If $$z^4+1=\sqrt{3}$$i then?

z represents vertices of a square of side $$2^{\dfrac{1}{4}}$$

z represents vertices of a square of side $$2^{\dfrac{3}{4}}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 11

An example for twin primes is _____.

0%
$5, 11$
0%
$3, 5$
0%
$11, 17$
0%
$3, 7$

Explanation

Two continuous prime number are called twin prime numbers.

Only

$3$ and

$5$ are continuous prime numbers.

Hence, the answer is

$3,5$ .

$z = x+iy$ and

$w = \dfrac{(1-iz)}{(z-i)}$ , then

$|w| = 1$ implies that, in the complex plane

0%
$z$ lies on the imaginary axis
0%
$z$ lies on the real axis
0%
$z$ lies on the unit circle
0%
None of these

Explanation

Given

$|w|$

$=1$

$1=|\dfrac{1-iz}{z-i}|$

$\Rightarrow |z-i|=|1-iz|$ .....(i)

Now let

$z=x+iy$

$\Rightarrow iz=-y+ix$

Put in (i)

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

$\Rightarrow (y+1)^{2}-(y-1)^{2}=0$

$\Rightarrow (2y)(2)=0$

$\Rightarrow y=0$ .

This is the equation of the x-axis.

$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }=$

0%
$-2$
0%
$2$
0%
$-1$
0%
$1$

Explanation

We have

$\displaystyle \frac { \sqrt { 3 } +i }{ 2 } =\frac { i\sqrt { 3 } +{ i }^{ 2 } }{ 2i }$ (Here we multiplied the numerator and denominator by

$(-i)$ on the LHS)

We get:

$-i\left( \frac { -1+\sqrt { 3 } i }{ 2 } \right) =i\omega$

Simliarly applying the same idea we get the steps below:

and

$\displaystyle \frac { i-\sqrt { 3 } }{ 2 } =\frac { { i }^{ 2 }-i\sqrt { 3 } }{ 2i } =-i\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) =-i{ \omega }^{ 2 }$

Hence

$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }={ \left( -i\omega \right) }^{ 6 }+{ \left( -i{ \omega }^{ 2 } \right) }^{ 6 }\\ \\ ={ i }^{ 6 }\left( { \omega }^{ 6 }+{ \omega }^{ 12 } \right) =-1\left( 1+1 \right) =-2$

$z_1$ and

$z_2$ are two complex numbers such that

$| z_1 | = | z_2 | + | z_1 z_2 |$ , then

$arg(z_1) - arg(z_2)$ is

0%
$0$
0%
$\pi$ /2
0%
$-\pi$ /2
0%
none of these

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

0%
$\displaystyle \frac{1}{2}(z+1)+i$
0%
$\displaystyle \frac{1}{2}(iz+1)+i$
0%
$\displaystyle \frac{1}{2}(z-1)+i$
0%
$\displaystyle \frac{1}{2}(z+1)+1$

$(\sqrt 3-i)^n=2^n, n\in N$ , then

$n$ is a multiple of

0%
$6$
0%
$10$
0%
$9$
0%
$12$

Explanation

$\sqrt3-i = 2\left(\dfrac{\sqrt{3}-i}{2}\right)$

$=2.e^{-i\frac{\pi}{6}}$
Hence,

$\left(2.e^{-i\frac{\pi}{6}}\right)^{n}=2^{n}$

$2^{n}e^{-i\frac{n\pi}{6}}=2^{n}.e^{i2k\pi}$ ,

$k\in N$

$\because e^{i2k\pi} = 1$

Hence,

$-\dfrac{n\pi}{6}=2k\pi$

$-n=12k$
Now

$\cos(-x)=\cos x$
Hence,

$n=12k$
Where

$k$ is a natural number.

Evaluate in standard form:

$\dfrac {(2-3i)}{(2-2i)}$ , where

${i}^{2}=-1$ .

