Complex Numbers and Quadratic Equations Test 24

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Complex Numbers and Quadratic Equations Test 24

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Weekly Quiz Competition

Question 1
1 / -0

If $$(1 - i\sqrt {3})^{2} (z) (4i) = (1 + i\sqrt {3})$$, then $$Amp\ z$$ is

A
$$\pi$$

B
$$\dfrac {\pi}{2}$$

C
$$\dfrac {\pi}{3}$$

D
$$0$$

Solution

$$(1 - i\sqrt {3})^{2} (z) (4i) = (1 + i\sqrt {3})$$
$$\Rightarrow 04iz (1 + i\sqrt {3}) = \dfrac {1 + i\sqrt {3}}{2} \Rightarrow z = \dfrac {0 + i}{8} = \dfrac {i}{8}$$
$$\therefore$$ Amp of $$z$$ or argument of $$z = \tan^{-1} \left (\dfrac {1/8}{0}\right )$$
$$\therefore$$ Argument of $$z = \dfrac {\pi}{2}$$.
Question 2
1 / -0

If the equation $$x^2 - bx + 1 = 0$$ does not possess real roots, then

A
$$-3 < b < 3$$

B
$$-2 < b < 2$$

C
$$b > 2$$

D
$$b < -2$$

Solution

If equation $$x^2-bx+1=0$$ does not have real roots, then

$$\Delta <0\implies b^2-4<0$$

$$(b+2)(b-2)<0$$

$$\therefore -2<b<2$$
Question 3
1 / -0

If $$ z = \dfrac {-1}{2} + i \dfrac {\sqrt3}{2} $$, then $$ 8 + 10z + 7z^2 $$ is equal to :

A
$$ - \dfrac {1} {2} - i \dfrac {\sqrt3}{2} $$

B
$$ \dfrac {1} {2} + i \dfrac {\sqrt3}{2} $$

C
$$ - \dfrac {1} {2} + i \dfrac {3\sqrt3}{2} $$

D
$$ \dfrac {\sqrt3}{2} i $$

E
$$ - \dfrac {\sqrt3}{2} i $$

Solution

Given, $$ z = - \dfrac {1} + i \dfrac {\sqrt3}{2} $$
Therefore, $$z^2 = \dfrac {1}{4} - \dfrac {3}{4} - 2i \times \dfrac {1}{2} \dfrac {\sqrt3}{2} = - \dfrac {1}{2} - i \dfrac {\sqrt3}{2} $$
$$ \Rightarrow 8 + 10z + 7z^2 = 8 + 10 \left( - \dfrac {1}{2} + i \dfrac { \sqrt3}{2} \right) + 7 \left( - \dfrac {1}{2} + \dfrac { 3 \sqrt3}{2}i \right) $$
$$ = 8 - 5 + i 5 \sqrt3 - \dfrac {7}{2} - i \dfrac {7 \sqrt3}{2}$$
$$ = - \dfrac {1}{2} + \dfrac { 3 \sqrt3}{2} i$$
Question 4
1 / -0

If the complex numbers $$z_1, z_2$$ and $$z_3$$ denote the vertices of an isosceles triangle, right angled at $$z_1$$, then $$(z_1 - z_2)^2 + (z_1 - z_3)^2$$ is equal to

A
$$0$$

B
$$(z_2 + z_3)^2$$

C
$$2$$

D
$$3$$

E
$$(z_2 - z_3)^2$$

Solution

Since, the triangleis isosceles and right angled, we can say that
$$(z_3-z_1) = (z_2-z_1) (\cos 90^o + i \sin 90^o)$$
$$\Rightarrow (z_3-z_1) = i(z_2-z_1)$$
Squaring bothe sides, we get
$$(z_3-z_1)^2 = -(z_2-z_1)^2$$
$$\Rightarrow (z_3-z_1)^2 + (z_2-z_1)^2 = 0$$
Question 5
1 / -0

If $$z$$ is a complex number such that $$z + |z| = 8 + 12i$$, then the value of $$|z^{2}|$$ is

