Number Theory Test 50

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Number Theory Test 50

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Weekly Quiz Competition

Question 1
1 / -0

Dividing $$ f(z) $$ by $$ z - i$$ we obtain the remainder i and dividing it by $$ z + i$$ we get the remainder $$1 + i$$ then remainder upon the division of $$ f(z)$$ by $$ z^{2} + 1$$ is

A
$$\displaystyle \frac{1}{2}(z+1)+i$$

B
$$\displaystyle \frac{1}{2}(iz+1)+i$$

C
$$\displaystyle \frac{1}{2}(z-1)+i $$

D
$$\displaystyle \frac{1}{2}(z+1)+1$$

Solution
Question 2
1 / -0

Evaluate in standard form: $$\dfrac {(2-3i)}{(2-2i)}$$, where $${i}^{2}=-1$$.

A
$$\dfrac {5}{4}-\dfrac {i}{4}$$

B
$$\dfrac {5}{4}+\dfrac {i}{4}$$

C
$$-\dfrac {5}{4}-\dfrac {i}{4}$$

D
$$-\dfrac {5}{4}+\dfrac {i}{4}$$

Solution

Consider $$\dfrac {(2-3i)}{(2-2i)}$$
Rationalise and solve considering that $$i^2=-1$$ as shown below:
$$\Rightarrow \dfrac { 2-3i }{ 2-2i } \times \dfrac { 2+2i }{ 2+2i } =\dfrac { 4+4i-6i-{ 6i }^{ 2 } }{ 4-{ 4i }^{ 2 } }$$
$$ =\dfrac { 4-2i+6 }{ 4+4 } $$
$$=\dfrac { -2i+10 }{ 8 } $$
$$=\dfrac { -i }{ 4 } +\dfrac { 5 }{ 4 }$$
Question 3
1 / -0

Express in the form of a complex number $$a+bi$$
$$-(7-i)(-4-2i)(2-i)$$

A
$$60-10i$$

B
$$70-10i$$

C
$$28-14i$$

D
$$70-15i$$

Solution

Multiplying the given expression as follows:
$$-[(7-i)(-4-2i)(2-i)]$$
$$=-[(-28-14i+4i+2i^2)(2-i)]$$
$$=-[(-28-10i-2)(2-i)]$$
$$=-[(-30-10i)(2-i)]$$
$$=-[-60+30i-20i+10i^2]$$
$$=-[-60-10+10i]$$
$$=-[-70+10i]$$
$$=70-10i$$
Question 4
1 / -0

The numbers which have more than two factors are called ______.

A
Prime

B
Composite

C
Both $$(A)$$ and $$(B)$$

D
None of these
Question 5
1 / -0

In binary system the highest value of a 8-bit number is

A
255

B
256

C
253

D
none of these

Solution

In binary system the highest value of a 8-bit number is 255.

8 bits is 11111111, or 255 in decimal notation.

A byte can hold 2 (binary) ^ 8 numbers ranging from 0 to 2^8-1 = 255.

A binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically 0 (zero) and 1 (one).
Question 6
1 / -0

The $$3$$ bit operation code for ADD operation is $$001$$ and indirect memory address is $$23$$ then $$16$$-bit instruction code can be written as-

A
$$00001000000010111$$

B
$$1001000000010111$$

C
$$100010000000010111$$

D
None of the above

Solution

The 3-bit operation code for ADD operation is 001 and the indirect memory address is 23 then 16-bit instruction code can be written as 00001000000010111.
Indirect memory address:-
The effective address of the operand is the contents of a register or main memory location, location whose address appears in the instruction. Indirection is noted by placing the name of the register or the memory address given in the instruction in parentheses.
Question 7
1 / -0

Let $$z_1$$ = 18 + 83i, $$z_2$$ = 18 + 39i, ana $$z_3 $$= 78 + 99i. where i = $$\sqrt-1$$. Let z be a unique comlpex number with the properties that $$\dfrac{z_3 - z_1}{z_2 - z_1}$$ $$\cdot$$ $$\dfrac{z - z_2}{z - z_3}$$ is a real number and the imaginary part of the size z is the greatest possible.

A
$$Re (z) = 56$$

B
$$Re (z) = 61$$

C
$$Re (z) = 54$$

D
$$Re (z) = 59$$

Solution

Using the property that

if $$\dfrac{z_3-z_1}{z_2-z_1}.\dfrac{z-z_2}{z-z_3}$$is a real number then the four points are concyclic i.e. are present on a circle

since it is told the imaginary part is maximum so,it the top point of the circle
so we take the perpendicular bisector of any two line and take its intersection.
As shown in the figure,

Equation of perpendicular bisector of $$z_1z_2$$
$$y=61$$

Equation of perpendicular bisector of $$z_3z_2$$
$$y+x=117$$

Solving the equation we get,
$$x=56$$

So, $$Re(z)=56$$
Question 8
1 / -0

If $$z$$ is a complex number such that $$|z|\geq 2$$ then the minimum value of $$\left |z + \dfrac {1}{2}\right |$$ is

A
Is strictly greater than $$\dfrac {5}{2}$$

B
Is strictly greater than $$\dfrac 32$$ but less than $$\dfrac {5}{2}$$

C
Is equal to $$\dfrac {5}{2}$$

D
Lies in the interval $$(1, 2)$$

Solution

$$\left| z \right| \ge 2$$ is the region on or out side the circle whose center is $$\left( 0,0 \right)$$ and radius is $$2$$.

