Complex Numbers and Quadratic Equations Test 10

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Complex Numbers and Quadratic Equations Test 10

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Weekly Quiz Competition

Question 1
1 / -0

Discuss the nature of the roots of the equations $$\displaystyle (a+c-b)x^{2}+2cx+(b+c-a)=0$$

A
Real and equal

B
Real and unequal

C
Imaginary

D
None of these

Solution

$$\left( a+c-b \right) { x }^{ 2 }+2cx+\left( b+c-a \right) =0\\ D={ B }^{ 2 }-4AC\\ ={ \left( 2c \right) }^{ 2 }-4\left( b+c-a \right) \left( a+c-b \right) \\ =4{ c }^{ 2 }-4\left( ab+cb-{ b }^{ 2 }+ac+{ c }^{ 2 }-cb-{ a }^{ 2 }-ac+ab \right) \\ =4c^{ 2 }+4{ b }^{ 2 }+{ 4a }^{ 2 }-{ 4c }^{ 2 }-8ab\\ ={ 4a }^{ 2 }+{ 4b }^{ 2 }-8ab=4{ \left( a-b \right) }^{ 2 }>0$$
Hence roots are real and unequal
Question 2
1 / -0

If $$-2$$ is a root of the quadratic equation $$x^2 + px + 2 = 0$$ and the quadratic equation $$2x^2 + px+ k = 0$$ has equal roots, find the value of $$k$$.

A
$$\displaystyle k = \frac{8}{9}$$

B
$$\displaystyle k = -\frac{8}{9}$$

C
$$\displaystyle k = -\frac{9}{8}$$

D
$$\displaystyle k = \frac{9}{8}$$

Solution

Given, $$ -2 $$ is one of the root of first equation $$ x^2 + px + 2 = 0$$
Let's take $$ \alpha$$ as other root of the equation.
Now, product of the roots is given by $$ (-2).\alpha = 2 $$ $$\longrightarrow$$ $$ \alpha = -1$$ and from sum of roots we get $$ -2 + (-1) = -p $$ $$\longrightarrow$$ $$ p = 3 $$
Now in the second equation roots are equal, take it as $$\beta$$.
From sum of roots we get $$ 2.\beta = \dfrac{-p}{2} $$ $$\longrightarrow$$ $$\beta = \dfrac{-3}{4} $$ and from product of the roots we get $${\beta}^{2} = \dfrac{k}{2}$$
$$\Rightarrow$$ $$ k = 2.{\beta}^{2} $$
$$ \Rightarrow$$ $$ k = \frac{9}{8}$$
Question 3
1 / -0

If $$z = re^{i\theta}$$, then the value of $$|e^{iz}|$$ is equal to

A
$$e^{rcos\theta}$$

B
$$e^{-rcos\theta}$$

C
$$e^{rsin\theta}$$

D
$$e^{-rsin\theta}$$

Solution

Ginen,
$$ z=r{ e }^{ i\theta }$$
To find$$ \left| { e }^{ iz } \right| $$
solution,
for given,$$z=r{ e }^{ i\theta }\\ z=r(\cos\theta +i\sin\theta )\left\{ { e }^{ i\theta }=(\cos\theta +i\sin\theta ) \right\} \\ iz=ir(\cos\theta +i\sin\theta )\\ iz=r(i\cos\theta +{ i }^{ 2 }\sin\theta )\\ iz=r(-\sin\theta +i\cos\theta )\\ iz=-r\sin\theta +ir\cos\theta \\ { e }^{ iz }={ e }^{ -r\sin\theta +ir\cos\theta }\\ \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta } \right| \left| { e }^{ ir\cos\theta } \right| $$
We know that$$ \left| { e }^{ iz } \right| =1\\ \therefore \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta } \right| \longrightarrow \{ always\quad positive\} \\ { e }^{ iz }={ e }^{ -r\sin\theta }$$
Question 4
1 / -0

Find the value of $$p$$ for which the quadratic equation $$x^2 + p(4x + p - 1) + 2 = 0$$ has equal roots ?

