Compare List I with List II and choose the correct answer using codes given below: List I (Complex number) List II (Its modulus) $$(4-3i)$$ $$10$$ $$(8+6i)$$ $$\dfrac{1}{5}$$ $$\dfrac{1}{(3+4i)}$$ $$1$$ $$\dfrac{(3-4i)}{(3+4i)}$$ $$5$$

$$(i)-(s), (ii)-(p), (iii)-(q), (iv)-(r)$$

$$(i)-(p), (ii)-(s), (iii)-(r), (iv)-(q)$$

$$(i)-(s), (ii)-(p), (iii)-(r), (iv)-(q)$$

$$(i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)$$

MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 10

A number system with a base of two is referred as ______________.

0%
Unary number system
0%
Binary number system
0%
Octal number system
0%
None of these

If a register containing data (11001100)2 is subjected to arithmetic shift left operation, then the content of the register after 'ashl' shall be _____________.

0%
(11001100)2
0%
(1101100)2
0%
(10011001)2
0%
(10011000)2

C it refers to a _____________.

0%
computer language.
0%
CPU instruction.
0%
0 or 1 value.
0%
digital representation of an alphabetic character.

Which of the following is true?

0%
Byte is a single digit in a binary number
0%
Bit represents a grouping of digital numbers
0%
Eight-digit binary number is called a bit
0%
Eight-digit binary number is called a byte

State true(T) or false(F).
The sum of primes cannot be a prime.

0%
True
0%
False

Explanation

This statement is also false.

As consider, the numbers

$2$ and

$3$ then sum of

$2$ and

$3$ i.e. is

$5$ is also a prime number.

State true or false:
The product of primes cannot be a prime.

0%
True
0%
False

Explanation

The product of two primes can't be a prime because it violates the definition of the prime number as it will be divided by the prime numbers which are multiplied rather than

$1$ and the number itself. So the number is not prime it will become a composite number instead.

State true(T) or false(F).
Odd numbers cannot be composite.

0%
True
0%
False

Explanation

The given statement is false.

As, consider the number

$9$ , which is an odd number and it is a composite number too.

Since

$3$ divides the given number.

$3$ is a divisor of

$9$ other than

$1$ and

$9$ .

State true(T) or false(F).
An even number is composite.

0%
True
0%
False

Explanation

This statement is also false.

Since

$2$ is the number which is both even and prime.

Hence an even number may not be composite.

Mark the correct alternative of the following.
Which of the following numbers is prime?

0%
$23$
0%
$51$
0%
$38$
0%
$26$

The least prime is?

0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

The number

$2$ is the least prime. It is the only even prime also.

Mark the correct alternative of the following.
Which of the following are not twin-primes?

0%
$3, 5$
0%
$5, 7$
0%
$11, 13$
0%
$17, 23$

Mark the correct alternative of the following.
Which of the following numbers are twin primes?

0%
$3, 5$
0%
$5, 11$
0%
$3, 11$
0%
$13, 17$

Mark the correct alternative of the following.
The smallest number which is neither prime nor composite is?

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

Express the following complex numbers in the standard form

$a+ib$ :

$\left ( \dfrac{1}{1-4i}-\dfrac{2}{1+i} \right )\left ( \dfrac{3-4i}{5+i} \right )$

0%
$\dfrac{307}{442}+i \dfrac{599}{442}i$
0%
$\dfrac{307}{442}-i \dfrac{599}{442}i$
0%
$\dfrac{-307}{442}+i \dfrac{599}{442}i$
0%
None of the above

Express the following complex numbers in the standard form

$a+ib$ :

$\dfrac{\left ( 2+i \right )^{3}}{2+3i}$

0%
$\dfrac{37}{13}-\dfrac{16}{13}i$
0%
$\dfrac{-37}{13}+\dfrac{16}{13}i$
0%
$\dfrac{37}{13}+\dfrac{16}{13}i$
0%
None of the above

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:

$1-i$

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

Explanation

Let

$z=1-i$ . Then,

$\left | z \right |=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$ .

Let

$\alpha$ be the acute angle given by

$tan\, \alpha =\dfrac{\left | Im(z) \right |}{\left | Re(z) \right |}$ .