Minimum $$\left| z+\dfrac { 1 }{ 2 } \right|$$ is distance of $$z$$ which lie on circle $$\left| z \right| =2$$ from $$\left( -\dfrac { 1 }{ 2 } ,0 \right)$$

Thus minimum $$\left| z+\dfrac { 1 }{ 2 } \right| =$$ Distance between $$\left( -\dfrac { 1 }{ 2 } ,0 \right)$$ to $$\left( 0,0 \right)$$

$$= \sqrt { { \left( -\dfrac { 1 }{ 2 } +2 \right) }^{ 2 }{ +\left( 0-0 \right) }^{ 2 } } $$

$$=\sqrt { \left( -\dfrac { 1 }{ 2 } +2 \right) ^{ 2 } } $$

Apply radical rule $$\sqrt [ n ]{ a^{ n } } =a,$$ assuming $$a\ge \: 0$$

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

$$=\dfrac { 2 }{ 1 } -\dfrac { 1 }{ 2 } $$

$$=\dfrac { 2\cdot \: 2 }{ 2 } -\dfrac { 1 }{ 2 } $$

$$=\dfrac { 2\cdot \: 2-1 }{ 2 } $$

$$=\dfrac { 3 }{ 2 }$$

Hence, option B is correct.
Question 9
1 / -0

If $$z_ 1 = 2 \sqrt 2 (1 + i)$$ and $$z = 1 + i \sqrt 3$$, then $$z_1^2 z_2^3$$ is equal to

A
$$128 i$$

B
$$64 i$$

C
$$-64 i$$

D
$$- 128 i$$

E
$$256$$

Solution

Given, $$z_1=2\sqrt 2(1+i), z_2=1+i\sqrt 3$$
Therefore, $$z_1^2=8(1+i)^2$$
$$=8(1+2i-1)$$
$$=16i$$
and $$z_2^3=(1+i\sqrt 3)^3$$
$$=1+3\sqrt3i-9-3\sqrt3i$$
$$=-8$$
Thus, $$z_1^2z_2^3=-8(16i)$$ $$=-128i$$
Question 10
1 / -0

If $$z_1+ z_2 + z_3 = 0$$ and $$|z_1|=|z_2|=|z_3|= 1$$, then area of triangle whose vertices are $$z_1, z_2$$ and $$z_3$$ is:

A
$$\dfrac{3 \sqrt{3}}{4}$$

B
$$\dfrac{\sqrt{3}}{4}$$

C
$$1$$

D
$$2$$

Solution

$${|z^{}_{1}+z^{}_{2}|}^2+{|z^{}_{1}-z^{}_{2}|}^2=2({|z^{}_{1}|}^2+{|z^{}_{2}|}^2)$$ ....... $$(1)$$
$$z^{}_{1}+z^{}_{2}+z^{}_{3}=0$$
$$z^{}_{3}=-z^{}_{2}-z^{}_{1}$$
By taking modulus both sides
$$|-z^{}_{3}|=|z^{}_{2}+z^{}_{1}|$$ ......... $$(2)$$
Using $$(2)$$ in $$(1),$$ we have
$${|-z_{3}|}^2+{|z^{}_{1}-z^{}_{2}|^2}=2{(1+1)}$$
$$1+{|z^{}_{1}-z^{}_{2}|^2}=2{(1+1)}$$
$${|z^{}_{1}-z^{}_{2}|^2}=3$$
$$|z^{}_{1}-z^{}_{2}|=\sqrt{3}$$
Similarily $$|z^{}_{2}-z^{}_{3}|$$=$$|z^{}_{3}-z^{}_{1}|=\sqrt{3}$$
Hence sides of triangle formed are $$\sqrt{3},$$ $$\sqrt{3},$$ $$\sqrt{3}$$ and are equal.
So triangle formed is an equilateral triangle
Area of an equilateral triangle $$=\dfrac{\sqrt{3}}{4}s^2$$
where $$s$$ is side of equilateral triangle
So area of triangle formed by $$z^{}_{1} ,z^{}_{2},z^{}_{3}=3\dfrac{\sqrt{3}}{4}$$

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Number Theory Test 50

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Number Theory Test 50

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Question 1

$$\displaystyle \frac{1}{2}(z+1)+i$$

$$\displaystyle \frac{1}{2}(iz+1)+i$$

$$\displaystyle \frac{1}{2}(z-1)+i $$

$$\displaystyle \frac{1}{2}(z+1)+1$$

Solution

Question 2

$$\dfrac {5}{4}-\dfrac {i}{4}$$

$$\dfrac {5}{4}+\dfrac {i}{4}$$

$$-\dfrac {5}{4}-\dfrac {i}{4}$$

$$-\dfrac {5}{4}+\dfrac {i}{4}$$

Solution

Question 3

$$60-10i$$

$$70-10i$$

$$28-14i$$

$$70-15i$$

Solution

Question 4

Prime

Composite

Both $$(A)$$ and $$(B)$$

None of these

Question 5

255

256

253

none of these

Solution

Question 6

$$00001000000010111$$

$$1001000000010111$$

$$100010000000010111$$

None of the above

Solution

Question 7

$$Re (z) = 56$$

$$Re (z) = 61$$

$$Re (z) = 54$$

$$Re (z) = 59$$

Solution

Question 8

Is strictly greater than $$\dfrac {5}{2}$$

Is strictly greater than $$\dfrac 32$$ but less than $$\dfrac {5}{2}$$

Is equal to $$\dfrac {5}{2}$$

Lies in the interval $$(1, 2)$$

Solution

Question 9

$$128 i$$

$$64 i$$

$$-64 i$$

$$- 128 i$$

$$256$$

Solution

Question 10

$$\dfrac{3 \sqrt{3}}{4}$$

$$\dfrac{\sqrt{3}}{4}$$

$$1$$

$$2$$

Solution

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