Number Theory Test 41

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Number Theory Test 41

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Weekly Quiz Competition

Question 1
1 / -0

If$${ z }_{ 1 }$$ and $${ z }_{ 2 }$$ are two complex number such that Im$$({ z }_{ 1 }={ z }_{ 2 })$$= 0= Im $$({ z }_{ 1 }{ z }_{ 2 })$$, then

A
$${ z }_{ 1 }={ z }_{ 2 }$$

B
$${ z }_{ 1 }={ \overline { z } }_{ 2 }$$

C
$${ z }_{ 1 }={ -z }_{ 2 }$$

D
$${ z }_{ 1 }=-{ \overline { z } }_{ 2 }$$

Solution

As per given condition:

$$Img(z_1=z_2)=0=Im(z_1z_2)$$

$$\Rightarrow Im(z_1z_2)=0$$

Product of imaginary parts is zero, implies that the direction of each number is opposite in direction.

$$\therefore z_1=\bar{z_2}$$
Question 2
1 / -0

$$z$$ is a complex number. If $$a = | x | + | y |$$ and
$$b = \sqrt { 2 } | x + i y |$$ then which of the following is
true

A
$$a \leq b$$

B
$$a > b$$

C
none of these

D
$$a - b + 2$$

Solution
Question 3
1 / -0

If $$z=(3+4i)^6+(3-4i)^6,$$ where $$i=\sqrt { -1 }, $$ then $$Im(z)$$ equals to

A
$$-6$$

B
$$0$$

C
$$6$$

D
None of these

Solution

$$z={ \left( 3+4i \right) }^{ 6 }+{ \left( 3-4i \right) }^{ 6 }$$
$$={ \left[ 5\left( { e }^{ i\theta } \right) \right] }^{ 6 }+{ \left[ 5\left( { e }^{ -i\theta } \right) \right] }^{ 6 }$$ [where $$\theta ={ tan }^{ -1 }\left( 4/3 \right) $$]
$$={ 5 }^{ 6 }\left[ { e }^{ i6\theta }+{ e }^{ -i6\theta } \right] $$
$$={ 5 }^{ 6 }\left[ 2Re\left( { e }^{ i6\theta } \right) \right] $$
$$\therefore$$ In $$\left( z \right) =0$$ [B]
Question 4
1 / -0

Number of complex numbers $$z$$ such that $$|z|=1$$ and $$\left|\dfrac {z}{z}+\dfrac {\bar {z}}{z}\right|=1$$ is

A
$$4$$

B
$$1$$

C
$$8$$

D
$$more\ then\ 8$$

Solution

Since $$z$$ is equidistant from $$1,-1,i$$, we can say that $$z$$is center of circle with $$1,-1,i$$ lying on it's circumference and so there is one one circle possible passing through $$3$$ points. So, the answer is $$1$$.
Question 5
1 / -0

The modulus of $$\dfrac { \left( 3+2i \right) ^{ 2 } }{ \left( 4-3i \right) } $$ is:

A
$$\frac { 13 }{ 5 } $$

B
$$\frac { 11 }{ 5 } $$

C
$$\frac { 9 }{ 5 } $$

D
$$\frac { 7 }{ 5 } $$

Solution
Question 6
1 / -0

If a complex number z and $$z+\dfrac { 1 }{ z } $$ have same argument then-

A
z must be purely real

B
z must be purely imaginary

C
z cannot be imaginary

D
z must be raal

Solution

we have,

z is a complex number.