Then,

$tan\, \alpha =\dfrac{|-1|}{|1|}=1\Rightarrow \alpha =\dfrac{\pi }{4}$

Clearly, z lies in the fourth quadrant. Therefore,

$arg(z)= -\alpha =-\dfrac{\pi }{4}$ .

Express the following complex numbers in the standard form

$a+ib$ :

$\dfrac{3-4i}{\left ( 4-2i \right )\left ( 1+i \right )}$

0%
$\dfrac{1}{4}+\dfrac{3}{4}i$
0%
$\dfrac{1}{4}-\dfrac{3}{4}i$
0%
$\dfrac{-1}{4}-\dfrac{3}{4}i$
0%
None of the above

Express the following complex numbers in the standard from

$a+ib$ :

$\dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$

0%
$1-2\sqrt{2}i$
0%
$1+\sqrt{2}i$
0%
$1+2\sqrt{2}i$
0%
$1-\sqrt{2}i$

Explanation

$\dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$

Rationalizing the denominator, we get

$= \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i} \times \dfrac{1+\sqrt{2}i}{1+\sqrt{2}i}$

$= \dfrac{5+2i^2+(5+1)\sqrt{2}i}{1+2}$

$= \dfrac{3+6\sqrt{2}i}{1+2}$ ......................

$(\because i^2=-1)$

$= 1+2\sqrt{2}i$

The real part of

$(i - \sqrt{3})^{13}$ is

0%
$2^{-10}\sqrt3$
0%
$-2^{12}\sqrt3$
0%
$2^{-12}\sqrt3$
0%
$-2^{-12}\sqrt3$
0%
$-2^{10}\sqrt3$

Explanation

$w^2=\dfrac{-1-i\sqrt3}{2}$

$\Rightarrow iw^2=\dfrac{-i+\sqrt3}{2}$

$\Rightarrow -2iw^2=i-\sqrt3$

$\Rightarrow (-2iw^2)^{13}=(i-\sqrt3)^{13}$

$\Rightarrow -2^{13}i^{13}w^{26}=(i-\sqrt3)^{13}$

$\Rightarrow -2^{13}iw^2=(i-\sqrt3)^{13}$

$\Rightarrow -2^{13}\left({\dfrac{-i+\sqrt3}{2}}\right)=(i-\sqrt3)^{13}$

$\Rightarrow -2^{12}(-i+\sqrt3)=(i-\sqrt3)^{13}$

$\therefore (i-\sqrt3)^{13}=-2^{12}\sqrt3+i2^{12}$

Let z be a complex number such that

$\left|\dfrac{z-i}{z+2i}\right|=1$ and

$|z|=\dfrac{5}{2}$ . Then the value of

$|z+3i|$ is?

0%
$\dfrac{7}{2}$
0%
$\dfrac{15}{4}$
0%
$2\sqrt{3}$
0%
$\sqrt{10}$

Explanation

Given

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

$\Rightarrow |z - i| = |z + 2i|$

Let

$z = x + iy$

So,

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

$-2y + 1 = 4y + 4$

$6y = -3 \Rightarrow y = \dfrac{-1}{2}$

$x^2 + y^2 = \dfrac{25}{4} \Rightarrow x^2 = \dfrac{24}{4} = 6$

$z = \pm \sqrt{6} - \dfrac{i}{2}$

$|z + 3i| = \sqrt{6 + \dfrac{25}{4}} = \sqrt{\dfrac{49}{4}} = \dfrac{7}{2}$

$\therefore$

$\boxed{|2 + 3i| = \dfrac{7}{2}}.....Answer$

Hence option

$'A'$ is the answer.

Mark against the correct answer in each of the following .

$i^{91}=$ ?

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

Explanation

$i^{91}\\=(1^4)^{22}\times i^3 \\=1\times i^2 \times i\\=1\times (-1)\times i\\=-i$

$(1-\sqrt{-1})(1+\sqrt{-1})(5-\sqrt{-7})(5+\sqrt{-7})=?$

0%
$(25+7i)$
0%
$(32+5i)$
0%
$(29-3i)$
0%
$none\ of\ these$

Explanation

Given expression

$=(1-i)(1+i)(5-\sqrt 7 i)(5+\sqrt 7i)$

$=(1-i^2)(25-7i^2)\\=(1+1)(25+7)\\=(2\times 32)=64$ .

