Complex Numbers and Quadratic Equations Test 11

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Complex Numbers and Quadratic Equations Test 11

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Weekly Quiz Competition

Question 1
1 / -0

$$(1 + i)^8 + (1 -i)^8 =$$

A
$$16$$

B
$$-16$$

C
$$32$$

D
$$-32$$

Solution

$${ \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }$$
$$\Rightarrow { \left( 1+i \right) }^{ 8 }={ \left[ { { \left( 1+i \right) }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i+{ i }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i-1 } \right] }^{ 4 }$$
                     $$={ \left[ { { \left( 2i \right) }^{ 2 } } \right] }^{ 2 }$$
                     $$={ \left( -4 \right) }^{ 2 }$$
                     $$=16$$

Similarely $${ \left( 1-i \right) }^{ 8 }=16$$

$$\therefore { \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }$$

$$=16+16$$

$$=32.$$

Hence, the answer is $$32.$$
Question 2
1 / -0

If z = x + iy and |z 1 + 2i | = | z + 1 2i |,then the locus of z is

A
circle

B
parabola

C
straight line

D
None of these

Solution

We know that $$z=x+iy$$
$$\Rightarrow \left| z-1+2i \right| =\left| z+1-2i \right| ,$$ locus of $$x=?$$
$$\Rightarrow \left| x+iy-1+2i \right| =\left| x+iy+1-2i \right| $$
$$\Rightarrow \left| \left( x-1 \right) +i\left( y+2 \right) \right| =\left| \left( x+1 \right) +i\left( y-2 \right) \right| $$
$$\Rightarrow \sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y+2 \right) }^{ 2 } } =\sqrt { { \left( x+1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 } } $$
$$\Rightarrow x^2-2x+1+y^2+4y+4=x^2+2x+1+y^2-4y+4$$
$$\Rightarrow -2x+4y+5=2x-4y+5$$
$$\Rightarrow 4x=8y$$
$$\Rightarrow x=2y$$
$$\therefore x-2y=0$$
$$\therefore$$ The locus of $$z$$ is a straight line.
Hence, the answer straight line.
Question 3
1 / -0

Modulus of $$\displaystyle \frac{cos \theta- isin\theta }{sin\theta - icos\theta} is$$

A
0

B
2$$\theta$$

C
$$\pi - 2\theta$$

D
none of these

Solution

$$\displaystyle |\frac{\cos \theta- i\sin\theta }{\sin\theta - i\cos\theta}|$$ $$=\displaystyle \frac{|\cos \theta- i\sin\theta| }{|\sin\theta - i\cos\theta|}$$
$$= \displaystyle \frac{\sin^2 \theta +\cos^2 \theta}{\sin^2 \theta +\cos^2 \theta}=1$$
$$\therefore \displaystyle |\frac{\cos \theta- i\sin\theta }{\sin\theta - i\cos\theta}|=1$$
Hence, option D.
Question 4
1 / -0

Find the value of: $$i^2 + i^4 + i^6$$ +..... upto $$(2n +1)$$ terms.

A
$$i$$

B
$$-i$$

C
$$1$$

D
$$-1$$

Solution

$$i^2$$ + $$i^4$$ + $$i^6$$ + ...upto $$(2n + 1 )$$ terms

$$=(-1 +1)+ (-1 +1)+( -1 + $$ .....upto $$(2n +1))$$terms $$\Rightarrow$$ (odd terms)

$$= 0+ 0 + ....-1$$

$$= -1$$

Hence, option D is correct.
Question 5
1 / -0

$$i^n + i^{n + 1} + i^{n + 2}+ i^{n + 3} (n \in N) $$ is equal to

A
$$4$$

B
$$1$$

C
$$0$$

D
$$2$$

Solution

$$\textbf{Step-1: Expanding the terms}$$
$${{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}$$

$$= {{i}^{n}}+{{i}^{n}}.i+{{i}^{n}}.{{i}^{2}}+{{i}^{n}}.{{i}^{3}}$$
$$\textbf{Step-2: Taking } \mathbf{{i}^{n}} \textbf{ common}$$

$$= {{i}^{n}}(1+i+{{i}^{2}}+{{i}^{3}})$$

$$= {{i}^{n}}(1+i-1-i) $$ $$\mathbf{[\because {{i}^{2}}=-1,{{i}^{3}}=-i]}$$

$$= {{i}^{n}}(0)=0$$
$$\textbf{Hence the correct option is (C) 0}$$
Question 6
1 / -0

The argument of the complex number $$z = sin \alpha + i (1 - cos \alpha)$$ is

A
$$2 sin \displaystyle \frac{\alpha}{2}$$

B
$$\displaystyle \frac{\alpha}{2}$$

C
$$\alpha$$

D
none of these

Solution

$$z = sin \alpha + i (1 - cos \alpha)$$
$$\Rightarrow arg (z) = tan^{-1} \displaystyle \left ( \frac{1 - cos \alpha}{sin \alpha} \right ) = tan^{-1} \left (\frac{\displaystyle 2 sin^2 \frac{\alpha}{2}}{2 sin \displaystyle \frac{\alpha}{2} cos \frac{\alpha}{2}} \right )$$
$$= tan^{-1} \displaystyle \left ( tan \frac{\alpha}{2} \right ) = \frac{\alpha}{2}$$
Question 7
1 / -0