0%
$\dfrac {5}{4}-\dfrac {i}{4}$
0%
$\dfrac {5}{4}+\dfrac {i}{4}$
0%
$-\dfrac {5}{4}-\dfrac {i}{4}$
0%
$-\dfrac {5}{4}+\dfrac {i}{4}$

Explanation

Consider

$\dfrac {(2-3i)}{(2-2i)}$

Rationalise and solve considering that

$i^2=-1$ as shown below:

$\Rightarrow \dfrac { 2-3i }{ 2-2i } \times \dfrac { 2+2i }{ 2+2i } =\dfrac { 4+4i-6i-{ 6i }^{ 2 } }{ 4-{ 4i }^{ 2 } }$

$=\dfrac { 4-2i+6 }{ 4+4 }$

$=\dfrac { -2i+10 }{ 8 }$

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

$z_1, z_2, ..., z_n$ lie on

$|z|=r$ and

$Re\left(\displaystyle\sum_{j=1}^n\displaystyle\sum_{k=1}^n{\displaystyle\frac{z_j}{z_k}}\right) = 0$ , then

0%
$\displaystyle\sum_{j=1}^n{z_j}=0$
0%
$\left|\displaystyle\sum_{j=1}^n{z_j}\right|=0$
0%
$\displaystyle\sum_{j=1}^n{\displaystyle\frac{1}{z_j}}=0$
0%
None of these

Explanation

Let

${ z }_{ i }=r(cos{ \theta }_{ i }+i.sin{ \theta }_{ i })$

$Re\left(\displaystyle\sum_{j=1}^n\displaystyle\sum_{k=1}^n{\displaystyle\frac{z_j}{z_k}}\right) = 0$
Thus,

$\displaystyle ({ z }_{ 1 }+{ z }_{ 2 } + ... + { z }_{ n }) \times \left (\frac { 1 }{ { z }_{ 1 } } +\frac { 1 }{ { z }_{ 2 } } ...\frac { 1 }{ { z }_{ n } } \right )=0$
Thus,

$[(cos{ \theta }_{ 1 }+...cos{ \theta }_{ n })+i.(sin{ \theta }_{ 1 }+...s{ in\theta }_{ n })] \times [(cos{ \theta }_{ 1 }+...cos{ \theta }_{ n })-i.(sin{ \theta }_{ 1 }+...s{ in\theta }_{ n })]=0$
So,

${ (cos{ \theta }_{ 1 }+...cos{ \theta }_{ n }) }^{ 2 }+{ (sin{ \theta }_{ 1 }+...s{ in\theta }_{ n }) }^{ 2 }=0$
So,

${ (cos{ \theta }_{ 1 }+...cos{ \theta }_{ n }) }={ (sin{ \theta }_{ 1 }+...s{ in\theta }_{ n }) }=0$
Hence (a), (b), (c) are correct.

Given that

$i = \sqrt {-1}$ , find the multiplicative inverse of

$5 - i$ .

0%
$5 + i$
0%
$\dfrac {5 + i}{26}$
0%
$\dfrac {1}{5 + i}$
0%
$\dfrac {5 + i}{24}$
0%
$\dfrac {5 - i}{24}$

Explanation

Let the multiplicative inverse of

$5-i$ be

$a+bi$

Therefore we have

$(5-i)(a+bi) = 1$

$\Rightarrow 5a+b +i(5b-a) = 1$

$\Rightarrow 5a+b = 1$ and

$5b-a = 0$

$\Rightarrow a=5b$

Substitute this in

$5a+b =1$ , we get

$b = \dfrac {1}{26}$ and

$a = \dfrac {5}{16}$

Therefore, multiplicative inverse is

$\dfrac {(5+i)}{26}$ .

Write the complete number

$- 2 - 2i$ in polar form.

0%
$2(cos\dfrac{\pi}{4}+i\,sin\dfrac{\pi}{4})$
0%
$-2(cos\dfrac{\pi}{4}-i\,sin\dfrac{\pi}{4})$
0%
$2\sqrt{2}(cos\dfrac{3\pi}{4}+i\,sin\dfrac{3\pi}{4})$
0%
$2\sqrt{2}(cos\dfrac{7\pi}{4}+i\,sin\dfrac{7\pi}{4})$
0%
$2\sqrt{2}(cos\dfrac{5\pi}{4}+i\,sin\dfrac{5\pi}{4})$

Explanation

given complex number is $-2-2i$
the magnitude of given number is $2\sqrt2$
So the number can be written as $2\sqrt2(-1/\sqrt2 -i/\sqrt2)$
which gives $2\sqrt2(cos(225)+isin(225)) = 2\sqrt2 (cos\frac{5\pi}{4}+isin\frac{5\pi}{4})$

What is the product of the complex numbers

$\left( -3i+4 \right)$ and

$\left( 3i+4 \right)$ ?

