Number Theory Test 19

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Number Theory Test 19

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Weekly Quiz Competition

Question 1
1 / -0

If $$z = x + iy$$ and $$x^2 + y^2 = 16$$, then the range of $$\left|\left|x\right| - \left|y\right|\right|$$ is

A
[0, 4]

B
[0,2]

C
[2, 4]

D
none of these

Solution

$$z=x+iy$$
And $$|z|^{2}=16$$
$$|z|=4$$
If z is purely real, $$x=\pm4$$ and $$||x|-|y||=4$$ ...(i)
If z is purely imaginary $$y=\pm4i$$ and $$||x|-|y||=4$$ ...(ii)
Now if $$|x|=|y|$$
Thus
$$2x^{2}=2y^{2}=16$$
$$x=y=\pm2\sqrt{2}$$
Under such case
$$||x|-|y||=0$$
Hence the range of values of $$||x|-|y||$$ is $$[0,4]$$
Question 2
1 / -0

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

Statement 1: $$z_1z_2$$ is unimodular.

Statement 2: Both $$z_1$$ and $$z_2$$ are unimodular.

A
Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

B
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

C
Statement 1 is true and Statement 2 is false.

D
Statement 1 is false and Statement 2 is true.

Solution

$$|z_1+z_2|=\left|\displaystyle\frac { z_1+z_2 }{ z_1z_2 } \right|\\ \Longrightarrow |z_1+z_2|(1-\displaystyle\frac { 1 }{ |z_1z_2| } )=0\\ since\quad |z_1+z_2|\neq 0,\\ so,1=\displaystyle\frac { 1 }{ |z_1z_2| } \\ \Longrightarrow |z_1z_2|=1$$

Hence the statement 1 is correct but statement 2 is false
Question 3
1 / -0

The complex number z satisfies the condition $$\displaystyle \left | z- \frac{25}{z} \right | = 24$$. The maximum distance from the origin of co-ordinates to the point z is

A
25

B
30

C
32

D
none of these

Solution

The maximum value would be when
$$|z-\dfrac{25}{z}|$$
$$=|z|-\dfrac{25}{|z|}=24$$
$$=|z|^{2}-25=24|z|$$
$$|z|^{2}-24|z|-25=0$$
$$|z|^{2}-25|z|+|z|-25=0$$
$$|z|(|z|-25)+1(|z|-25)=0$$
$$(|z|+1)(|z|-25)=0$$
$$|z|=-1$$ not possible
And $$|z|=25$$
Question 4
1 / -0

The roots of the equation $$z^n = (z + 1)^n$$ on the complex plane lie on the line

A
$$2x + 1 = 0$$

B
$$2x - 1 = 0$$

C
$$x + 1 = 0$$

D
$$x - 1 = 0$$

Solution

$$|\dfrac{z+1}{z}|^{n}=1$$
Let $$n=1$$,
$$|z+1|=|z|$$
Let $$z=x+iy$$
$$(x+1)^{2}=x^{2}$$
$$(x+1)^{2}-x^{2}=0$$
$$(2x+1)(1)=0$$
$$2x+1=0$$
Hence answer is A.
Question 5
1 / -0

If $$\left|z\right |^2 - 3 = 3\left|z\right|$$, then the value of $$\left|z\right|$$ is

A
$$1$$

B
$$\displaystyle \frac{3 + \sqrt{21}}{2}$$

C
$$\displaystyle \frac{\sqrt{21}- 3}{2}$$

D
none of these

Solution

$$|z|^{2}-3=3|z|$$
$$|z|^{2}-3|z|-3=0$$
$$|z|= \displaystyle \frac{3\pm\sqrt{21}}{2}$$
$$|z|=\displaystyle \frac{3+\sqrt{21}}{2}$$
Question 6
1 / -0

Find the complex numbers z which simultaneously satisfy the equation $$\displaystyle \left | \frac{z - 12}{z - 8 i} \right | = \frac{5}{3}$$ and $$\displaystyle \left | \frac{z - 4}{z - 8} \right | = 1$$.

A
6 + 8 i or 6 + 17 i

B
6 + 8 i or 6 - 17 i

C
6 - 8 i or 6 + 17 i

D
6 - 8 i or 6 - 17 i

Solution

Put $$z=x+iy$$
$$\displaystyle \left| \frac { z-4 }{ z-8 } \right| =1\Rightarrow \left| \frac { x-4+iy }{ x-8+iy } \right| =1\\ \Rightarrow { \left( x-4 \right) }^{ 3 }+{ y }^{ 2 }={ \left( x-8 \right) }^{ 2 }+{ y }^{ 2 }\\ \Rightarrow x=6$$

With $$\displaystyle x=6,\left| \frac { z-12 }{ z-8i } \right| =\frac { 5 }{ 3 } $$
$$\displaystyle \Rightarrow \left| \frac { -6+iy }{ 6+i\left( y-8 \right) } \right| =\frac { 5 }{ 3 } \\ \Rightarrow 9\left( 36+{ y }^{ 2 } \right) =25\left[ 36+\left( y-8 \right) 2 \right] \\ \Rightarrow { y }^{ 2 }-25y+136=0\\ \Rightarrow y=17,8$$
Hence the required numbers are
$$z=6+17i,6+8i$$
Question 7
1 / -0

If $$z$$ is a complex number such that $$-\pi / 2 \le$$ arg $$ z \le \pi / 2,$$ then which of the following inequality is true?

