Number Theory Test 49

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Number Theory Test 49

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$\displaystyle\ Re(\omega_{1}\omega_{2})=1$$

B
$$\displaystyle\ Im(\omega_{1}\omega_{2})=1$$

C
$$\displaystyle\ Rm(\omega_{1}\omega_{2})=0$$

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

Solution

$$z_1 = a + ib$$ and $$z_2 = p + iq$$
Given that $$Im(z_1z_2) = 1$$, so $$aq + bp = 1$$
Now, $$\omega_1 = a + ip, \omega_2 = b + iq$$
$$\Rightarrow \omega_1\omega_2 = ab - pq + i(aq + bp)$$
$$\Rightarrow Im(\omega_1\omega_2) = 1$$
Question 2
1 / -0

If $$z = x+iy$$ and $$w = \dfrac{(1-iz)}{(z-i)}$$, then $$|w| = 1$$ implies that, in the complex plane

A
$$z$$ lies on the imaginary axis

B
$$z$$ lies on the real axis

C
$$z$$ lies on the unit circle

D
None of these

Solution

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.
Question 3
1 / -0

If $$(\sqrt 3-i)^n=2^n, n\in N$$, then $$n$$ is a multiple of

A
$$6$$

B
$$10$$

C
$$9$$

D
$$12$$

Solution

$$\sqrt3-i = 2\left(\dfrac{\sqrt{3}-i}{2}\right)$$
$$=2.e^{-i\frac{\pi}{6}}$$
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.
Question 4
1 / -0

Given that $$i = \sqrt {-1}$$, find the multiplicative inverse of $$5 - i$$.

A
$$5 + i$$

B
$$\dfrac {5 + i}{26}$$

C
$$\dfrac {1}{5 + i}$$

D
$$\dfrac {5 + i}{24}$$

E
$$\dfrac {5 - i}{24}$$

Solution

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 $$\dfrac {(5+i)}{26}$$.
Question 5
1 / -0

Write the complete number $$- 2 - 2i$$ in polar form.

A
$$2(cos\dfrac{\pi}{4}+i\,sin\dfrac{\pi}{4})$$

B
$$-2(cos\dfrac{\pi}{4}-i\,sin\dfrac{\pi}{4})$$

C
$$2\sqrt{2}(cos\dfrac{3\pi}{4}+i\,sin\dfrac{3\pi}{4})$$

D
$$2\sqrt{2}(cos\dfrac{7\pi}{4}+i\,sin\dfrac{7\pi}{4})$$

E
$$2\sqrt{2}(cos\dfrac{5\pi}{4}+i\,sin\dfrac{5\pi}{4})$$
Solution
- 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 $$2\sqrt2(cos(225)+isin(225)) = 2\sqrt2 (cos\frac{5\pi}{4}+isin\frac{5\pi}{4})$$
Question 6
1 / -0

An example for twin primes is _____.

A
$$5, 11$$

B
$$3, 5$$

C
$$11, 17$$

D
$$3, 7$$

Solution

Two continuous prime number are called twin prime numbers.
Only $$3$$ and $$5$$ are continuous prime numbers.
Hence, the answer is $$3,5$$.
Question 7
1 / -0

What is the product of the complex numbers $$\left( -3i+4 \right) $$ and $$\left( 3i+4 \right) $$?

A
$$1$$

B
$$7$$

C
$$25$$

D
$$-7+24i$$

E
$$7+24i$$

Solution

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.
Question 8
1 / -0

Add and express in the form of a complex number $$a+bi$$
$$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$$

A
$$-4+9i$$

B
$$-5+9i$$

C
$$2+9i$$

D
$$-5+8i$$

Solution

Adding the given expression as follows:
$$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$$
$$=(2+3i)+(-4+5i)-(3-i)$$
$$=2+3i-4+5i-3+i$$
$$=-5+9i$$
Question 9
1 / -0

What is the next-highest prime number after 67?

A
68

B
69

C
71

D
73

E
76

Solution
Question 10
1 / -0

If $$\alpha $$ and $$\beta$$ are two different complex numbers with $$|\beta|=1$$, then $$\left | \dfrac{\beta -\alpha}{1-\bar{\alpha }\beta } \right |$$ is equal to.

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$-1$$

Solution

Let $$t = \left|\cfrac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right|$$
Multiply the numerator and denominator with $$\overline { \beta } $$
$$\Rightarrow t = \left|\cfrac { 1-\alpha \overline { \beta } }{ \overline { \beta } -\overline { \alpha } } \right| = \left|\cfrac { \overline { 1- \alpha \overline { \beta } } }{ \overline { \overline{\beta} -\overline{\alpha} } } \right| = \left|\cfrac { 1-\overline { \alpha } \beta }{ \beta -\alpha } \right|=\cfrac{1}{t}$$
$$\Rightarrow {t}^{2} = 1$$
$$\Rightarrow t = 1$$

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Number Theory Test 49

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Number Theory Test 49

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Question 1

$$\displaystyle\ Re(\omega_{1}\omega_{2})=1$$

$$\displaystyle\ Im(\omega_{1}\omega_{2})=1$$

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

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

Solution

Question 2

$$z$$ lies on the imaginary axis

$$z$$ lies on the real axis

$$z$$ lies on the unit circle

None of these

Solution

Question 3

$$6$$

$$10$$

$$9$$

$$12$$

Solution

Question 4

$$5 + i$$

$$\dfrac {5 + i}{26}$$

$$\dfrac {1}{5 + i}$$

$$\dfrac {5 + i}{24}$$

$$\dfrac {5 - i}{24}$$

Solution

Question 5

$$2(cos\dfrac{\pi}{4}+i\,sin\dfrac{\pi}{4})$$

$$-2(cos\dfrac{\pi}{4}-i\,sin\dfrac{\pi}{4})$$

$$2\sqrt{2}(cos\dfrac{3\pi}{4}+i\,sin\dfrac{3\pi}{4})$$

$$2\sqrt{2}(cos\dfrac{7\pi}{4}+i\,sin\dfrac{7\pi}{4})$$

$$2\sqrt{2}(cos\dfrac{5\pi}{4}+i\,sin\dfrac{5\pi}{4})$$

Solution

Question 6

$$5, 11$$

$$3, 5$$

$$11, 17$$

$$3, 7$$

Solution

Question 7

$$1$$

$$7$$

$$25$$

$$-7+24i$$

$$7+24i$$

Solution

Question 8

$$-4+9i$$

$$-5+9i$$

$$2+9i$$

$$-5+8i$$

Solution

Question 9

68

69

71

73

76

Solution

Question 10

$$0$$

$$1$$

$$\dfrac{1}{2}$$

$$-1$$

Solution

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