Number Theory Test 17

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Number Theory Test 17

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Weekly Quiz Competition

Question 1
1 / -0

lf $$\log_{\frac{1}{2}}|\mathrm{z}-2|>\log_{\frac{1}{2}}|z|$$ then

A
$$x>1$$

B
$$x<1$$

C
$$x<2$$

D
$$x>2$$

Solution

$$ Let \quad z= x+iy $$
$$\log_{\frac{1}{2}}|\mathrm{z}-2|>\log_{\frac{1}{2}}|z|$$
$$\Rightarrow |\mathrm{z}-2| < |z|$$
$$\Rightarrow |x - 2 + iy| < |x + iy|$$
$$\Rightarrow \sqrt{x^2 - 4x + 4 + y^2} < \sqrt{x^2 + y^2}$$
$$\Rightarrow -4x + 4 < 0 $$
$$\Rightarrow x > 1.$$
Question 2
1 / -0

The minimum value of $$f(z)=|\mathrm{z}|+|\mathrm{z}-1|+|z+2|$$ is

A
3

B
4

C
5

D
6

Solution

Taking $$\left | z-1 \right |+\left | z+2 \right |$$

We know

$$\left | z_{1} \right |+\left | z_{2} \right |\geqslant \left | z_{1}+z_{2} \right |$$

$$\left | z-1 \right |+\left | z+2 \right |=\left | -z+1 \right |+\left | z+2 \right

|\geqslant \left | 1-\not{z}+\not{z}+2 \right |$$
$$\geqslant 3$$

And clearly the value 3 occurs
When z=0, -2, 1

Now, $$\left | z \right | is\ minimum\ at \ z=0$$

Hence minimum value $$ = $$ 3

Similarly

Taking $$\left | z \right |+\left | z-1 \right |$$

$$\Rightarrow \left | z \right |+\left | z-1 \right |= \left | z \right |+\left | 1-z \right |\geqslant \left | \not{z}+1-\not{z} \right |$$
$$=1$$

And value 1

occurs at$$ z=0, z=1 or z= z=\frac{1}{2}$$

But $$ \left | z+2 \right |$$ is minimum at $$z=0$$

Hence minimum value $$= i+ \left | z+2 \right |$$
$$=3 at z=0$$

Similarly
Taking $$\left | z \right |+\left | z+2 \right |$$

$$\Rightarrow \left | z \right |+\left | z+2 \right |= \left | z \right |+\left | 2-z \right |\geqslant \left | \not{z}-2-\not{z} \right |$$

$$=2$$

But value 2 occurs at o, -1 and -2

But $$ \left | z-1 \right |$$ is minimum at z$$=0$$
Hence minimum

value $$ = 2+ \left | z-1 \right |$$

$$=3$$
Question 3
1 / -0

The real value of $$\theta$$ for which the expression, $$\displaystyle \frac{1+i\cos\theta}{1-2i\cos\theta}$$ is real number is

A
$$n\displaystyle \pi\pm\frac{\pi}{2}$$

B
$$n\displaystyle \pi-\frac{\pi}{2}$$

C
$$n\displaystyle \pi+\frac{\pi}{2}$$

D
$$2n\displaystyle \pi\pm \cfrac {\pi}{2}$$

Solution

Let $$z=\cfrac{1+i\cos\theta}{1-2i\cos\theta}$$
$$=\cfrac{(1+i\cos\theta)(1+2i\cos\theta)}{(1-2i\cos\theta)(1+2i\cos\theta)}$$
$$=\cfrac{1-2\cos^{2}\theta+i3\cos\theta}{1+4\cos^{2}\theta}$$
$$=\cfrac{1-2\cos^{2}\theta}{1+4\cos^{2}\theta}+i\left (\dfrac{3\cos\theta}{1+4\cos^{2}\theta}\right)$$
If the above complex number is purely real, then
$$Im(z)=0$$
$$3\cos\theta=0$$
$$\theta=\cfrac{(2n\pm1)\pi}{2}$$, where $$n\epsilon N$$
Question 4
1 / -0

The region represented by z such that $$\left | \dfrac{\mathrm{z}-a}{z+a} \right |=1({\rm Im} (a) = 0)$$ is

A
$$y=0$$

B
$$x=0$$

C
$$x+y=0$$

D
$$x-y=0$$

Solution

Let $$z=x+iy$$

$$\Rightarrow

\left | (x-a)+iy \right |=\left | (x+a)+iy \right |$$
$$\Rightarrow

(x-a)^{2}+\not{ y^{2}} =(x+a)^{2}+\not{ y^{2}}$$
$$\Rightarrow

\not{ x^{2}} +\not{ a^{2}} -2ax =\not{ x^{2}} +\not{ a^{2}} +2ax$$
$$\Rightarrow 0$$
Question 5
1 / -0