0%
$1$
0%
$7$
0%
$25$
0%
$-7+24i$
0%
$7+24i$

Explanation

Consider the product of the complex numbers as shown below:

$(-3i+4)(3i+4)=(-3i)(3i)-3i(4)+4(3i)+4(4)$

$=-9i^2-12i+12i+16$

$=-9(-1)+16=9+16$

$=25$

Hence, option C is the correct answer.

Add and express in the form of a complex number

$a+bi$

$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$

0%
$-4+9i$
0%
$-5+9i$
0%
$2+9i$
0%
$-5+8i$

Explanation

Adding the given expression as follows:

$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$

$=(2+3i)+(-4+5i)-(3-i)$

$=2+3i-4+5i-3+i$

$=-5+9i$

What is the next-highest prime number after 67?

0%
68
0%
69
0%
71
0%
73
0%
76

$z$ is a complex number such that

$|z|\geq 2$ then the minimum value of

$\left |z + \dfrac {1}{2}\right |$ is

0%
Is strictly greater than $\dfrac {5}{2}$
0%
Is strictly greater than $\dfrac 32$ but less than $\dfrac {5}{2}$
0%
Is equal to $\dfrac {5}{2}$
0%
Lies in the interval $(1, 2)$

Explanation

$\left| z \right| \ge 2$ 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.

$\alpha$ and

$\beta$ are two different complex numbers with

$|\beta|=1$ , then

$\left | \dfrac{\beta -\alpha}{1-\bar{\alpha }\beta } \right |$ is equal to.

0%
$0$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$-1$

Explanation

Let

$t = \left|\cfrac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right|$

Multiply the numerator and denominator with

$\overline { \beta }$

$\Rightarrow t = \left|\cfrac { 1-\alpha \overline { \beta } }{ \overline { \beta } -\overline { \alpha } } \right| = \left|\cfrac { \overline { 1- \alpha \overline { \beta } } }{ \overline { \overline{\beta} -\overline{\alpha} } } \right| = \left|\cfrac { 1-\overline { \alpha } \beta }{ \beta -\alpha } \right|=\cfrac{1}{t}$

$\Rightarrow {t}^{2} = 1$

$\Rightarrow t = 1$

$z_ 1 = 2 \sqrt 2 (1 + i)$ and

$z = 1 + i \sqrt 3$ , then

$z_1^2 z_2^3$ is equal to

0%
$128 i$
0%
$64 i$
0%
$-64 i$
0%
$- 128 i$
0%
$256$

Explanation

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,

$z_1^2z_2^3=-8(16i)$

$=-128i$

Express in the form of a complex number

$a+bi$

$-(7-i)(-4-2i)(2-i)$

0%
$60-10i$
0%
$70-10i$
0%
$28-14i$
0%
$70-15i$

Explanation

Multiplying the given expression as follows:

$-[(7-i)(-4-2i)(2-i)]$

$=-[(-28-14i+4i+2i^2)(2-i)]$

$=-[(-28-10i-2)(2-i)]$

$=-[(-30-10i)(2-i)]$

$=-[-60+30i-20i+10i^2]$

$=-[-60-10+10i]$

$=-[-70+10i]$

$=70-10i$

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

0%
Prime
0%
Composite
0%
Both $(A)$ and $(B)$
0%
None of these

If z be any complex number such that

$|3z-2|+|3z+2|=4$ , then locus of z is

0%
An ellipse
0%
A circle
0%
A line-segment
0%
None of these

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

0%
255
0%
256
0%
253
0%
none of these

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-

0%
$00001000000010111$
0%
$1001000000010111$
0%
$100010000000010111$
0%
None of the above

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.