Then,

Let $$z=r\cos \theta +i\sin \theta $$

Now, we know that,

$$\overline{z}=r\left( \cos \theta -i\sin \theta \right)$$

So,

$$z\overline{z}={{r}^{2}}$$

Then,

$$\dfrac{1}{z}=\dfrac{\overline{z}}{z\overline{z}}=\dfrac{\overline{z}}{{{r}^{2}}}$$

Now,

According to given question,

$$ z+\dfrac{1}{z}=\left| z+\dfrac{1}{z} \right|=1\,\,\,\left( \text{z}\,\text{and}\,\text{z+}\dfrac{\text{1}}{\text{z}}\,\text{same argument} \right) $$

$$ =\left| r\left( \cos \theta +i\sin \theta \right)+\dfrac{r\left( \cos \theta -i\sin \theta \right)}{{{r}^{2}}} \right|=1 $$

$$ =\left| \left( r+\dfrac{1}{r} \right)\cos \theta +i\left( r-\dfrac{1}{r} \right)\sin \theta \right|=1 $$

Now, squaring both side and we get,

$$\left| \left( {{r}^{2}}+\dfrac{1}{{{r}^{2}}}+2 \right){{\cos }^{2}}\theta +{{i}^{2}}\left( {{r}^{2}}+\dfrac{1}{{{r}^{2}}}-2 \right){{\sin }^{2}}\theta \right|=1$$

So,

$$ \left( {{r}^{2}}+\dfrac{1}{{{r}^{2}}}+2 \right){{\cos }^{2}}\theta =1 $$

$$ {{r}^{2}}+\dfrac{1}{{{r}^{2}}}+2\cos 2\theta =1 $$

$$ {{r}^{2}}+\dfrac{1}{{{r}^{2}}}=1-2\cos 2\theta $$

Now,

$$ 1-2\cos 2\theta =0 $$

$$ \Rightarrow \cos 2\theta =\dfrac{1}{2} $$

$$ \Rightarrow \cos 2\theta =cos\dfrac{\pi }{3} $$

$$ \Rightarrow 2\theta =\dfrac{\pi }{3} $$

$$ \Rightarrow \theta =\dfrac{\pi }{6} $$
Then, Z is purely imaginary.

Hence, this is the answer.
Question 7
1 / -0

A value of $$\theta $$ for which$$\dfrac { 2+3isin\theta }{ 1-2isin\theta } $$ is purely imaginary, is:

A
$${ sin }^{ -1 }\left( \dfrac { 1 }{ \sqrt { 3 } } \right) $$

B
$$\dfrac { \pi }{ 3 } $$

C
$$cos^{-1}\sqrt-1$$

D
$$None of these$$

Solution

We have,

$$\dfrac{2+3i\sin \theta }{1-2i\sin \theta }$$

On rationalize and we get,

$$ \dfrac{2+3i\sin \theta }{1-2i\sin \theta }\times \dfrac{1+2i\sin \theta }{1+2i\sin \theta } $$

$$ \Rightarrow \dfrac{2+3i\sin \theta +4i\sin \theta +6{{i}^{2}}{{\sin }^{2}}\theta }{{{1}^{2}}-4{{i}^{2}}{{\sin }^{2}}\theta } $$

$$ \Rightarrow \dfrac{2-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$

$$ \Rightarrow \dfrac{1+1-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$

$$ \Rightarrow \dfrac{1+{{\cos }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$

$$ \Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+\dfrac{7i\sin \theta }{1+4{{\sin }^{2}}\theta } $$

$$ \Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+i\dfrac{7\sin \theta }{1+4{{\sin }^{2}}\theta } $$
Now,

Then purely imaginary is

Real part of equal to zero.

$$ \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=0 $$

$$ \Rightarrow 1+{{\cos }^{2}}\theta =0 $$

$$ \Rightarrow {{\cos }^{2}}\theta =-1 $$

$$ \Rightarrow \cos \theta =\sqrt{-1} $$

$$ \Rightarrow \theta ={{\cos }^{-1}}\sqrt{-1} $$
Hence, this is the answer.
Question 8
1 / -0

If $$z$$ satisfies $$|z - 2+ 2i| \le 1$$

A
$$|z|_{least} = 2\sqrt{2} - 1$$

B
$$|z|_{least} = 2\sqrt{2} + 1$$

C
$$|z - 1| \le 1$$

D
None of these

Solution
Question 9
1 / -0

If $$\dfrac { z+1 }{ z+i }$$ is purely imaginary, then z lies on a

A
straight lone

B
circle

C
Circle with radius 1

D
circle passing through (1, 1).