Mark against the correct answer in each of the following .

$i^{326}=$ ?

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

Explanation

$i^{326}\\=(i^4)^{81}\times i^2 =(1)^{81}\times (-1)\\=1\times (-1)=-1$

$\mid z^2 - 3\mid = 3\mid z\mid$ then the maximum value of

$\mid z\mid$ is

0%
1
0%
$\dfrac{3+\sqrt {21}}{2}$
0%
$\dfrac{\sqrt {21} - 3}{2}$
0%
none of these

Explanation

$\mid z^{2} - 3 \mid \geq \mid z \mid^{2} - 3$

$\Rightarrow 3 \mid z \mid \geq \mid z \mid ^{2} - 3$

$\Rightarrow \mid z \mid ^{2} - 3 \mid z \mid -3 \leq 0$

$\Rightarrow 0 < \mid z \mid \leq \dfrac{3 + \sqrt{21}}{2}$

$z_1$ and

$z_2$ are any two complex numbers then

$|z_1 +\sqrt {z_1^2 -z_2^2}| + |z_1 -\sqrt {z_1^2 -z_2^2}|$ is equal to

0%
$|z_1|$
0%
$|z_2|$
0%
$|z_1 + z_2|$
0%
$None\ of\ these$

Explanation

We know that,

$|z_1+z_2|^2+|z_1-z_2|^2=2[|z_1|^2+|z_2|^2]$ ---- ( 1 )

Now,

$\left[|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|\right]^2$

$=\left|z_1+\sqrt{z_1^2-z_2^2}\right|^2+\left|z_1-\sqrt{z_1^2-z_2^2}\right|^2+2\left|z_1^2-(z_1^2-z_2^)\right|$

$=2|z_1|^2+2|z_1^2-z_2^2|+2|z_2^2|$ [ From ( 1 ) ]

$=2|z_1|^2+2|z_2|^2+2|z_1^2-z_2^2|$

$=|z_1+z_2|^2+|z_1-z_2|^2+2|z_1+z_2||z_1-z_2|$

$=\left(\left|z_1+z_2\right|+\left|z_1-z_2\right|\right)^2$

Taking square root on both sides we get,

$|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|$

$=\left|z_1+z_2\right|+\left|z_1-z_2\right|$

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

0%
$(6+17i)$
0%
$(6-17i)$
0%
$(-6+17i)$
0%
$none\ of\ these$

Explanation

$(2-3i)(-3+4i)\\=(-6+8i+9i-12i^2)\\=(-6+17i+12)\\=(6+17i)$ .

Mark against the correct answer in each of the following .

$i^{273}=$ ?

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

Explanation

$i^{273}=(i^4)^{68}\times i=(1)^{68}\times i=(1\times i)=i$

Compare List I with List II and choose the correct answer using codes given below:

List I (Complex number)	List II (Its modulus)
$(4-3i)$	$10$
$(8+6i)$	$\dfrac{1}{5}$
$\dfrac{1}{(3+4i)}$	$1$
$\dfrac{(3-4i)}{(3+4i)}$	$5$

0%
$(i)-(p), (ii)-(s), (iii)-(r), (iv)-(q)$
0%
$(i)-(s), (ii)-(p), (iii)-(q), (iv)-(r)$
0%
$(i)-(s), (ii)-(p), (iii)-(r), (iv)-(q)$
0%
$(i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)$

Which of the following is a composite number?

0%
$23$
0%
$29$
0%
$32$
0%
none of these

Explanation

(a)

$23$
Since it cannot be broken into factors
Hence

$23$ is not a composite number
(b)

$29$
Since it cannot be broken into factors
Hence

$29$ is not a composite number
(c)

$32$
Since it can be broken into factors i.e.

$2 \times 2 \times 2 \times 2 \times 2$
Hence

$32$ is a composite number
Option (c) is the correct answer

State whether the following statements are true (T) or false (F) :
There are infinitely many prime numbers.