A
$$228$$

B
$$144$$

C
$$121$$

D
$$169$$

E
$$189$$

Solution

Let $$z = x + iy$$
Then, we have $$z + |z| = 8 + 12i$$
$$\Rightarrow(x + iy) + |x + iy| = 8 + 12l$$
$$\Rightarrow (x + \sqrt {x^{2} + y^{2}}) + iy = 8 + 12i$$
On comparing the real and imaginary part, we get
$$y = 12$$
and $$x + \sqrt {x^{2} + y^{2}} = 8$$
$$\Rightarrow \sqrt {x^{2} + 144} = 8 - x$$
On squaring both sides, we get
$$x^{2} + 144 = 64 + x^{2} - 16x$$
$$\Rightarrow 16x = -80$$
$$\Rightarrow x = -5$$
$$\therefore z = x + iy = -5 + i\cdot 12$$
Then, $$|z| = \sqrt {25 + 144} = \sqrt {169} = 13$$
$$\Rightarrow |z|^{2} = 169$$
$$\Rightarrow |z^{2}| = 169 (\because |z^{n}| = |z|^{n})$$.
Question 6
1 / -0

Which of the given alternatives represent a point in Argand plane, equidistant from roots of the equation $$(z+1)^4= 16z^4$$?

A
$$(0,0)$$

B
$$\left(-\dfrac{1}{3},0\right)$$

C
$$\left(\dfrac{1}{3},0\right)$$

D
$$\left(0,\dfrac{2}{\sqrt5}\right)$$

Solution

Consider the given equation $$(z+1)^4=16z^4$$
$$ \Rightarrow (z+1)^4=16z^4$$

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

$$ \Rightarrow |z+1|=2|z|$$

We know that $$z=x+iy$$

Therefore $$|x+iy+1|=2|x+iy|$$

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

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

$$ \Rightarrow x^2+2x+1+y^2=4x^2+4y^2$$

$$ \Rightarrow 3x^2+3y^2-2x-1=0$$

Divide throughout by 3 we get,
$$ \Rightarrow x^2+y^2-\dfrac{2}{3}x-\dfrac{1}{3}=0$$, which represents a circle.

We know that for the circle equation of the form $$x^2+y^2+2gx+2hy+c=0$$ the center of the circle is given by $$(-g,-h)$$

We have $$3x^2+3y^2-2x-1=0$$ where $$g=-\dfrac{1}{3}, h=0$$.

Hence the center is $$(\dfrac{1}{3},0)$$ which is equidistant from the root of the equation.
Question 7
1 / -0

The number of values k for which $$[x^2- (k -2)x + k^2$$][$$x^2+kx+(2k-1)]$$ is a perfect square is

A
$$2$$

B
$$1$$

C
$$0$$

D
None of these

Solution

$$[x^2-(k-2)x+k^2][x^2+kx+(2k-1)]$$

For the expression to be a perfect square ,ther are two possible way
$$(i)$$ When both the quadratic expression are perfect square for a particular value.For this to happen,
$$k-2=2k\rightarrow k=-2$$
Now,for $$k=-2$$in the second expression ,we get,

$$x^2-2x-5$$,which is not a perfect square.

$$(ii)$$The other way is when both the quadratic equations are same.
$$-(k-2)=k\rightarrow k=1+k^2=2k-1\rightarrow k=1$$
At $$k=1$$,the expression is a perfect square.

Thus, minimum values of $$ k\quad is$$ 1
Question 8
1 / -0

Let$$ z$$ = $$\cos\theta + i \sin\theta$$. Then the value of $$\sum\limits_{m=1}^15Im( z^{2m-1})$$ at $$\theta = 2^0$$ is

A
$$\dfrac{1}{sin2^0}$$

B
$$\dfrac{1}{3sin2^0}$$

C
$$\dfrac{1}{2sin2^0}$$

D
$$\dfrac{1}{4sin2^0}$$

Solution

using the sum of series of sine
$$\sum_{m=1}^{n}sin(a+k.d)=\dfrac{sin\dfrac{n\times d}{2}}{sin\dfrac{d}{2}}\times sin(\dfrac{2a+(n-1).d}{2})$$
Given,
$$\sum_{m=1}^{15}Im(z^{2m-1})\rightarrow$$

