Complex Numbers and Quadratic Equations Test 31

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Complex Numbers and Quadratic Equations Test 31

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Weekly Quiz Competition

Question 1
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
$$\dfrac {5\pi}{6}$$

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

D
$$\dfrac {2\pi}{5}$$

Solution

$$\sin{\dfrac{6\pi}{5}}+i\left(1+\cos{\dfrac{6\pi}{5}}\right)$$ is equal to $$\sin{\left(\dfrac{\pi}{5}\right)}+i\left(1-\cos{\left(\dfrac{\pi}{5}\right)}\right)$$
Lies in the second quardent of complex plane and its argument is
$$arg \left(n+iy\right)=\pi-{\tan}^{-1}{\left|\dfrac{y}{x}\right|}$$
$$=\pi-{\tan}^{-1}{\left|\dfrac{1-\cos{\left(\dfrac{\pi}{5}\right)}}{ \sin{\left(\dfrac{\pi}{5}\right)}}\right|}$$
$$=\pi-{\tan}^{-1}{\left|\dfrac{2{\sin}^{2}{\left(\dfrac{\pi}{10}\right)}}{ 2\sin{\left(\dfrac{\pi}{10}\right)\cos{\left(\dfrac{\pi}{10}\right)}}}\right|}$$
$$=\pi-{\tan}^{1}{\left|\dfrac{\sin{\left(\dfrac{\pi}{10}\right)}}{\cos{\left(\dfrac{\pi}{10}\right)}}\right|}$$
$$=\pi-{\tan}^{-1}{\left|\tan{\left(\dfrac{\pi}{10}\right)}\right|}$$
$$=\pi-{\tan}^{-1}{\left(\tan{\left(\dfrac{\pi}{10}\right)}\right)}$$
$$=\pi-\dfrac{\pi}{10}$$
$$=\dfrac{9\pi}{10}$$
Question 2
1 / -0

Let $$a, b, c\epsilon R_{0}$$ and $$1$$ be a root of the equation $$ax^{2} + bx + c = 0$$, then the equation $$4ax^{2} + 3bx + 2c = 0$$ has

A
Imaginary roots

B
Real and equal roots

C
Real and unequal roots

D
Rational roots

Solution
Question 3
1 / -0

If z is a complex number such that $$|z|\ge 2$$, then the minimumm value of $$\left|z+\dfrac{1}{2}\right|$$:

A
is equal to $$\dfrac{5}{2}$$

B
lies in the interval $$(1,2)$$

C
is strictly greater then $$\dfrac{5}{2}$$

D
is strictly greater than $$\dfrac{3}{2}$$ but less than $$\dfrac{5}{2}$$

Solution
Question 4
1 / -0

The complex no. $$\dfrac{1+2i}{1-i}$$ lies in which quadrant of the complex plane

A
first

B
second

C
third

D
fourth

Solution
Question 5
1 / -0

If $$z \neq 0$$, then $$ \overset{100}{\underset{0}{\int}}arg(-|z|)dx =$$

A
$$0$$

B
Not defined

C
$$100$$

D
$$100\pi$$

Solution

Let z is a complex number. Then,
⇒ z=x+iy and
∣z∣= $$(X^2+ Y^2)^{1/2}$$
∣z∣ is a distance of a complex number from the origin which is always real number and positive.
We know, in complex number line x−axis= Real number and y−axis= imaginary number.
If ∣z∣ lies of real axis then it will make $$0^0 $$angle with respect to the origin.
∴ arg∣z∣=0 Then,
⇒ $$arg(-∣z∣)=\pi$$

$$\int^{100} _0 arg(-∣z∣)dx=\int^{100} _0 \pi dx=100\pi$$

hence D is the correct answer
Question 6
1 / -0

If $$z$$ is purely real and $$Re(z)<0$$, then $$arg(x)$$ is

A
$$0$$

B
$${\pi}$$

C
$$-{\pi}$$

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

Solution

Given $$z$$ is purely real is $$z = a + i(0) $$ for some $$a \in R$$

$$\Rightarrow z = a$$

given $$Re(z) < 0$$

$$\Rightarrow a < 0$$

$$\therefore z$$ can be represented as Ref. image

$$\therefore Arg (z) = \pi$$
Question 7
1 / -0

In Argand diagram, O, P, Q represents the origin, $$z$$ and $$z+iz$$
respectively. then $$\angle OPQ = $$

