Number Theory Test 40

Result

Number Theory Test 40

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Weekly Quiz Competition

Question 1
1 / -0

If $$\left| {z + 2 - i} \right| = 5$$ then the maximum value of $$\left| {3z + 9 - 7i} \right|$$ is

A
18

B
19

C
20

D
8

Solution
Question 2
1 / -0

Let $$A = \left\{ {z \in c:\left| z \right|} \right. = 2\left. 5 \right\}$$ and $$B = \left\{ {z \in c:\left| {z + 5 + 12i} \right|} \right. = 4$$. Then the minimum value of $$\left| {z - w} \right|$$ for $$Z \in A$$ and $$\omega \in B$$ is :

A
7

B
8

C
9

D
6

Solution
Question 3
1 / -0

Solve $$i^{57}+\dfrac{1}{i^{125}}$$

A
$$0$$

B
$$2i$$

C
$$-2i$$

D
$$2$$

Solution

We know that $${i}^{2}=-1$$
$${i}^{3}={i}^{2}.i=-i$$
$${i}^{4}={i}^{2}\times {i}^{2}=-1\times -1=1$$
Thus, $${i}^{57}={\left({i}^{4}\right)}^{14}\times i={1}^{14}\times i=i$$
$${i}^{125}={\left({i}^{4}\right)}^{31}\times i={1}^{4}\times i=i$$
$$\dfrac{1}{{i}^{125}}=\dfrac{1}{i}$$
$$\dfrac{1}{i}=\dfrac{1}{i}\times\dfrac{i}{i}=\dfrac{i}{{i}^{2}}=-i$$
Now $${i}^{57}+\dfrac{1}{{i}^{125}}=i+\left(-i\right)=i-i=0$$
Question 4
1 / -0

If $${z_1}$$ and $${z_2}$$ be complex numbers such that $${z_1} \ne {z_2}$$ and $$\left| {{z_1}} \right| = \left| {{z_2}} \right|$$. If $${z_1}$$ has positive real part and $${z_2}$$ has negative imarinary part, then $$\frac{{\left( {{z_1} + {z_2}} \right)}}{{\left( {{z_1} - {z_2}} \right)}}$$ may be

A
Purely imaginary

B
Real and positive

C
Real and negative

D
zero

Solution
Question 5
1 / -0

If $$|z_{1}+z_{2}|=|z_{1}-z_{2}|$$, then the different in the amplitudes of $$z_{1}$$ and $$z_{2}$$ is

A
$$\dfrac {\pi}{4}$$

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

C
$$\dfrac {\pi}{2}$$

D
$$0$$

Solution

$$|z_{1}+ z_{2}|= |z_{1}- z_{2}|$$ then we can say that the long of diagonals of the parallelogram is equal hence the parallelogram is a rectangle so angle between $$z_{1}$$ & $$z_{2}$$ is $$90$$ degree
Question 6
1 / -0

The value of $$2x^{4}+5x^{3}+7x^{2}-x+41$$, when $$x=-2-\sqrt{3i}$$ is:

A
-4

B
4

C
-6

D
6

Solution

$$ 2x^{4}+5x^{3}+4x^{2}-x+41 $$ ---(1)
when $$ x = -2-\sqrt{3}i = -(2+\sqrt{3}i) $$---(2)
$$ x^{2} = (2+\sqrt{3}i)^{2} $$
$$ = 4+3(i)^{2}+4\sqrt{3}i $$
$$ =4-3+4\sqrt{3}! $$ $$ (\because i^{2} = -1) $$
$$ x^{2} = 1+4\sqrt{3}i $$ ---(3)
$$ x^{3} = x^{2}.x $$
$$ = (1+4\sqrt{3}i)[-(2+\sqrt{3}i)] $$
$$ = -(2+\sqrt{3}i+8\sqrt{3}i+12!^{2}) $$
$$ = -(2-12+9\sqrt{3}i) $$
$$ x^{3} = 10-9\sqrt{3}i $$ ---(4)
$$ x^{4} = (x^{2})^{2} $$
$$ = (1+4\sqrt{3}i)^{2} $$
$$ = 1+48!^{2}+8i\sqrt{3} $$
$$ = 1-48+8i\sqrt{3} $$ $$ (i^{2} = -1) $$
$$ x^{4} = -47+8!^{3} $$---(5)
From (1),(2),(3),(4) and (5)
$$ 2(-47+8i\sqrt{3})+5(10-9\sqrt{3}i)+7(1+4\sqrt{3}i)-(-2\sqrt{3}i)+41$$ $$ -94+16i\sqrt{3}+50-45\sqrt{3}i+7+28\sqrt{3}i+2+\sqrt{3}i+41 $$
$$ = 6 $$ d)
Question 7
1 / -0

If the six solutions of $$x^6 = -64$$ are written in the form $$a + bi$$, where $$a$$ and $$b$$ are real, then the product those solution with $$a < 0$$, is

A
$$4$$

B
$$8$$

C
$$16$$

D
$$64$$
Question 8
1 / -0

The value of the sum $$\displaystyle\sum^{13}_{n=1}\left(i^n+i^{n+1}\right)$$, where $$i=\sqrt{-1}$$, is?

A
i

B
$$i-1$$

C
$$-i$$

D
$$0$$

Solution

$$ = \displaystyle\sum_{n=1}^{13}(i^{n}+i^{n+1}) $$
$$ =\displaystyle \sum_{n=1}^{13}(l^{n}.(i+1)) $$
$$ = (i+1)\displaystyle\sum_{n=1}^{13}(i^{n}) $$
$$ = (i+1)[i+i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+$$
$$i^{10}+i^{11}+i^{12}+i^{13}] $$
$$ = (i+1)[i+(-1)+(-i)+1+i+(-1)+(-i)+1+1$$
$$+(-1)+(-i)+1+i] $$
$$ = (i+1)(i) $$
$$ =l^{2}+i $$
$$ = -1+i $$
$$ = i-1 $$
Question 9
1 / -0

If $$z_{1} and z_{2} are on straight line$$ $$\left| \frac { 1 } { 2 } \left( z _ { 1 } + z _ { 2 } \right) + \sqrt { z _ { 1 } z _ { 2 } } \right| + \left| \frac { 1 } { 2 } \left( z _ { 1 } + z _ { 2 } \right) - \sqrt { z _ { 1 } z _ { 2 } } \right| =$$

A
$$\left| z _ { 1 } + z _ { 2 } \right|$$

B
$$\left| z _ { 1 } - z _ { 2 } \right|$$

C
$$\left| z _ { 1 } \right| + \left| z _ { 2 } \right|$$

D
$$\left| z _ { 1 } \right| - \left| z _ { 2 } \right|$$

Solution
Question 10
1 / -0

If $$z_{1}$$ and $$z_{2}$$ two non-zero complex number such that $$|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$$, then $$arg z_{1}-arg z_{2}$$ is equal to

A
$$-p$$

B
$$p/2$$

C
$$-p/2$$

D
$$0$$

Solution

$$\left| z_1+z_2\right|=\left| z_1\right| +\left| z_2\right|$$
$$\Rightarrow z_1 \;\&\; z_2$$ are collinear and in same direction.
$$\Rightarrow arg \left(z_1\right)-arg\left(z_2\right)=0.$$
Hence, the answer is $$0.$$