$$sin\theta+sin3\theta+sin5\theta+.......sin29\theta=$$

$$\rightarrow\dfrac{sin\dfrac{15\times 2\theta}{2}}{sin\dfrac{2\theta}{2}}\times sin(\dfrac{2\theta+(14\times2\theta)}{2})$$

$$\rightarrow\dfrac{sin^215\theta}{sin\theta}$$

put $$\theta = 2^{\circ}$$

$$\rightarrow\dfrac{sin^230^{\circ}}{sin2^{\circ}}$$

$$\rightarrow\dfrac{1}{4.sin2^{\circ}}$$
Question 9
1 / -0

If $$z_1, z_2$$ are two complex numbers such that $$arg(z_1+z_2)=0$$ and $$Im(z_1z_2)=0$$, then.

A
$$z_1=-z_2$$

B
$$z_1=z_2$$

C
$$z_1=\vec{z_2}$$

D
none of these

Solution

Let $$ z_1=a+ib$$
and $$z_2=c+id$$
Now $$z_1+z_2=(a+c)+i(b+d)$$
Given $$arg(z_1+z_2)=0$$
$$\Rightarrow tan^{-1}[\dfrac{b+d}{a+c}]=0$$
$$\Rightarrow b+d=0$$.........$$(1)$$
Now,
$$z_1z_2=(ac-bd)+i(ad+bc)$$
Given $$Im(z_1z_2)=0$$
$$\Rightarrow ad+bc=0$$.........$$(2)$$
Putting $$b=-d$$ from $$(1)$$ in $$(2)$$ we get $$a=c$$
Hence $$z_1=a+ib=c-id= $$conjugate of $$z_2=z_2^{\rightarrow}$$
Hence option C is correct.
Question 10
1 / -0

The principal value of $$arg(z)$$ lies in the interval:

A
$$\left[ 0,\cfrac { \pi }{ 2 } \right] $$

B
$$(-\pi ,\pi ]$$

C
$$\left[ 0,\pi \right] $$

D
$$(-\pi ,0]$$

Solution

For a complex number $$z=re^{\phi i}$$ the principal value of the square root is: $$pv\: \sqrt{z}=\sqrt{r}e^{i\phi/2}$$ with argument $$-\pi < 0 \leq \pi$$
Thus the principal value of $$arg(z)$$ lies in the interval: $$(-\pi,\pi]$$.

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Complex Numbers and Quadratic Equations Test 24

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Complex Numbers and Quadratic Equations Test 24

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SHARING IS CARING

Question 1

$$\pi$$

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

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

$$0$$

Solution

Question 2

$$-3 < b < 3$$

$$-2 < b < 2$$

$$b > 2$$

$$b < -2$$

Solution

Question 3

$$ - \dfrac {1} {2} - i \dfrac {\sqrt3}{2} $$

$$ \dfrac {1} {2} + i \dfrac {\sqrt3}{2} $$

$$ - \dfrac {1} {2} + i \dfrac {3\sqrt3}{2} $$

$$ \dfrac {\sqrt3}{2} i $$

$$ - \dfrac {\sqrt3}{2} i $$

Solution

Question 4

$$0$$

$$(z_2 + z_3)^2$$

$$2$$

$$3$$

$$(z_2 - z_3)^2$$

Solution

Question 5

$$228$$

$$144$$

$$121$$

$$169$$

$$189$$

Solution

Question 6

$$(0,0)$$

$$\left(-\dfrac{1}{3},0\right)$$

$$\left(\dfrac{1}{3},0\right)$$

$$\left(0,\dfrac{2}{\sqrt5}\right)$$

Solution

Question 7

$$2$$

$$1$$

$$0$$

None of these

Solution

Question 8

$$\dfrac{1}{sin2^0}$$

$$\dfrac{1}{3sin2^0}$$

$$\dfrac{1}{2sin2^0}$$

$$\dfrac{1}{4sin2^0}$$

Solution

Question 9

$$z_1=-z_2$$

$$z_1=z_2$$

$$z_1=\vec{z_2}$$

none of these

Solution

Question 10

$$\left[ 0,\cfrac { \pi }{ 2 } \right] $$

$$(-\pi ,\pi ]$$

$$\left[ 0,\pi \right] $$

$$(-\pi ,0]$$

Solution

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