Solution

$$\dfrac { z+1 }{ z+i } $$ is purely imaginary

$$\Rightarrow Re\left( \dfrac { z+1 }{ z+i } \right) =0$$

Say $$z=x+iy$$

then,

$$Re\left( \dfrac { x+iy+1 }{ x+iy+i } \right) =0$$

$$\Rightarrow Re\left( \dfrac { \left( x+1 \right) +iy }{ x+i\left( y+1 \right) } \right) =0$$

$$\Rightarrow Re\left[ \dfrac { \left[ \left( x+1 \right) +iy \right] \left[ x-i\left( y+1 \right) \right] }{ { x }^{ 2 }+{ \left( y+1 \right) }^{ 2 } } \right] =0$$

$$\Rightarrow Re\left[ \dfrac { x\left( x+1 \right) +y\left( y+1 \right) +i\left( xy-\left( x+1 \right) \left( y+1 \right) \right) }{ { x }^{ 2 }+{ \left( y+1 \right) }^{ 2 } } \right] =0$$

$$\therefore$$ $$\dfrac { x\left( x+1 \right) +y\left( y+1 \right) }{ { x }^{ 2 }+{ \left( y+1 \right) }^{ 2 } } =0$$

$$\Rightarrow { x }^{ 2 }+x+{ y }^{ 2 }+y=0$$

$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+x+y=0$$

$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+2.\dfrac{1}{2}x+2.\dfrac{1}{2}y=0$$

The above is a circle with centre $$\left( \dfrac { -1 }{ 2 } ,\dfrac { -1 }{ 2 } \right) $$ and radius $$\sqrt { \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 } } =\sqrt { \dfrac { 1 }{ 2 } } $$ units

Option B.
Question 10
1 / -0

Purely imaginary then find the sum of statement i $$a,b$$

A
$$\dfrac {5\pi}{6}$$

B
$$\pi$$

C
$$\dfrac {3\pi}{4}$$

D
$$\dfrac {2\pi}{3}$$

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Number Theory Test 41

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Number Theory Test 41

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Question 1

$${ z }_{ 1 }={ z }_{ 2 }$$

$${ z }_{ 1 }={ \overline { z } }_{ 2 }$$

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

$${ z }_{ 1 }=-{ \overline { z } }_{ 2 }$$

Solution

Question 2

$$a \leq b$$

$$a > b$$

none of these

$$a - b + 2$$

Solution

Question 3

$$-6$$

$$0$$

$$6$$

None of these

Solution

Question 4

$$4$$

$$1$$

$$8$$

$$more\ then\ 8$$

Solution

Question 5

$$\frac { 13 }{ 5 } $$

$$\frac { 11 }{ 5 } $$

$$\frac { 9 }{ 5 } $$

$$\frac { 7 }{ 5 } $$

Solution

Question 6

z must be purely real

z must be purely imaginary

z cannot be imaginary

z must be raal

Solution

Question 7

$${ sin }^{ -1 }\left( \dfrac { 1 }{ \sqrt { 3 } } \right) $$

$$\dfrac { \pi }{ 3 } $$

$$cos^{-1}\sqrt-1$$

$$None of these$$

Solution

Question 8

$$|z|_{least} = 2\sqrt{2} - 1$$

$$|z|_{least} = 2\sqrt{2} + 1$$

$$|z - 1| \le 1$$

None of these

Solution

Question 9

straight lone

circle

Circle with radius 1

circle passing through (1, 1).

Solution

Question 10

$$\dfrac {5\pi}{6}$$

$$\pi$$

$$\dfrac {3\pi}{4}$$

$$\dfrac {2\pi}{3}$$

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