0%
$Re (z) = 56$
0%
$Re (z) = 61$
0%
$Re (z) = 54$
0%
$Re (z) = 59$

Explanation

Using the property that

$\dfrac{z_3-z_1}{z_2-z_1}.\dfrac{z-z_2}{z-z_3}$ 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,

$Re(z)=56$

What is

${ i }^{ 1000 }+{ i }^{ 1001 }+{ i }^{ 1002 }+{ i }^{ 1003 }$ equal to (where

$i=\sqrt { -1 }$ )?

0%
$0$
0%
$i$
0%
$-i$
0%
$1$

Explanation

$i^{1000} + i^{100} + i^{1002} + i^{1003}$

$i^{2} = -1, i^{3} = -1, i^{4} = 1$

$i^{1000} = (i^{4})^{250} = 1\ i^{1002} = i^{2} - i^{1000} = -1$

$i^{1001} = i\cdot i^{1000} = i\ i^{1003} = i^{3}\cdot i^{1000} = -i$

$\therefore i^{1000} + i^{1001} + i^{1002} + i^{1003} = 1 - 1 + i - i = 0$

$z_1+ z_2 + z_3 = 0$ and

$|z_1|=|z_2|=|z_3|= 1$ , then area of triangle whose vertices are

$z_1, z_2$ and

$z_3$ is:

0%
$\dfrac{3 \sqrt{3}}{4}$
0%
$\dfrac{\sqrt{3}}{4}$
0%
$1$
0%
$2$

Explanation

${|z^{}_{1}+z^{}_{2}|}^2+{|z^{}_{1}-z^{}_{2}|}^2=2({|z^{}_{1}|}^2+{|z^{}_{2}|}^2)$ .......

$(1)$

$z^{}_{1}+z^{}_{2}+z^{}_{3}=0$

$z^{}_{3}=-z^{}_{2}-z^{}_{1}$

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

$\sqrt{3},$

$\sqrt{3}$ 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

$z^{}_{1} ,z^{}_{2},z^{}_{3}=3\dfrac{\sqrt{3}}{4}$

The least positive integer

$n$ for which

$\left(\dfrac {1+i}{1-i}\right)^{n}=\dfrac {2}{\pi} \sin^{-1}\left(\dfrac {1+x^{2}}{2x}\right)$ , where

$x>0$ , is

0%
$2$
0%
$4$
0%
$8$
0%
$12$

${z}_{1},{z}_{2},..{z}_{n}$ lie on the circle

$|z|=2$ then the value of

$|{z}_{1},{z}_{2},..{z}_{n}|-4|\dfrac {1}{{z}_{1}}+\dfrac {1}{{z}_{2}}++\dfrac {1}{{z}_{n}}|=$

0%
$0$
0%
$n$
0%
$-n$
0%
$1$

${z}_{1},{z}_{2}$ and

${z}_{3}$ be three complex numbers such that

$\left| { z }_{ 1 }+1 \right| \le 1,\left| { z }_{ 2 }+2 \right| \le 2$ and

$\left| { z }_{ 3 }+4 \right| \le 4$ , then the maximum value of

$\left| { z }_{ 1 } \right| +\left| { z }_{ 2 } \right| +\left| { z }_{ 3 } \right|$ is ?

0%
$7$
0%
$10$
0%
$12$
0%
$14$

$Z_{1},Z_{2}$ are two complex numbers satisfying

$|\dfrac{Z_{1}-3Z_{2}}{3-Z_{1}Z_{2}}|=1|z_{1}|\neq 3$ then

$|z_{2}|=$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

A complex number z is said to be unimodular if

$|z| =$ . Suppose

$z_1$ and

$z_2$ are complex numbers such that

$\frac{z_1-2z_2}{2-z_1\overline {z}_2}$ is unimolecolar and

$z_2$ is not unimodular. Then the point

$z_1$ lies on a:

0%
straight line parallel to y-axis
0%
circle of radius 2.
0%
Circle of radius $\sqrt2$ .
0%
Staright line parallel to x-axis.