0%
True
0%
False

State whether the following statements are true (T) or false (F):
A natural number is called a composite number if it has at least one more factor other than

$1$ and the number itself.

0%
True
0%
False

Which of the following is an odd composite number ?

0%
$7$
0%
$9$
0%
$11$
0%
$12$

Explanation

$9 = 3 \times 3$ , is an odd composite number.

The modulus of

$\overline { 6+{ i }^{ 3 } } +\overline { 6+{ i } }+\overline { 6+{ i }^{ 2 } }$ is

0%
$17$
0%
$\sqrt{533}$
0%
$\sqrt{456}$
0%
$49$

Given

$z$ is a complex number with modulus

$1$ . Then the equation in

$a$ ,

$\left(\dfrac{1+ia}{1-ia}\right )^4=z$ has

0%
all roots real and distinct.
0%
two real and two imaginary.
0%
three roots real and one imaginary.
0%
one root real and three imaginary.

Explanation

$\left (\dfrac{1+ia}{1-ia}\right )^4 = z$

$\left (\dfrac{-i(i-a)}{-i(i+a)}\right )^4 = z$

$\left (\dfrac{a-i}{a+i}\right )^4= z$

$\left |\dfrac{a-i}{a+i}\right |^4= |z| =1$

$|a-i| ^4 = |a+i|^4$

$|a-i| = |a+i|$

$\therefore a$ lies on the perpendicular bisector of

$i$ and

$-i$ .

$a$ lies on real axis. Hence the roots all are real and distinct.

$\displaystyle z_{0}=\frac{1-i}{2}$ , then

$\displaystyle \left (1+z_{0} \right )\left (1+z_{0}^{{2}^{1}} \right )\left (1+z_{0}^{{2}^{2}} \right ).......... \left (1+z_{0}^{2^n} \right )$ must be

0%
$(1-i)(1+\dfrac{1}{2^{2^n}})$ for $n>1$
0%
$(1-i)(1-\dfrac{1}{2^{2^n}})$ for $n>1$
0%
$\dfrac{1+i}{2}$ for $n>1$
0%
$(1-i)(1-\dfrac{1}{2^{2^{n+1}}})$ for $n>1$

Explanation

Let

$P=(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$

$\Rightarrow (1-z_0)P=(1-z_0)(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$

$\Rightarrow (1-z_0)P=(1-{z_0}^{2^2})(1+{z_0}^{2^2}).......(1+{z_0}^{2^n})$

$\Rightarrow (1-z_0)P=1-{z_0}^{2^{n+1}}$

$\Rightarrow P=\dfrac{1-{z_0}^{2^{n+1}}}{1-{z_0}}$

$\Rightarrow P=\dfrac{1-({z_0}^2)^{2n}}{1-z_0}$

$n>1$

Since,

$z_0=\dfrac{1-i}{2}$

${z_0}^2=\dfrac{1-1-2i}{4}$

$\Rightarrow {z_0}^2=\dfrac{-i}{2}$

$\therefore \ P=\dfrac{1-\left(\dfrac{-i}{2}\right)^{2^n}}{1-z_0}$

$\Rightarrow P=\dfrac{1-\dfrac{1}{2^{2^n}}}{\dfrac{1+i}{2}}=\dfrac{\left(1-\dfrac{1}{2^{2^n}}\right)2(1-i)}{1+1}$

