Complex Numbers and Quadratic Equations Test 33

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Complex Numbers and Quadratic Equations Test 33

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Weekly Quiz Competition

Question 1
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 2
1 / -0

If $$z=(3+7i)(p+iq)$$ where $$p,q\in I-\left\{ 0 \right\} $$, is purely imaginary then minimum value of $${ \left| z \right| }^{ 2 }$$ is

A
$$0$$

B
$$58$$

C
$$\dfrac{3364}{3}$$

D
$$3364$$

Solution
Question 3
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 4
1 / -0

If $$Re(\dfrac{z+2i}{z+4})=0$$ then z lies on a circle with center:

A
(-2,-1)

B
(-2,1)

C
(2,-1)

D
(2,1)

Solution

Let $$z=x+ iy$$
Then

$$w=\dfrac{z+2i}{z+4}=\dfrac{x+i(2+y)}{(x+4)+iy}$$
Rationalizing we get:

$$w=\dfrac{z+2i}{z+4}=\dfrac{x+i(2+y)}{(x+4)-iy}\times \dfrac{x+4-iy}{x+4+iy}=\dfrac{x^2+4x+y^2+2y+i(xy+2x+4y+xy+8)}{(x+4)^2+y^2}$$
Since real part of $$w$$ is 0, hence

$$(x+2)^2+(y+1)^2-5=0$$

Hence centre of circle is $$(-2,-1)$$
Question 5
1 / -0

If $$z_{1}=8 +4i,\ z_{2}=6+4i$$ and $$arg \left(\dfrac {z-z_{1}}{z-z_{2}}\right)=\dfrac {\pi}{4}$$, then $$z$$ satisfy

A
$$|z-7-4i|=1$$

B
$$|z-7-5i|=\sqrt {2}$$

C
$$|z-4i|=8$$

D
$$|z-7i|=\sqrt {18}$$
Question 6
1 / -0

The argument of the complex number $$\sin \dfrac{{6\pi }}{5} + i\left( {1 + \cos \dfrac{{6\pi }}{5}} \right)$$ is

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

B
$$\dfrac{{5\pi }}{5}$$

C
$$\dfrac{{9\pi }}{10}$$

D
$$\dfrac{{7\pi }}{10}$$

Solution

Given that

$$\sin { \cfrac { 6\pi }{ 5 } } +i\left( 1+\cos { \cfrac { 6\pi }{ 5 } } \right) $$

Notice the point $$\sin { \left( \cfrac { 6\pi }{ 5 } \right) } +i\left( 1+\cos { \cfrac { 6\pi }{ 5 } } \right) $$ or

$$-\sin { \left( \cfrac { 6\pi }{ 5 } \right) } +i\left( 1-\cos { \cfrac { \pi }{ 5 } } \right) $$ lies in the second quadrant of complex plane hence its argument is given as

$$arg\left( z \right) =\pi -\tan ^{ -1 }{ \left| y/x \right| } \quad $$ ($$\because z=x+iy$$)

$$\left( \forall x<0,y\ge 0 \right) $$

$$arg(z)=\pi -\tan ^{ -1 }{ \left| \cfrac { 1-\cos { \cfrac { \pi }{ 5 } } }{ \sin { \cfrac { \pi }{ 5 } } } \right| } $$

$$=\pi -\tan ^{ -1 }{ \left| \cfrac { 2\sin ^{ 2 }{ \cfrac { \pi }{ 10 } } }{ 2\sin { \cfrac { \pi }{ 10 } } \cos { \cfrac { \pi }{ 10 } } } \right| } =\pi -\tan ^{ -1 }{ \left| \cfrac { \sin { \cfrac { \pi }{ 10 } } }{ \cos { \cfrac { \pi }{ 10 } } } \right| } $$

$$=\pi -\tan ^{ -1 }{ \left| \tan { \cfrac { \pi }{ 10 } } \right| } \left( \because \tan ^{ -1 }{ \cfrac { \pi }{ 10 } } >1 \right) $$

$$=\pi -\cfrac { \pi }{ 10 } \left( \because -\cfrac { \pi }{ 2 } \le \tan ^{ -1 }{ x } \le \cfrac { \pi }{ 2 } \right) $$

$$arg(z)=\cfrac{9\pi}{10}$$
Question 7
1 / -0

If $$a, b \notin R$$, then $$|e^{a + ib}| $$ is equal to

A
$$e^a$$

B
$$e^b$$

C
$$1$$

D
None of these

Solution

$$|e^{a + ib}| = |e^a . e^{ib}|$$

$$= |e^a . (\cos b + i \sin b)|$$

$$= e^a |(\cos b + i \sin b)|$$

$$= e^a . \sqrt{(\cos b)^2 + (\sin b)^2}$$

$$= e^a . \sqrt{1}$$

$$= e^a$$
Question 8
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 9
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 $$0.$$
Question 10
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

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Re-Attempt Weekly Quiz Competition

Complex Numbers and Quadratic Equations Test 33

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Complex Numbers and Quadratic Equations Test 33

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Question 1

$$0$$

$$2i$$

$$-2i$$

$$2$$

Solution

Question 2

$$0$$

$$58$$

$$\dfrac{3364}{3}$$

$$3364$$

Solution

Question 3

-4

4

-6

6

Solution

Question 4

(-2,-1)

(-2,1)

(2,-1)

(2,1)

Solution

Question 5

$$|z-7-4i|=1$$

$$|z-7-5i|=\sqrt {2}$$

$$|z-4i|=8$$

$$|z-7i|=\sqrt {18}$$

Question 6

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

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

$$\dfrac{{9\pi }}{10}$$

$$\dfrac{{7\pi }}{10}$$

Solution

Question 7

$$e^a$$

$$e^b$$

$$1$$

None of these

Solution

Question 8

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

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

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

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

Solution

Question 9

$$-p$$

$$p/2$$

$$-p/2$$

$$0$$

Solution

Question 10

$$a \leq b$$

$$a > b$$

none of these

$$a - b + 2$$

Solution

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