A
$$\displaystyle -1, \frac{2}{3}$$

B
$$3 , 5$$

C
$$\displaystyle -1, \frac{4}{3}$$

D
$$\displaystyle \frac{3}{4}, 2$$

Solution

Given equation is $$ x^2 + p(4x + p - 1) + 2 = 0$$.
Comparing it with the $$ax^2+bx+c$$, we get
$$a=1, b=4p, c=(p^2-p+2)$$
Therefore, $$D=b^2-4ac$$
$$ x^2 + p(4x + p - 1) + 2 = 0$$.
$$\Rightarrow x^2+4xp+p^2-p+2=0$$
$$\Rightarrow (4p)^2-4(p^2-p+2)=0$$
$$\Rightarrow 16p^2-4p^2+4p-8=0$$
$$\Rightarrow 12p^2+4p-8=0$$
$$\Rightarrow 3p^2+p-2=0$$
$$\Rightarrow 3p^2+3-2p-2=0$$
$$\Rightarrow 3p(p+1)-2(p+1)=0$$
$$\Rightarrow (p+1)(3p-2)=0$$
$$\Rightarrow p=-1 $$and $$p=\dfrac {2}{3}$$
Question 5
1 / -0

Find the modulus and amplitude of $$-2i$$

A
$$|z|=2; amp(z)=-\dfrac {3\pi}{2}$$

B
$$|z|=2i; amp(z)=\dfrac {\pi}{2}$$

C
$$|z|=2; amp(z)=\dfrac {\pi}{2}$$

D
$$|z|=2; amp(z)=-\dfrac {\pi}{2}$$

Solution

$$z=-2i$$ $$|z|=2$$

$$=2(-i)$$

$$=2e^{i\frac{-\pi}{2}}$$

$$=|z|(e^{i\arg(z)})$$

Hence

$$|z|=2$$ and $$arg(z)=\dfrac{-\pi}{2}$$
Question 6
1 / -0

Find the modulus and amplitude of $$-2 + 2 \sqrt 3i$$

A
$$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {\pi}{3}$$

B
$$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {2\pi}{3}$$

C
$$|z|=4; amp(z)=\dfrac {2\pi}{3}$$

D
$$|z|=4; amp(z)=\dfrac {\pi}{3}$$

Solution

Given $$z=-2+2\sqrt{3}i$$
Hence, $$|z|=2\sqrt{1+3}\ \Rightarrow 2\sqrt{4}$$
$$\Rightarrow 2\times2=4$$
And let $$\alpha$$ be the argument of the complex number.
Then
$$tan\alpha=-\sqrt{3}$$
Now $$Re(z)<0$$ and $$Im(z)>0$$
$$\therefore $$ the number lies in the II quadrant
Hence,
$$\alpha=\pi-\dfrac{\pi}{3}$$

$$=\dfrac{2\pi}{3}$$
Question 7
1 / -0

Find the modulus and amplitude of $$-\sqrt 3-i$$

A
$$|z|=2; amp(z)=-\dfrac {5\pi}{6}$$

B
$$|z|=4; amp(z)=\dfrac {5\pi}{6}$$

C
$$|z|=4; amp(z)=-\dfrac {\pi}{6}$$

D
none of these

Solution

$$z=-\sqrt{3}-i$$ $$|z|=2$$

$$=2(\dfrac{-\sqrt{3}-i}{2})$$

$$=2e^{i(\frac{\pi}{6}-\pi)}$$

$$=2e^{i\frac{-5\pi}{6}}$$

$$=|z|.e^{iarg(z)}$$

Hence

$$|z|=2$$ and $$arg(z)=\dfrac{-5\pi}{6}$$.
Question 8
1 / -0

The amplitude of

$$\displaystyle sin \frac{\pi}{5} + i \left ( 1 - cos \frac{\pi}{5} \right )$$

A
$$\displaystyle \frac{\pi}{5}$$

B
$$\displaystyle \frac{2\pi}{5}$$

C
$$\displaystyle \frac{\pi}{10}$$

D
$$\displaystyle \frac{\pi}{15}$$

Solution

Given,
$$\Rightarrow\displaystyle \sin\frac { \pi }{ 5 } +i\left( 1-\cos\frac { \pi }{ 5 } \right)$$

$$ =2\sin\dfrac { \pi }{ 10 } \cos\dfrac { \pi }{ 10 } +i2\sin^{ 2 }\dfrac { \pi }{ 10 } $$ $$[\because \sin2\theta=2\sin\theta \cos\theta\quad 2\sin^2\theta=1-\cos\theta]$$

$$\displaystyle =2\sin\frac { \pi }{ 10 } \left( \cos\frac { \pi }{ 10 } +i\sin\frac { \pi }{ 10 } \right) $$