$\Rightarrow P=(1-i)\left(1-\dfrac{1}{2^{2^n}}\right)$

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

Explanation

$\textbf{Step -1: Using remainder theorem to get the following equations.}$

$\text{Given, Dividing }f(z) \text{ by } z - i \text{, we get remainder }i$

$\text{Therefore by remainder theorem, }f(i) = i$

$\dots \text{(i)}$

$\text{Given, Dividing }f(z) \text{ by } z + i \text{, we get remainder }1 + i$

$\text{Therefore by remainder theorem, }f(-i) = 1+i$

$\dots \text{(ii)}$

$\text{Let }R(x) \text{ be the remainder when }f(z)\text{is divided by }z^2 + 1$

$\text{degree of R(z) is 1, hence }R(z) = az+b$

$\text{we can simplify }z^2 + 1 = (z -i) (z+i)$

$\therefore f(z) = Q(z)(z-i)(z+i) + R(z)$

$\textbf{Step -2: Solving for a and b.}$

$\text{From (i) and (ii)}$

$f(i) = i = ai + b$

$\Rightarrow b = i - ai$

$f(-i) = 1+i = -ai +b$

$\Rightarrow -ai + i - ai = 1+i$

$\Rightarrow -2ai = 1$

$\Rightarrow a = -\dfrac{1}{2i}$

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

$= -\dfrac{i}{2(-1)}$

$\therefore a = \dfrac{i}{2}$

$b = i - ai$

$\Rightarrow b = i - \dfrac{i}{2} (i)$

$\Rightarrow b = i - \dfrac{(-1)}{2}$

$\Rightarrow b = i + \dfrac{1}{2}$

$\therefore R(z) = az+b$

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

$R(z) = \dfrac{1}{2}(iz + 1) + i$

$\textbf{Hence option B is correct}$

The number of solutions of

$log _{\frac{1}{5}}log_{\frac{1}{2}}(\left | z \right |^{2}+4\left | z \right |+3)< 0$ is/are?

0%
$0$
0%
$2$
0%
$4$
0%
infinite

Explanation

The glven inequality becomes

$log_{\frac{1}{2}}|\mathrm{z}|^{2}+4|\mathrm{z}|+3>1$ , since

$\displaystyle \frac{1}{5}<1$

$\displaystyle \Rightarrow|\mathrm{z}|^{2}+4|\mathrm{z}|+3<\frac{1}{2}$

$\Rightarrow(|\mathrm{z}|+2)^{2}$ + 1

$<\displaystyle \frac{1}{2}$

$(|\displaystyle \mathrm{z}|+2)^{2}<\frac{-1}{2}$
This is not possible. Hence, there is no solution.

Let

$z$ be a complex number and

$c$ be a real number

$\geq$ 1 such that z +

$c\left | z+1 \right |+i=0 ,$ then

$c$ belongs to

0%
$[2, 3]$
0%
$(3, 4)$
0%
$[1,\sqrt{2}]$
0%
None of these

Explanation

Since

$c$

$\left | z+1 \right |$ is real

$z + i$ is real
let

$z-i = x$ , where

$x$ is real

$z+i=-c$

$\left | z+1 \right |$ (given)

$x^{2}=c^{2}\left \{ x+1 \right \}^{2}+1^{2})$

$(c^{2}-1)x^{2}+2c^{2}x+2c^{2}=0$
As

$x$ is real

$4c^{4}-8c^{2}(c^{2}-1)\geq 0$

$4c^{2}-(c^{2}-2)\leq 0$

$c^{2}\leq 2$

$c\leq \sqrt{2}$
But

$c\geq1$

$\Rightarrow c\, \, \epsilon [1,\sqrt{2}]$

$\displaystyle \log_{\tan 30^{\circ}}\left(\frac{2|Z|^{2}+2|Z|-3}{|z|+1}\right) <-2$ then

0%
$|\displaystyle \mathrm{z}|<\frac{3}{2}$
0%
$|z|>\displaystyle \frac{3}{2}$
0%
$|z|>2$
0%
$|z|<2$

Explanation

we have,

$\Rightarrow log_{(\dfrac{1}{\sqrt{3}})}(\dfrac{2|z|^{2}+2|z|-3}{|z|+1})<-2$

$\Rightarrow \dfrac{2|z|^{2}+2|z|-3}{|z|+1}>(\dfrac{1}{\sqrt{3}})^{-2}=(\sqrt{3})^{2}=3$ [as

$\dfrac{1}{\sqrt{3}}<1$ so equality sign changes]