A
$$\left|z -\overline{z}\right| \le \left|z\right|$$(arg $$ z $$- arg $$\overline{z})$$

B
$$\left|z -\overline{z}\right| \ge \left|z\right|$$(arg $$z$$ - arg $$\overline{z})$$

C
$$\left|z -\overline{z}\right| < \left|z\right|$$(arg $$z$$ - arg $$\overline{z})$$

D
none of these

Solution

Let $$z=\left| z \right| \left( \cos { A } +i\sin { A } \right) \quad \quad \quad where\quad A=arg\left( z \right) ,\quad -\frac { \Pi }{ 2 } \le A\le \frac { \Pi }{ 2 } $$

$$\left| z-\bar { z } \right| =\left| z \right| \left| \cos { A } +i\sin { A } -\left( \cos { A } -i\sin { A } \right) \right| =2\left| z \right| \left| i\sin { A } \right| $$

$$\Longrightarrow \left| z-\bar { z } \right| =2\left| z \right| \left| \sin { A } \right| $$

$$for\quad -\frac { \pi }{ 2 } \le A\le \frac { \pi }{ 2 } \quad \quad \left| \sin { A } \right| \le |A| $$

$$\Longrightarrow \left| z-\bar { z } \right| \le 2\left| z \right| \left| A \right| =\left| z \right| \left| A-\left( -A \right) \right| $$

$$\Longrightarrow \left| z-\bar { z } \right| \le \left| z \right| \left| argz-arg\bar { z } \right| $$

Ans: A
Question 8
1 / -0

If $$i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0$$, then

A
$$\left| z \right| <1$$

B
$$\left| z \right| >1$$

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

D
$$\left| z \right| =0$$

Solution

Given $$i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0\Rightarrow i{ z }^{ 2 }\left( z-i \right) -\left( z-i \right) =0$$
$$\Rightarrow \left( z-i \right) \left( i{ z }^{ 2 }-1 \right) =0\Rightarrow z=i$$ or $$\displaystyle { z }^{ 2 }=\frac { 1 }{ i } =-i$$
Now $$z=i\Rightarrow \left| z \right| =\left| i \right| =1$$
and $${ z }^{ 2 }=-i\Rightarrow \left| { z }^{ 2 } \right| =\left| -i \right| \Rightarrow \left| { z }^{ 2 } \right| =1\Rightarrow \left| z \right| =1$$
Thus, in both cases $$\left| z \right| =1$$
Question 9
1 / -0

Locate the complex number $$z = x + iy$$ for which $$log_{1/3} \{ log_{1/2} (|z|^2 + 4 |z| + 3) \} < 0$$

A
Empty set

B
circle of radius 6 and center at origin

C
circle of radius 3 and center at origin

D
circle of radius 2 and center at origin

Solution

$$\log_{ 1/3 }\{ \log_{ 1/2 }(|z|^{ 2 }+4|z|+3)\} <0$$
$$\Rightarrow \log_{ 1/2 }(|z|^{ 2 }+4|z|+3)>0$$
$$\Rightarrow |z|^{ 2 }+4|z|+3<0$$
$$\Rightarrow \left( \left| z \right| +1 \right) \left( \left| z \right| +3 \right) <0$$
Since, $$\left| z \right| +1>0$$ and $$\left| z \right| +3>0$$
Therefore, $$z$$ belongs to empty set.

Ans: A
Question 10
1 / -0

Number of roots of the equation $$z^{10} - z^5 - 992 = 0$$ where real parts are negative is

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

We have,
$$z^{10} - z^5 - 992 = 0$$
Put $$z^5=u$$
$$\Rightarrow u^2-u-992=0$$
$$\Rightarrow u = \dfrac{1\pm\sqrt{1+ 4\times 992}}{2}$$
$$\quad =\dfrac{1\pm 63}{2}=32, -31$$
$$\Rightarrow z^5 =32, (-1)^531$$
$$\Rightarrow z=(32)^{1/5}, -(31)^5$$
Hence number of roots of $$z$$ having negative real parts is $$5$$
corresponding to $$z^5 =-31$$

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Number Theory Test 19

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Number Theory Test 19

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Question 1

[0, 4]

[0,2]

[2, 4]

none of these

Solution

Question 2

Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

Statement 1 is true and Statement 2 is false.

Statement 1 is false and Statement 2 is true.

Solution

Question 3

25

30

32

none of these

Solution

Question 4

$$2x + 1 = 0$$

$$2x - 1 = 0$$

$$x + 1 = 0$$

$$x - 1 = 0$$

Solution

Question 5

$$1$$

$$\displaystyle \frac{3 + \sqrt{21}}{2}$$

$$\displaystyle \frac{\sqrt{21}- 3}{2}$$

none of these

Solution

Question 6

6 + 8 i or 6 + 17 i

6 + 8 i or 6 - 17 i

6 - 8 i or 6 + 17 i

6 - 8 i or 6 - 17 i

Solution

Question 7

$$\left|z -\overline{z}\right| \le \left|z\right|$$(arg $$ z $$- arg $$\overline{z})$$

$$\left|z -\overline{z}\right| \ge \left|z\right|$$(arg $$z$$ - arg $$\overline{z})$$

$$\left|z -\overline{z}\right| < \left|z\right|$$(arg $$z$$ - arg $$\overline{z})$$

none of these

Solution

Question 8

$$\left| z \right| <1$$

$$\left| z \right| >1$$

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

$$\left| z \right| =0$$

Solution

Question 9

Empty set

circle of radius 6 and center at origin

circle of radius 3 and center at origin

circle of radius 2 and center at origin

Solution

Question 10

$$3$$

$$4$$

$$5$$

$$6$$

Solution

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