Then

$$Amplitude=\tan\theta=\dfrac yx$$

$$\displaystyle \tan\theta =\frac { \sin\frac { \pi }{ 10 } }{ \cos\frac { \pi }{ 10 } } =\tan\frac { \pi }{ 10 }$$

$$ \Rightarrow \theta =\dfrac { \pi }{ 10 } $$
Question 9
1 / -0

If $$z_1, z_2, \varepsilon C$$ are such that $$|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$$ then $$\displaystyle \frac{z_1}{z_2}$$ is

A
zero

B
purely real

C
purely imaginary

D
complex

Solution

$$\because |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$$

$$\Rightarrow |z_1|^2 + |z_2|^2 + 2 R (z_1 \bar z_2) = |z_1|^2 + |z_2|^2$$

$$\Rightarrow R(z_1 \bar z_2) = 0 \Rightarrow z_1 \bar z_2$$ is imaginary

Now $$\displaystyle \frac{z_1}{z_2} = \frac{z_1 \bar z_2}{z_2 \bar z_2} = \frac{\text{imaginary}}{|z_2|^2} = $$ imaginary
Question 10
1 / -0

Find the value of $$x^3 + 7x^2 -x + 16$$, where $$x = 1 + 2i$$

A
$$-17 + 24i$$

B
$$24 + 17i$$

C
$$17 - 24i$$

D
$$24 - 17i$$

Solution

$$f\left(x\right)={x}^{3}+7{x}^{2}-x+16$$
$$f(1+2i)=(1+2i)^3+7(1+2i)^2-(1+2i)+16$$
$$=\left[1+8{i}^{3}+6i\left(1+2i\right)\right]+7\left[1+4{i}^{2}+4i\right]-1-2i+16$$
$$=\left[1-8i+6i-12\right]+7\left[1-4+4i\right]-2i+15$$
$$=\left[-11-2i\right]+7\left[4i-3\right]-2i+15$$
$$=-11-2i+28i-21-2i+15$$
$$=\left(-32+15\right)+24i$$
$$=-17+24i$$

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Complex Numbers and Quadratic Equations Test 10

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Question 1

Real and equal

Real and unequal

Imaginary

None of these

Solution

Question 2

$$\displaystyle k = \frac{8}{9}$$

$$\displaystyle k = -\frac{8}{9}$$

$$\displaystyle k = -\frac{9}{8}$$

$$\displaystyle k = \frac{9}{8}$$

Solution

Question 3

$$e^{rcos\theta}$$

$$e^{-rcos\theta}$$

$$e^{rsin\theta}$$

$$e^{-rsin\theta}$$

Solution

Question 4

$$\displaystyle -1, \frac{2}{3}$$

$$3 , 5$$

$$\displaystyle -1, \frac{4}{3}$$

$$\displaystyle \frac{3}{4}, 2$$

Solution

Question 5

$$|z|=2; amp(z)=-\dfrac {3\pi}{2}$$

$$|z|=2i; amp(z)=\dfrac {\pi}{2}$$

$$|z|=2; amp(z)=\dfrac {\pi}{2}$$

$$|z|=2; amp(z)=-\dfrac {\pi}{2}$$

Solution

Question 6

$$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {\pi}{3}$$

$$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {2\pi}{3}$$

$$|z|=4; amp(z)=\dfrac {2\pi}{3}$$

$$|z|=4; amp(z)=\dfrac {\pi}{3}$$

Solution

Question 7

$$|z|=2; amp(z)=-\dfrac {5\pi}{6}$$

$$|z|=4; amp(z)=\dfrac {5\pi}{6}$$

$$|z|=4; amp(z)=-\dfrac {\pi}{6}$$

none of these

Solution

Question 8

$$\displaystyle \frac{\pi}{5}$$

$$\displaystyle \frac{2\pi}{5}$$

$$\displaystyle \frac{\pi}{10}$$

$$\displaystyle \frac{\pi}{15}$$

Solution

Question 9

zero

purely real

purely imaginary

complex

Solution

Question 10

$$-17 + 24i$$

$$24 + 17i$$

$$17 - 24i$$

$$24 - 17i$$

Solution

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