$\Rightarrow 2|z|^{2}+2|z|-3>3|z|+3$

$\Rightarrow 2|z|^{2}-|z|-6>0$

$\Rightarrow 2|z|^{2}-4|z|+3|z|-6>0$

$\Rightarrow 2|z| (|z|-2)+3 (|z|-2)>0$

$\Rightarrow (2|z|+3) (|z|-2)>0$

$\Rightarrow |z|>2$

$x = 2 + 5i($ where

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

$2(\displaystyle \frac{1}{1! 9!} + \frac{1}{3! 7!}) + \frac{1}{5! 5!} = \frac{2^{a}}{b!}$ then

$x^{3}-5x^{2}+33x-10 =$

0%
$a + b$
0%
$b - a$
0%
$a-b$
0%
$-a-b$
0%
$(a - b)(a + b)$

Explanation

$\displaystyle \frac{10! 2^{a}}{b!} = 2[^{10}C_{1} + ^{10}C_{3}] + ^{10} C_{5}$

$\displaystyle \frac{10! 2^{a}}{b!} = ^{10}C_{1} + ^{10}C_{3} + ^{10}C_{5} + ^{10}C_7 + ^{10}C_{9} = 2^{9}$

$\therefore a = 9, b = 10$

$x = 2 + 5i$

$(x - 2)^{2} = - 25$

$x^{2} - 4x + 29 = 0$

$\therefore x^{3} - 5x^{2} + 33x - 10 = (x - 1)(x^{2} - 4x + 29) + 19$

$\Rightarrow 19 = a + b$

$\left | \log_{\sqrt{3}} \frac{\left | z \right |^{2}-\left | z \right |+1}{2+{}\left | z \right |}\right |< 2$ , then

0%
$|\displaystyle \mathrm{z}|<\frac{1}{3}$
0%
$|\mathrm{z}|=1$
0%
$|\mathrm{z}|=5$
0%
$1<|\mathrm{z}|<5$

Explanation

$\left|log_{{\sqrt{3}}}(\dfrac{|z|^{2}-|z|+1}{|z|+2})\right|<2$

$\Rightarrow -2<\log_{{\sqrt{3}}}(\dfrac{|z|^{2}-|z|+1}{|z|+2})<2$

$\Rightarrow \dfrac{1}{3}<\dfrac{|z|^{2}-|z|+1}{|z|+2}<3$

$\Rightarrow |z|+2<3|z|^{2}-3|z|+3$

$\Rightarrow 3|z|^{2}-4|z|+1>0$

$\Rightarrow (3|z|-1)(|z|-1)>0$

Hence

$|z|<\dfrac{1}{3}$ and

$|z|>1$ ...(i)

Similarly

$\dfrac{|z|^{2}-|z|+1}{|z|+2}<3$

$\Rightarrow |z|^{2}-|z|+1<3|z|+6$

$\Rightarrow |z|^{2}-4|z|-5<0$

$\Rightarrow (|z|+1)(|z|-5)<0$

$\Rightarrow 1<|z|<5$ ...(ii)

From (i) and (ii), we get

$1<|z|<5$

If z be a complex number satisfying

$\displaystyle\ z^{4}+z^{3}+2z^{2}+z+1=0$ then

$\displaystyle\ |z|$ is

0%
$\displaystyle\ \frac{1}{2}$
0%
$\displaystyle\ \frac{3}{4}$
0%
$\displaystyle\ 1$
0%
None of these

Explanation

${ z }^{ 4 }+{ z }^{ 3 }+2{ z }^{ 2 }+z+1=0$

$\left[ { z }^{ 4 }+2{ z }^{ 2 }+1 \right] +\left[ { z }^{ 3 }+z \right] =0$

$\left[ { z }^{ 2 }+1 \right] \left[ { z }^{ 2 }+1+z \right] =0$

${ z }^{ 2 }+1=0\Longrightarrow z=\pm i$

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

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

$z=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } =\cfrac { -1\pm \sqrt { 3i } }{ 2 }$

$\left| z \right| =\sqrt { \cfrac { 1 }{ 4 } +\cfrac { 3 }{ 4 } } =1$

$\therefore \boxed { \left| z \right| =1 }$

If the expression

${(1+ir)}^{3}$ is of the form of

$s(1+i)$ for some real

$s$ where

$r$ is also real and

$i=\sqrt{-1}$ , then the value of

$r$ can be

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$\cot{\cfrac{\pi}{8}}$
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$\sec{\pi}$
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$\tan{\cfrac{\pi}{12}}$
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$\tan{\cfrac{5\pi}{12}}$

