Number Theory Test 16

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Number Theory Test 16

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Question 1
1 / -0

The modulus of $$(1 + i) (3 + 4i) =$$

A
$$\sqrt{50}$$

B
$$\sqrt{25}$$

C
$$10$$

D
$$10 \sqrt{2}$$

Solution

$$z_{1},z_{2}$$ being two complex numbers,
$$|z_{1}.z_{2}|=|z_{1}|.|z_{2}|$$
$$\therefore |(1+i)(3+4i)|$$$$=|1+i|\cdot|3+4i|$$
$$=\sqrt{1^{2}+1^{2}}\times \sqrt{3^{2}+4^{2}}$$
$$=\sqrt{2}\sqrt{25}$$
$$=\sqrt{50}$$
Question 2
1 / -0

If $$|\mathrm{z}-4|<|\mathrm{z}-2|$$ then

A
Re z $$>0$$

B
Re z $$<0$$

C
Re z $$>2$$

D
Re z $$>3$$

Solution

$$|z - 4| < |z - 2|$$
Let $$z = x + iy$$
$$\therefore$$ $$|x - 4 + iy| < |x - 2 + iy|$$
$$\therefore$$ $$\sqrt {x^2 - 8x + 16 + y^2} < \sqrt{x^2 - 4x + 4 + y^2}$$
$$\therefore$$ $$x^2 - 8x + 16 + y^2 < x^2 - 4x + 4 + y^2$$
$$\therefore$$ $$8x - 4x > 16 - 4$$
$$\therefore$$ $$4x > 12$$
Thus, $$Re(z) > 3.$$
Question 3
1 / -0

If $$z_1$$, $$z_2$$, $$z_3$$ are complex numbers such that $$\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =\left| \dfrac { 1 }{ { z }_{ 1 } } +\dfrac { 1 }{ { z }_{ 2 } } +\dfrac { 1 }{ { z }_{ 3 } } \right| =1$$, then $$|z_1+z_2+z_3|$$ is:

A
Equal to 1

B
Less than 1

C
Greater than 3

D
Equal to 3

Solution

Let $$z_1 = cis \theta_1$$, $$z_2 = cis\theta_2 $$ and $$z_3 =cis\theta_3 $$
We have,
$$\left | \dfrac{1}{z_1} + \dfrac{1}{z_2} +\dfrac{1}{z_3} \right| = 1 $$

$$| cis(-\theta_1) + cis(-\theta_2) + cis (-\theta_3) | = 1 $$

Hence,
$$ | (\cos \theta_1 + \cos \theta_2 + \cos \theta_3 ) - i ( \sin \theta_1 + \sin \theta_2 + \sin \theta_3 ) | = 1$$

Hence,
$$ | (\cos \theta_1 + \cos \theta_2 + \cos \theta_3 ) + i ( \sin \theta_1 + \sin \theta_2 + \sin \theta_3 ) | =1 $$

Hence,
$$ |z_1+ z_2+ z_3 | =1 $$
Question 4
1 / -0

If z is a complex number satisfying $$|z^2+1|=4|z|$$, then the minimum value of $$|z|$$ is

A
$$2\sqrt{5}+4$$

B
$$2\sqrt{5}-4$$

C
$$\sqrt{5}-2$$

D
None of these

Solution

Given:
$$|z^2+1|=4|z|$$
$$\Rightarrow |z+\cfrac{1}{z}|=4$$
We know,
$$\left ||z|-\cfrac{1}{|z|}\right |\leq \left|z+\cfrac{1}{z}\right|\leq |z|+\cfrac{1}{|z|}$$
$$\therefore \pm(|z|-\cfrac{1}{|z|})\leq 4\leq |z|+\cfrac{1}{|z|}$$
Equations we get:
$$1.$$ $$|z|^2-4|z|-1=0$$
$$2.$$ $$|z|^2+4|z|-1=0$$
$$3.$$ $$|z|^2-4|z|+1=0$$
$$[|z|$$ always $$\geq 0]$$
Solving eq (1):
$$|z|=\sqrt{5}+2$$
Solving eq (2):
$$|z|=\sqrt{5}-2$$
Solving eq (3):
$$|z|=2\pm \sqrt{3}$$
$$\therefore$$ Minimum $$|z|=\sqrt{5}-2$$
Question 5
1 / -0