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

B
$$\dfrac{\pi }{6}$$

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

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

Solution

$$z=x+iy $$

$$z+iz=(x+iy)+i(x+iy)$$

$$=x+iy+xi-y$$

$$=(x-y)+(x+y)i $$

Slope of line $$PQ = \dfrac{x+y-y}{x-y-x} = \dfrac{-x}{y}=m_{1}$$

Slope of $$OP = \dfrac{y}{x}=m_{2}$$

$$m_{1}m_{2}=-1$$

lines are perpendicular. $$\angle OPQ=90^{\circ}=\dfrac{\pi }{2}$$

C is correct.
Question 8
1 / -0

If $$Z$$ is a complex number such that $$|z| \ge 2$$,
then the minimum value of $$\left|z + \dfrac{1}{2}\right|$$

A
Is equal to $$\dfrac{5}{2}$$

B
Lies in the interval $$(1, 2)$$

C
Is strictly grater than $$\dfrac{5}{2}$$

D
Is strictly greater than $$\dfrac{3}{2}$$ but less than $$\dfrac{5}{2}$$

Solution

R.E.F image

graph of $$|z|\geq 2 $$

everything outside the

circle and on the circle is $$|z|\geq 2 $$

Minimum distance bet$$^{n}$$ $$ |z+1/2| $$

$$ \Rightarrow $$ distance $$ AB $$

By using distance formula

$$ \Rightarrow \dfrac 32 $$

$$ \therefore $$ Option B is correct
Question 9
1 / -0

If $$\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = 1$$ and $$\arg \left( {{z_1}{z_2}} \right) = 0$$ , then

A
$${z_1} = {z_2}$$

B
$${\left| {{z_2}} \right|^2} = {z_1}{z_2}$$

C
$${z_1}{z_2} = 1$$

D
$${ z_{ 1 } }=-{ z_{ 2 } }$$

Solution

If $$\left|\dfrac{z_1}{z_2}\right|=1$$ and $$arg(z_1z_2)=0$$

$$z_1=x_1+iy, z_2=x_2+iy_2$$

then if $$\left|\dfrac{z_1}{z_2}\right|=1$$

$$\sqrt{x_1^2+y_1^2}=\sqrt{x_2^2+y_2^2}$$

$$x_1^2+y_1^2=x_2^2+y_2^2$$

$$arg(z_1z_2)=arg(x_1x_2-y_1y_2+i(x_1y_2+y_1x_2))=0$$

$$\dfrac{x_1y_2+y_1x_2}{x_1x_2-y_1y_2}=0$$

$$x_1y_2=-y_1x_2$$

$$\dfrac{x_1}{x_2}=\dfrac{-y_1}{y_2}$$

$$\dfrac{x_1}{y_2}=\dfrac{-x_2}{y_2}$$

Both $$z_1z_2$$ make $$180^o$$ angle

So $$z_1+z_2=0$$

if both have an equal modulus

$$z_1+z_2=0$$

$$z_1=-z_2$$

$$D$$ is correct
Question 10
1 / -0

If $$|z|=1$$ and $$|\omega -1| =1$$ where $$z, \omega \in C$$, then the largest set of values of $$|2z - 1|^2 + | 2\omega -1|^2$$ equals

A
$$[1, 9]$$

B
$$[2, 6]$$

C
$$[2, 12]$$

D
$$[2, 18]$$

Solution

$$|z|=1$$ implies that Z in Q circle with radius with a center (0,0) in the complex plane.
$$|w-1|=1$$ implies that it in a circle with the radius unit, center $$(1,0)$$ in the complex plane.
$$|2z-1|^2+|2w-1|^2=4\left(\left|z-\dfrac{1}{2}\right|^2+\left|w-\dfrac{1}{2}\right|^2\right)$$
It means it is the range of $$4 $$ times the sum of the square of the distance of points from $$(0.5,)$$ from the respective circles of low of the z,w
The point $$(0.5,0)$$ in symmetric with the both circles on the center of a radius on the X-axis so both of them get equal ranges so it simply becomes $$8$$ times of annual circle range.
The min or max possible values of the distance from the point become the distance b/w the point and the vertices of the diameter on choice it lies that is
$$\rightarrow \dfrac{1}{2},\dfrac{3}{2}$$
$$=8\times \left[\left(\dfrac{1}{2}\right)^2,\left(\dfrac{3}{2}\right)^2\right]=[2,18]$$