$\sqrt{3}+i(a+ib)(c+id)$ , then

$\tan^{-1}\left(\dfrac{b}{a}\right)+\tan^{-1}\left(\dfrac{d}{c}\right)$ has the value

0%
$\dfrac{\pi}{3}+2n\pi, n \notin 1$
0%
$n\pi+\dfrac{\pi}{6}, n \in 1$
0%
$n\pi-\dfrac{\pi}{3}, n \in 1$
0%
$2n\pi-\dfrac{\pi}{3}, n \in 1$

$z$ is a complex number such that

$|z|\ge\ 2$ , then the minimum value of

$\left| z+\dfrac { 1 }{ 2 } \right|$

0%
is equal to $\dfrac{5}{2}$
0%
is strictly greater than $\dfrac{5}{2}$
0%
lies in the interval $(1,2)$
0%
is strictly greater than $\dfrac{3}{2}$ but less than $\dfrac{5}{2}$

Explanation

minimum value of

$|z|+|z-2|$

$\Rightarrow$ if

$z=x+iy$

Then

$z-2=(x-2)+iy$

$\Rightarrow \sqrt{x^2+y^2}+\sqrt{(x-2)^2+y^2}$

$\because121=\sqrt{x^2+y^2}$

$12-21=\sqrt{(x-2)^2+y^2}$

$y>0$ and

$y<0$ Then

$|z|$ and

$|z-2|$

Not getting minimum value so y=have only possiability

$y=0$

$\Rightarrow \sqrt{x^2}+\sqrt{(x-2)^2}$

$\ge|x|+|x-2|$

$\ge |x|+|x-2|$

This function having minimum value at

$n=0$

$\ge 2$

So Minimum value is

$2$

The value of

$(z+3) (\overline{z} +3)$ is eqquivalent to

0%
$|z +3 | ^2$
0%
$| z- 3|$
0%
$z^2+3$
0%
None of these

Explanation

$(2+3)(\bar{2}+3)$

We know that

$2.\bar{2}=|2|^{2}$

$(2+3).(\overline{2+3})$

$=12+31^{2}$

The imaginary part of

$(z - 1)(\cos \, \alpha - i \, \sin \, \alpha) + (z - 1)^{-1} \times (\cos \, \alpha + i \, \sin \, \alpha )$ is zero, if

0%
$|z - 1 | = 2$
0%
$\text{arg} \, (z - 1) = 2 \alpha$
0%
$\text{arg} \, (z - 1) = \alpha$
0%
$|z | = 1$

$| z _ { 1 } | < 1 \text { and } | \frac { z _ { 1 } - z _ { 2 } } { 1 - \overline { z } _ { 1 } z _ { 2 } } | < 1 , \text { then } | z _ { 2 } | > 1$

0%
True
0%
False

Explanation

Given.

$|z _i|<1$

$\left|\frac{z_{1}-z_{2}}{1-z_{1} z_{2}}\right|<1 \\$

$\left|z_{1}-z_{2}\right|<\left|1-\bar{z}_{1} z_{2}\right| \\$

$\left|z_{1}-z_{2}\right|^{2}<\left|1-\bar{z}_{1} z_{2}\right|^{2}$

$\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)<\left(1-\bar{z}_{1} z_{2}\right)\left(1-z_{1} \bar{z}_{2}\right)$

$z_{1} \bar{z}_{1}-z_{2} \bar{z}_{1}-\bar{z}_{2} z_{1}+z_{2} \bar{z}_{2} \quad<1-z_{1} \bar{z}_{2}-\bar{z}_{1} z_{2}+\left(z_{1} \bar{z}_{1}\right)\left(z_{2} \bar{z}_{2}\right)$

$\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2} \quad\left\langle\left.\quad1+| z_{1}\right|^{2}\left|z_{2}\right|^{2}\right.$

$\left|z_{1}\right|^{2}+\left.|z_{2}\right|^{2}+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-1<0$

$\left.\left|z_{2}\right|^{2}\left((1-\left|z_{1}\right|^{2}\right)-1\left(1-z_{1}\right)^{2}\right)< 0$

