information Interchange (ASCII), uses a 7-bit binary code to represent text and other characters within computers, communications equipment, and other devices. Each letter or symbol is assigned a number from 0 to 127.

  • Question 10
    1 / -0
    The real part of $$(1-\cos\theta +2i \sin\theta)^{-1}$$ is?
    Solution
    $$(1-\cos\theta +2i \sin\theta)^{-1}$$
    $$=\dfrac{1}{(1-\cos\theta +2i \sin\theta)}$$
    $$=\dfrac{1}{(2\sin^2\dfrac{\theta}{2} +4i \sin\dfrac{\theta}{2}.\cos \dfrac{\theta}{2})}$$
    $$=\dfrac{1}{2\sin\dfrac{\theta}{2}}\dfrac{1}{(\sin\dfrac{\theta}{2} +2i \cos\dfrac{\theta}{2})}$$
    $$=\dfrac{1}{2\sin\dfrac{\theta}{2}}\dfrac{\left(\sin\dfrac{\theta}{2}-2i\cos \dfrac{\theta}{2}\right)}{(\sin^2\dfrac{\theta}{2} +4 \cos^2\dfrac{\theta}{2})}$$
    $$=\dfrac{1}{\sin\dfrac{\theta}{2}}\dfrac{\left(\sin\dfrac{\theta}{2}-2i\cos \dfrac{\theta}{2}\right)}{(2\sin^2\dfrac{\theta}{2} +8 \cos^2\dfrac{\theta}{2})}$$
    $$=\dfrac{\left(1-2i\cot \dfrac{\theta}{2}\right)}{\{1-\cos \theta+4(1+\cos \theta)\}}$$
    $$=\dfrac{\left(1-2i\cot \dfrac{\theta}{2}\right)}{5+3\cos \theta}$$
    So the real part is $$\dfrac{1}{5+3\cos \theta}$$..
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