Find the value of

$x$ such that

$\displaystyle \frac{(x + \alpha)^2 - (x + \beta)^2}{ \alpha + \beta} = \frac{sin 2 \theta}{sin^2 \theta}$ . when

$\alpha$ and

$\beta$ are the roots of

$t^2 - 2t + 2 = 0$

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$x = icot \, \, \, \theta - 1$
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$x = -(icot \, \, \, \theta + 1$ )
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$x = icot \, \, \, \theta$
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$x = itan \, \, \, \theta - 1$

Explanation

$t^{2}-2t+2=0$

$(t-1)^{2}-1+2=0$

$(t-1)^{2}=-1$

$t=1\pm i$
Hence

$\alpha+\beta=2$

$\alpha-\beta=2i$
Now let

$n=2$
We get

$\dfrac{2x(\alpha-\beta)+\alpha^{2}-\beta^{2}}{2}=\dfrac{sin2\theta}{sin^{2}\theta}$
Hence

$\dfrac{4ix+2i-(-2i)}{2}=\dfrac{2sin\theta.cos\theta}{sin^{2}\theta}$

$2ix+2i=2cot\theta$

$ix=cot\theta-i$

$-x=icot\theta+1$

$x=-(1+icot\theta)$

$z_1, z_2$ be two non zero complex numbers satisfying the equation

$\displaystyle \left | \frac{z_1 + z_2}{z_1 - z_2} \right | = 1$ then

$\displaystyle \frac{z_1}{z_2} + \left ( \frac{z_1}{z_2} \right )$ is

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zero
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1
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purely imaginary
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2

Explanation

let

$\frac{z_{1}}{z_{2}} =x$

given

$|\frac{x+1}{x-1}| = 1$

$\Rightarrow |x+1|=|x-1|$

Let

$x=a+ib$

We get

$|a+1+ib|=|a-1+ib|$

$\Rightarrow a=0$

Therefore

$x=ib$ and

$\overline { x } =-ib$

Therefore

$x+\overline { x } =ib-ib=0$

So the correct option is

$A$

Find the range of real number

$\alpha$ for which the equation

$z + \alpha |z - 1| + 2i = 0; z= x + iy$ has a solution. Find the solution.

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$\displaystyle x = 5/2, y = - 2$
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$\displaystyle x = -2, y = 5/2$
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$\displaystyle x = -5/2, y = 2$
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$\displaystyle x = 2, y = -5/2$

Explanation

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

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

$x^{2}-(y+2)^{2}+i2x(y+2)=\alpha^{2}((x-1)^{2}+y^{2})$
Real part

$x^{2}-\alpha(x-1)^{2}-\alpha(y^{2})=0$ and
Imaginary part

$2x(y+2)=0$

$x(y+2)=0$
Either

$x=0$ or

$y=-2$ ...if the above equations have a solution.
Considering y=-2,
Hence

$\alpha\epsilon[-\dfrac{\sqrt{5}}{2},\dfrac{\sqrt{5}}{2}]$
Hence

$x=\dfrac{5}{2}$ if

$|\alpha|\neq1$ .

If n is a natural number

$\geq 2$ , such that

$z^n=(z+1)^n$ , then

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Roots of equation lie on a straight line parallel to the $y-axis$
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Roots of equation lie on a straight line parallel to the $x-axis$
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Sum of the real parts of the roots is $-[(n-1)/2]$
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None of these

Explanation

Given,

$z^n=(z+1)^n\rightarrow |z^n|=|(z+1)^n|$

$\therefore |z|^n = |z+1|^n\Rightarrow |z|=|z+1|$

$\Rightarrow |z|^2=|z+1|^2$

$\Rightarrow x^2+y^2=(x+1)^2+y^2$ , where

$z=x-iy$

$\Rightarrow x=-\dfrac {1}{2}$
Hence,

$z$ lies on the line

$x=-1/2$ .
Hence, sum of real parts of the roots is

$-(n-1)/2$ (since equation has

$n-1$ roots).