lf $$(x+iy)(2+cos\theta+isin\theta)=3$$ then $$x^{2}+y^{2}-4x+3$$ is

A
$$0$$

B
$$1$$

C
$$3$$

D
$$4$$

Solution

$$(x+iy)(2+cos\theta+isin\theta)=3$$

$$2x+x(cos\theta+isin\theta)+i2y+iy(cos\theta+isin\theta)=3$$

$$2x+xcos\theta-ysin\theta+i(xsin\theta+2y+ycos\theta)=3$$

Therefore, equating the real and imaginary parts of LHS and RHS gives us

$$2x+xcos\theta-ysin\theta=3$$

$$xsin\theta+ycos\theta+2y=0$$

$$\Rightarrow xcos\theta-ysin\theta=3-2x$$

Squaring both sides give us

$$x^{2}cos^{2}\theta+y^{2}sin^{2}\theta-2xycos\theta.sin\theta=(2x-3)^{2}$$ ...(i)

And

$$xsin\theta+ycos\theta=-2y$$

$$x^{2}sin^{2}\theta+y^{2}cos^{2}\theta+2xysin\theta.cos\theta=4y^{2}$$ ...(ii)

Adding both i and ii gives us

$$x^{2}+y^{2}=4y^{2}+(2x-3)^{2}$$

$$x^{2}+y^{2}=4y^{2}+4x^{2}-12x+9$$

$$3x^{2}+3y^{2}-12x+9=0$$

$$x^{2}+y^{2}-4x+3=0$$
Question 6
1 / -0

If $$|{z}|-z=1+2i$$ then $$z=$$

A
$$2-\displaystyle \frac{3}{2}i$$

B
$$\displaystyle \frac{3}{2}-2i$$

C
$$3+4i$$

D
$$3-4i$$

Solution

Given,
$$|z|-z=1+2i$$

Let $$z=x+iy$$

So, $$|z|-z=1+2i$$

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

$$\Rightarrow y=-2 --(1)$$

So, $$\sqrt{x^{2}+y^{2}}-x=1$$

$$\Rightarrow \sqrt{x^{2}+4}-x=1$$

$$\Rightarrow x^{2}+4=(1+x)^{2}=x^{2}+2x+1$$

$$\Rightarrow 2x=3$$

$$\Rightarrow x=\dfrac{3}{2} --(2)$$

Hence,
$$z=\dfrac{3}{2}-2i$$
Question 7
1 / -0

If $$z=2-i\sqrt{3}$$ then $$z^{4}-4z^{2}+8z+35$$ is :

A
$$6$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

Given $$z = 2 - i \sqrt3,$$ $$\therefore$$ $$z - 2 = - i \sqrt3$$
Squaring both sides, $$\therefore$$ $$z^2 - 4z + 4 = -3$$
$$\therefore$$ $$z^2 - 4z = -7$$
Now $$z^4 - 4z^2 + 8z + 35 = (z^2 - 4z + 7)(z^2 + 4z + 5)$$
Thus $$z^4 - 4z^2 + 8z + 35 = 0.$$
Question 8
1 / -0

For any two non-zero complex numbers $$Z_1$$ and $$Z_2$$, the value of $$\left(\left|{ Z }_{ 1 }\right|+\left|{ Z }_{ 2 }\right|\right)\left|\dfrac{{ Z }_{ 1 }}{\left|{Z}_{1}\right|}+\dfrac{{Z}_{2}}{\left|{Z}_{2}\right|}\right|$$ is:

A
less than $$2(|Z_1|+|Z_2|)$$

B
greater than $$2(|Z_1|+|Z_2|)$$

C
greater than or equal to $$2(|Z_1|+|Z_2|)$$

D
less than or equal to $$2(|Z_1|+|Z_2|)$$

Solution

$$\left|\dfrac{Z_1}{|Z_1|}+\dfrac{Z_2}{|Z_2|}\right|\le\left|\dfrac{Z_1}{|Z_1|}\right|+\left|\dfrac{Z_2}{|Z_2|}\right|\le2$$
$$\therefore \left|\dfrac{Z_1}{|Z_1|}+\dfrac{Z_2}{|Z_2|}\right|\le2$$
Multiplying both sides by $$(|Z_1|+|Z_2|)$$, we get
$$(|Z_1|+|Z_2|)\left|\dfrac{Z_1}{|Z_1|}+\dfrac{Z_2}{|Z_2|}\right|\le2(|Z_1|+|Z_2|)$$
Question 9
1 / -0

lf $$z_{1},\ z_{2}$$ are roots of equation $$z^{2}-az+a^{2}=0$$, then $$|\displaystyle \frac{z_{1}}{z_{2}}|=$$

A
$$a^{2}$$

B
a

C
2a

D
1

Solution

$$z_{1}$$ and $$z_{2}$$ are roots of $$z^{2}-az+a^{2}=0$$
$$\therefore z_{1},z_{2}=\dfrac{a\pm \sqrt{a^{2}-4a^{2}}}{2}$$
$$=a(\dfrac{1\pm i\sqrt{3}}{2})$$
$$\therefore z_{1} , z_{2}$$ are multiple of complex cube roots of unity.
$$\therefore |\dfrac{z_{1}}{z_{2}}|=|\dfrac{\omega ^{2}}{\omega }|=|\omega |=1$$
Question 10
1 / -0

I. lf $$\left |z-\displaystyle \frac{2}{z}\right |=2$$ then the greatest value of $$|z|$$ is $$\sqrt{3}+1$$
II. $$\left | z+1 \right |-\left | z-1\right |=\frac{3}{2}$$ then the least value of $$\left | z \right |=\frac{3}{4}$$

A
Only I is true

B
Only II is true

C
both I and II are true

D
Neither I nor II are true

Solution

I. We have $$|z|=|z+\frac{2}{z}-\frac{2}{z}|$$ $$\leq |z|+\frac{2}{z}+(\frac{-2}{z})|$$
$$\leq |z+\frac{2}{z}|+|-\frac{2}{z}|$$
$$\therefore |z|\leq 2+\frac{2}{|z|}$$
$$\therefore |z|^{2}-2|z|-2\leq 0$$
$$\therefore (|z|-1+\sqrt{3})(|z|-1-\sqrt{3})\leq 0$$
$$1-\sqrt{3}\leq |z|\leq 1+\sqrt{3}$$
Hence max value of z is $$1+\sqrt{3}$$

II. $$(1+x)-(1-x)=\frac{3}{2}$$
$$2x=\frac{3}{2}$$
$$x=\frac{3}{4}$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 16

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Number Theory Test 16

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SHARING IS CARING

Question 1

$$\sqrt{50}$$

$$\sqrt{25}$$

$$10$$

$$10 \sqrt{2}$$

Solution

Question 2

Re z $$>0$$

Re z $$<0$$

Re z $$>2$$

Re z $$>3$$

Solution

Question 3

Equal to 1

Less than 1

Greater than 3

Equal to 3

Solution

Question 4

$$2\sqrt{5}+4$$

$$2\sqrt{5}-4$$

$$\sqrt{5}-2$$

None of these

Solution

Question 5

$$0$$

$$1$$

$$3$$

$$4$$

Solution

Question 6

$$2-\displaystyle \frac{3}{2}i$$

$$\displaystyle \frac{3}{2}-2i$$

$$3+4i$$

$$3-4i$$

Solution

Question 7

$$6$$

$$0$$

$$1$$

$$2$$

Solution

Question 8

less than $$2(|Z_1|+|Z_2|)$$

greater than $$2(|Z_1|+|Z_2|)$$

greater than or equal to $$2(|Z_1|+|Z_2|)$$

less than or equal to $$2(|Z_1|+|Z_2|)$$

Solution

Question 9

$$a^{2}$$

a

2a

1

Solution

Question 10

Only I is true

Only II is true

both I and II are true

Neither I nor II are true

Solution

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