$\left(\left.| z_2\right|^{2}-1\right)\left(1-\left.|z_{1}\right|^{2}\right)<0$

$\left.\left.(|z_2|^{2}-1)\right(|z_2|^{2}-1\right)>0$

$|z_1|^{2}-1<0$

$\Rightarrow\left|\left.z_{2}\right|^{2}-1<0\right.$

$\left|z_{2}\right|^{2}<1$

$\left|z_{2}\right|<1$

option

$B$ is Correct

$\left| z \right| =1$ and

$\left| \omega -1 \right| =1$ where

$z,\omega \in C$ then the largest set of values of

${ \left| 2z-1 \right| }^{ 2 }+{ \left| 2\omega -1 \right| }^{ 2 }$ equals

0%
$\left[1,9\right]$
0%
$\left[2,6\right]$
0%
$\left[2,12\right]$
0%
$\left[2,18\right]$

$\dfrac {3+2i \sin x}{1-2i \sin x}$ is purely imaginary then

$x=$ ?

0%
$n \pi \pm \dfrac {\pi}{6}$
0%
$n \pi \pm \dfrac {\pi}{3}$
0%
$2n \pi \pm \dfrac {\pi}{3}$
0%
$2n \pi \pm \dfrac {\pi}{6}$

Explanation

A complex number is said to be purely imaginary if $z+\overline { z } =0$

If $z=\dfrac { 3+2isinθ }{ 1−2isinθ }$

then $\overline { z } =\overline { \dfrac { 3+2isinθ }{ 1−2isinθ } } =\dfrac { 3-2isinθ }{ 1+2isinθ }$

$z+\overline { z } =\dfrac { 3+2isinθ }{ 1−2isinθ } +\dfrac { 3-2isinθ }{ 1+2isinθ }$

So, $\dfrac { 3+2isinθ }{ 1−2isinθ } +\dfrac { 3−2isinθ }{ 1+2isinθ } =0$

$\dfrac { (3+2isinθ)(1+2isinθ)+(3−2isinθ)(1−2isinθ) }{ (1−2isinθ)(1+2isinθ) } =0$

$3+6isinθ+2isinθ−4{ sin }^{ 2 }θ+3−6isinθ−2isinθ−4{ sin }^{ 2 }θ=0$

$6−{ 8sin }^{ 2 }θ=0$

${ sin }^{ 2 }θ=\dfrac { 3 }{ 4 }$

$sinθ=\dfrac { \sqrt { 3 } }{ 2 } =sin\dfrac { \pi }{ 3 }$

$θ=nπ+{ (−1) }^{ n }\left( \dfrac { \pi }{ 3 } \right)$

$sinθ=\dfrac { \sqrt { 3 } }{ 2 } =sin−\dfrac { \pi }{ 3 }$

$θ=nπ+{ (−1) }^{ n }\left( \dfrac { -\pi }{ 3 } \right) =nπ+{ (−1) }^{ n+1 }\left( \dfrac { \pi }{ 3 } \right)$

$\alpha$ and

$\beta$ are different complex number with

$|\beta|=1$ , then

$\left |\dfrac {\beta-\alpha}{1-\overline {\alpha }\beta}\right|$ is equal to

0%
$0$
0%
$1/2$
0%
$1$
0%
$2$

Which of the following pairs are twin primes?

0%
$(19,21)$
0%
$(29,31)$
0%
$(39,41)$
0%
$(49,51)$

Explanation

Twin primes are prime numbers that differ by $2$ .

For example 11 and 13 are twin primes.

In a sense they are “right next to each other” in the sense that(with the exception of the prime number $2)$ ,

two primes can’t be closer together.

$21$ is not prime though. so $19$ and $21$ are not twin primes.

$19$ and $17$ are twin primes, but the next prime after $19$ is $23$ which is more than $2$ greater.

$z^4+1=\sqrt{3}$ i then?