Let

$\displaystyle\ z_{1}= a+ib, z_{2}= p+iq$ be two unimodular complex numbers such that

$\displaystyle\ Im(z_{1}z_{2})=1$ . If

$\displaystyle\ \omega_{1}= a+ip, \omega_{2}=b+iq$ then

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$\displaystyle\ Re(\omega_{1}\omega_{2})=1$
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$\displaystyle\ Im(\omega_{1}\omega_{2})=1$
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$\displaystyle\ Rm(\omega_{1}\omega_{2})=0$
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$\displaystyle\ Im(\omega_{1}\bar{\omega_{2}})=0$

Explanation

$z_1 = a + ib$ and

$z_2 = p + iq$
Given that

$Im(z_1z_2) = 1$ , so

$aq + bp = 1$
Now,

$\omega_1 = a + ip, \omega_2 = b + iq$

$\Rightarrow \omega_1\omega_2 = ab - pq + i(aq + bp)$

$\Rightarrow Im(\omega_1\omega_2) = 1$

Find the regions of the z-plane for which

$\displaystyle \left | \frac{z - a}{z + \overline a} \right | < 1, = 1$ or

$> 1$ . when the real part of a is positive.

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The required regions are the right half of the z-pane, the imaginary axis and the left half of the z-plane respectively.
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The required regions are the left half of the z-pane, the imaginary axis and the right half of the z-plane respectively.
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The required regions are the right half of the z-pane the real axis and the left half of the z-plane respectively.
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The required regions are the left half of the z-pane the real axis and the right half of the z-plane respectively.

Explanation

Let

$z=x+iy$ and

$a$ be a fixed complex number

$(a+ib)$ .
Hence

$|\dfrac{z-a-ib}{z+a-ib}|=1$

$|(x-a)+i(y-b)|=|(x+a)+i(y-b)|$

$(x-a)^{2}=(x+a)^{2}$

$2x(a)=0$

$x=0$ ... imaginary axis.
If

$|\dfrac{z-a-ib}{z+a-ib}|<1$
Then

$(x-a)^{2}<(x+a)^{2}$
Or

$x>0$ ... positive real axis.
If

$|\dfrac{z-a-ib}{z+a-ib}|>1$

$(x-a)^{2}>(x+a)^{2}$

$x<0$ ... negative real axis.

Find all complex numbers satisfying the equation

$2|z|^2 + z^2 - 5 + i \sqrt{3} = 0$

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$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} + \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} + \frac{3}{2} i \right )$
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$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{2}{\sqrt{6}} - \frac{3}{2} i \right )$
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$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{3}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{2} i \right )$
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$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{\sqrt{2}} i \right )$

Explanation

$Let\quad z=a+ib\\ Thus:\quad 2({ a }^{ 2 }+{ b }^{ 2 })+({ a }^{ 2 }-{ b }^{ 2 }+2abi)-5+i\sqrt { 3 } =0\\ =>3{ a }^{ 2 }+{ b }^{ 2 }-5=0\quad and\quad 2ab+\sqrt { 3 } =0=>a=-\frac { \sqrt { 3 } }{ b } \\ Substituting\quad in\quad the\quad first:\quad \frac { 9 }{ 4{ b }^{ 2 } } +{ b }^{ 2 }=5=>{ b }^{ 2 }=\frac { 9 }{ 2 } \quad or\quad \frac { 1 }{ 2 } \\ =>b=\pm \frac { 3 }{ \sqrt { 2 } } \quad or\quad \pm \frac { 1 }{ \sqrt { 2 } } =>a=\mp \frac { 1 }{ \sqrt { 6 } } \quad or\quad \mp \frac { \sqrt { 3 } }{ \sqrt { 2 } } =\mp \frac { \sqrt { 6 } }{ 2 } \\ =>z=\pm \left( \frac { \sqrt { 6 } }{ 2 } -\frac { i }{ \sqrt { 2 } } \right) \quad or\quad \pm \left( \frac { 1 }{ \sqrt { 6 } } -\frac { 3i }{ \sqrt { 2 } } \right) \\$
Hence, (d) is correct.

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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