If x and y are real then which one of the following is true

A
|x - y| = |x| - |y|

B
|x + y| $$\leq$$ |x| - |y|

C
|x - y| $$\geq$$ |x| - |y|

D
|x + y| = |x| + |y|

Solution

We know that $$|x+y|\leq |x|+|y|$$
$$\Rightarrow |x-y+y|\leq |x-y|+|y|$$
$$\Rightarrow |x|-|y|\leq |x-y|$$
$$\therefore |x-y|\geq |x|-|y|$$
Hence, option C.
Question 8
1 / -0

Solve:
$$\displaystyle \left|(1 + i)\frac{(2+i)}{(3 + i)}\right| $$

A
$$\dfrac{1}{2}$$

B
$$\dfrac{-1}{2}$$

C
$$1$$

D
$$-1$$

Solution

Given, $$\displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}| $$=$$\displaystyle |(1 + i)|\frac{|(2+i)|}{|(3 + i)|}$$

Since, $$|Z_{1}Z_{2}|=|Z_{1}||Z_{2}|$$ and $$\displaystyle|\frac{Z_{1}}{Z_{2}}|=\frac{|Z_{1}|}{|Z_{2}|}$$

$$=\displaystyle \sqrt{2}\cdot \frac{\sqrt{5}}{\sqrt{10}}=1$$

$$\therefore \displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}| = 1$$

Hence, option C.
Question 9
1 / -0

The modulus of (1 + i) (1 + 2i) (1 + 3i) is equal to

A
$$\sqrt10$$

B
$$\sqrt 5$$

C
5

D
10

Solution

$$|(1 + i) (1 + 2i) (1 + 3i)|$$ = $$|1+i||1+2i||1+3i|$$ ($$\because |Z_{1}Z_{2}|=|Z_{1}||Z_{2}|$$)
$$=\sqrt {2}\cdot\sqrt {5}\cdot\sqrt {10}$$
$$\therefore |(1 + i) (1 + 2i) (1 + 3i)|=10$$
Hence, option D.
Question 10
1 / -0

If $$|z -2 + i| = |z-3-i|$$, then the locus of z is

A
$$2x -4y -5 = 0$$

B
$$2x + 4y -5 = 0$$

C
$$x -2y + 5 = 0$$

D
none of these

Solution

Given: $$ |z-2+i|=|z-3-i|$$

To find the locus of: $$ z$$.

Solution:

Let, $$z=x+iy$$

$$|z-2+i|=|z-3-i|$$

$$|x+iy-2+i|=|x+iy-3-i|$$

$$ |(x-2)+i(y+1)|=|(x-3)+i(y-1)|$$

On squaring both sides, we get

$$ \Rightarrow { \left( \sqrt { { (x-2) }^{ 2 }+{ (y+1) }^{ 2 } } \right) }^{ 2 }={ \left( \sqrt { { (x-3) }^{ 2 }+{ (y-1) }^{ 2 } } \right) }^{ 2 } \quad ...\left\{ |x+iy|=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right\} $$

$$ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+1+2y$$$$ = { x }^{ 2 }+9-6x+{ y }^{ 2 }+1-2y$$

$$ \Rightarrow 2x+4y-5=0$$

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Re-Attempt Weekly Quiz Competition

Complex Numbers and Quadratic Equations Test 11

Result

Complex Numbers and Quadratic Equations Test 11

Score

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SHARING IS CARING

Question 1

$$16$$

$$-16$$

$$32$$

$$-32$$

Solution

Question 2

circle

parabola

straight line

None of these

Solution

Question 3

0

2$$\theta$$

$$\pi - 2\theta$$

none of these

Solution

Question 4

$$i$$

$$-i$$

$$1$$

$$-1$$

Solution

Question 5

$$4$$

$$1$$

$$0$$

$$2$$

Solution

Question 6

$$2 sin \displaystyle \frac{\alpha}{2}$$

$$\displaystyle \frac{\alpha}{2}$$

$$\alpha$$

none of these

Solution

Question 7

|x - y| = |x| - |y|

|x + y| $$\leq$$ |x| - |y|

|x - y| $$\geq$$ |x| - |y|

|x + y| = |x| + |y|

Solution

Question 8

$$\dfrac{1}{2}$$

$$\dfrac{-1}{2}$$

$$1$$

$$-1$$

Solution

Question 9

$$\sqrt10$$

$$\sqrt 5$$

5

10

Solution

Question 10

$$2x -4y -5 = 0$$

$$2x + 4y -5 = 0$$

$$x -2y + 5 = 0$$

none of these

Solution

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