0%
$z^3$ is purely real
0%
z represents vertices of a square of side $2^{\dfrac{1}{4}}$
0%
$z^9$ is purely imaginary
0%
z represents vertices of a square of side $2^{\dfrac{3}{4}}$

$a,b,c$ are non-zero numbers and the equation

$ax^{2}+bx+c+i=0$ has purely imaginary roots, then

0%
$a=bc$
0%
$a=b^{2}c$
0%
$a=\sqrt {bc}\ if\ b > 0$
0%
$none\ of\ these$

If the roots

$a^2x^2 +2bx+c^2 = 0$ are imaginary then the roots of

$b(x^2+1)+2acx = 0$ are

0%
complex number
0%
real and unequal
0%
real and equal
0%
none

$z$ is a complex number of unit modulus and argument

$\theta$ , then the real part of

$\dfrac { z\left( 1-\bar { z } \right) }{ \bar { z } \left( 1+2 \right) }$ is:

0%
$1+cos\dfrac { \theta }{ 2 }$
0%
$1-sin{ \dfrac { \theta }{ 2 } }$
0%
$-2\sin { ^{ 2 } } \dfrac { \theta }{ 2 }$
0%
$2cos^{ 2 }\dfrac { \theta }{ 2 }$

$a > b > c$ ,

$a\neq 0$ and the system of equations

$ax+by+cz=0$ ,

$bx+cy+az=0$ ,

$cx+ay+bz=0$ has non-trivial solutions, then the roots of the quadratic equation

$at^2+bt+c=0$ .

0%
Are imaginary
0%
Are real and equal
0%
Are real and distinct
0%
May be real of imaginary

The value of

$(sin\frac{\pi }{8}+i\cos \frac{\pi }{8})^{8}{(sin\frac{\pi }{8}-i cos \frac{\pi }{8})^{8}}$ is

0%
-1
0%
0
0%
1
0%
2i

Explanation

${\left( {\sin \pi /8 + i\cos \pi /8} \right)^8}{\left( {\sin \frac{\pi }{8} - 2\cos \frac{\pi }{8}} \right)^8}$

$\Rightarrow \left( {\sin \pi + i\cos \pi } \right)\left( {\sin \pi - i\cos \pi } \right)$

$\Rightarrow - {i^2}\left( {\cos \pi - i\sin \pi } \right)\left( {\cos \pi + 2\sin \pi } \right)$

$\Rightarrow 1 \times {e^{ - \pi i}} \cdot {e^{\pi 2}}\,\,\,\,\,\left[ {\because {i^{ - 2}} = - i} \right]$

$\Rightarrow {e^0} = 1\,\,\,\,Ans.$

The equation

$x(x+2)(x^{2}-x)=-1$ , has

0%
All roots imaginary
0%
All roots negative
0%
Two roots real and two roots imaginary
0%
All roots real

$\overline { \Delta } =\begin{vmatrix} -1 & 2-3i & 5+4i \\ 2+3i & 8 & 1-i \\ 5-4i & 1+i & 3 \end{vmatrix}$ then

$\Delta =$

0%
Purely real
0%
Purely imaginary
0%
Complex
0%
$0$

$\theta$ real then the modulus of

$\dfrac{1}{1+\cos\theta+i\sin\theta}$ is

0%
$\dfrac{1}{2}\sec\dfrac{\theta}{2}$
0%
$\dfrac{1}{2}\cos\dfrac{\theta}{2}$
0%
$\sec\dfrac{\theta}{2}$
0%
$\cos\dfrac{\theta}{2}$

The function of imaginary roots of the equation

$(x-1)(x-2)(3x+1)=32$ is

0%
$0$
0%
$1$
0%
$2$
0%
$4$

The number of imaginar roots of the equation

$(x-1)(x-2)(3x-2)(3x+1)=32$ is

0%
$Zero$
0%
$1$
0%
$2$
0%
$4$

$\mathrm{{z} _ { 1 }} = 10 + 6\mathrm{i} , \mathrm{{ z } _ { 2 }}= 4 + 6 \mathrm { i }$ and

$\mathrm{ z}$ is a complex number such that

$\operatorname { amp } \left( \dfrac { \mathrm { z } - \mathrm { z } _ { 1 } } { \mathrm { z } - \mathrm { z } _ { 2 } } \right) = \dfrac { \pi } { 4 }$ , then the value of

$\left| \mathrm{z} - 7 - 9 \mathrm { i } \right|$ is equal to

0%
$\sqrt { 2 }$
0%
$2\sqrt { 2 }$
0%
$3\sqrt { 2 }$
0%
$2\sqrt { 3 }$

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

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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