lf $$\displaystyle \log_{(\frac{1}{3})}|z+1|>\log_{(\frac{1}{3})}|z-1|$$, then

A
$${\rm Re}(z)\geq 0$$

B
$${\rm Re}(z)<0$$

C
$${\rm Im}(z)>0$$

D
$${\rm Im}(z)\leq 0$$

Solution

$$\displaystyle \log(\frac{1}{3})|z+1|>\log(\frac{1}{3})|z-1|$$
$$\Rightarrow |z + 1| < |z - 1|$$
$$\Rightarrow (x^2 + 2x + 1 + y^2) < (x^2 - 2x + 1 + y^2)$$
$$\Rightarrow | 4x | < 0$$
Thus, $$Re(z) < 0.$$
Question 6
1 / -0

$$ \left | \displaystyle \frac{1}{(1-i)^{2}}-\frac{1}{(1+i)^{2}}\right |=$$

A
4

B
3

C
2

D
1

Solution

$$\left |

\dfrac{1}{(1-i)^{2}}-\dfrac{1}{(1+i)^{2}} \right |=\left |

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

\dfrac{\not{1}+2i-\not{1}-(\not{1}-2i)-\not{1} }{(1^{2}+1^{2})^{2}}\right |$$
$$=\left |

\dfrac{4i}{4} \right |=1$$
Question 7
1 / -0

Number of solutions of the equation $$|z|^{2}+7{z}=0$$ is

A
$$1$$

B
$$2$$

C
$$4$$

D
$$6$$

Solution

Let $$ \ z=x+iy$$
Then,$$\left | z \right |^{2}+7z=x^{2}+y^{2}+7(x+iy)=0$$
$$\Rightarrow x^{2}+y^{2}+7x=0$$ & $$7y=0$$
$$\Rightarrow x^{2}+ 7x=0$$ and $$ y=0$$
$$\Rightarrow x=-7\ or\ x=0$$ and $$y=0$$
So, solutions are $$z=-7$$ or $$z=0$$
Hence, two solutions.
Question 8
1 / -0

lf $$\displaystyle \log_{\sqrt{3}}\frac{|z^{2}|-|z|+1}{2+|z|}<2$$, then locus of $${z}$$ is

A
$$|z|<1$$

B
$$0<|z|<5$$

C
$$|z|>1$$

D
$$|\mathrm{z}|>5$$

Solution

Given, $$\displaystyle \log_{\sqrt{3}}\frac{|z^{2}|-|z|+1}{2+|z|}<2$$
Therefore, $$\dfrac{|z|^{2}-|z|+1}{2+|z|}< \sqrt{3}^2$$
$$\Rightarrow |z^{2}|-|z|+1<6+3|z|$$
$$|z|^2 -4|z| -5<0$$
$$-1<|z|<5$$
But $$|z|<0$$ is not possible.
Question 9
1 / -0

The given figure represents a multiplication operation, where each alphabet represents a different number, then what is the value of A.

A
0

B
3

C
2

D
4

Solution

As $$A+7=9$$

$$A=9-7$$

$$A=2$$

Hence option- $$C$$
Question 10
1 / -0

lf $$Z_{1},Z_{2}$$ are two unimodular Complex numbers then $$ \left |\displaystyle \frac{1}{Z_{1}}+\frac{1}{Z_{2}} \right|=$$

A
$$1$$

B
$$|Z_{1}+Z_{2}|$$

C
$$|Z_{1}|+|Z_{2}|$$

D
$$2$$

Solution

$$\left |\dfrac{1}{z_1}+\dfrac{1}{z_2}\right|$$
$$=\left|\dfrac{z_1+z_2}{z_1z_2}\right|$$
$$=\dfrac{\left|z_1+z_2\right|}{\left|z_1\right|\left|z_2\right|}$$
$$= \dfrac{|z_1+z_2| }{1\times 1}$$
$$=|z_1+z_2|$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 17

Result

Number Theory Test 17

Score

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Rank

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SHARING IS CARING

Question 1

$$x>1$$

$$x<1$$

$$x<2$$

$$x>2$$

Solution

Question 2

3

4

5

6

Solution

Question 3

$$n\displaystyle \pi\pm\frac{\pi}{2}$$

$$n\displaystyle \pi-\frac{\pi}{2}$$

$$n\displaystyle \pi+\frac{\pi}{2}$$

$$2n\displaystyle \pi\pm \cfrac {\pi}{2}$$

Solution

Question 4

$$y=0$$

$$x=0$$

$$x+y=0$$

$$x-y=0$$

Solution

Question 5

$${\rm Re}(z)\geq 0$$

$${\rm Re}(z)<0$$

$${\rm Im}(z)>0$$

$${\rm Im}(z)\leq 0$$

Solution

Question 6

4

3

2

1

Solution

Question 7

$$1$$

$$2$$

$$4$$

$$6$$

Solution

Question 8

$$|z|<1$$

$$0<|z|<5$$

$$|z|>1$$

$$|\mathrm{z}|>5$$

Solution

Question 9

0

3

2

4

Solution

Question 10

$$1$$

$$|Z_{1}+Z_{2}|$$

$$|Z_{1}|+|Z_{2}|$$

$$2$$

Solution

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