Complex Numbers and Quadratic Equations Test -7

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Complex Numbers and Quadratic Equations Test -7

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Weekly Quiz Competition

Question 1
1 / -0

If $z=2-3i$ then $z^2-4z+13=$

A
$0$

B
$1$

C
$2$

D
$3$

Solution

$z=2-3i$
$z^{2}=2^{2}-3^{2}-12i$
$=-5-12i$
$\therefore z^{2}-4z+13$
$=(-5-12i)-4(2-3i)+13$
$=-5-12i-8+12i +13$
$=-13+13$
$=0$
Question 2
1 / -0

The complex number $\displaystyle \frac{1+2i}{1-i}$ lies in the quadrant :

A
I

B
II

C
III

D
IV

Solution

Let $z =\dfrac{1+2i}{1-i}$

$\Rightarrow z =\dfrac{(1+2i)}{1-i}\times \dfrac{1+i}{1+i}$

$= \dfrac{1+2i+i+2i^2}{1-i^2}$

$=\dfrac{1+3i-2}{1+1}$ ............ $[\because i^2 = -1]$

$\Rightarrow z =\dfrac{-1+3i}{2}$

$=-\dfrac{1}{2}+\dfrac{3}{2}i$
$=x+iy$
Clearly $x<0$ and $y>0$
Hence $z$ lies in $\text{II}$ quadrant
Question 3
1 / -0

$\sqrt{-3}\sqrt{-75}=$

A
$15$

B
$15i$

C
$-15$

D
$-15i$

Solution

$\sqrt{-3}\times\sqrt{-75}=\sqrt{3\times(-1)}\sqrt{75\times(-1)}$
$=\sqrt{3}i\times\sqrt{75}i$
$=\sqrt{225}i^{2}$
$= - 15$
Question 4
1 / -0

The sum of two complex numbers $a + ib$ and $c +id$ is a real number if

A
$a + c = 0$

B
$b + d = 0$

C
$a + b= 0$

D
$b + c = 0$

Solution

It is given that

$z_{1}=a+ib$ and

$z_{2}=c+id$

Then
$z_{1}+z_{2}=(a+c)+i(b+d)$

Now
$(z_{1}+z_{2})$ is purely real.

Then the imaginary part has to be $0$ .

Hence
$b+d=0$ .
Question 5
1 / -0

The locus of complex number z such that z is purely real and real part is equal to - 2 is

A
Negative y-axis

B
Negative x-axis

C
The point (-2, 0)

D
The point (2, 0)

Solution

$z=x+iy$
$(x,y)$
$z$ is purely real and the real part equals $-2$
$\therefore y=0$ & $x=-2$
$z=-2$
Hence, this would be represented by the point (-2,0) on the Argand Plane.
Question 6
1 / -0

$\dfrac{1}{i-1}+\dfrac{1}{i+1}$ is

A
positive rational number

B
purely imaginary

C
positive Integer

D
negative integer

Solution

Let $Z= \dfrac{1}{i-1}+\dfrac{1}{i+1}$

$=\dfrac{i+1+i-1}{(i-1)(i+1)}$

$=\dfrac{i+i}{(i^2-1^2)}$

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

$\therefore Z=-i$
Question 7
1 / -0

If $(x+iy)(2-3i)=4+i$ then (x, y) =

A
$\left ( 1,\dfrac{1}{13} \right )$

B
$\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )$

C
$\left ( \dfrac{5}{13},\dfrac{14}{13} \right )$

D
$\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )$

Solution

$(x+iy)(2-3i)=4+i$

$2x-(3x)i+(2y)i-3yi^{2}=4+i$

$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i$

Comparing the real & imaginary parts,
$2x+3y=4$ --------------------------(1)
$2y-3x=1$ ----------------------------(2)
Solving eq(1) & eq(2),
$4x+6y=8$
$-9x+6y=3$
$13x=5\Rightarrow x=\dfrac{5}{13}$
$y=\dfrac{14}{13}$
$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )$
Question 8
1 / -0

The argument of every complex number is

A
Double valued

B
Single valued

C
Many valued

D
Triple valued

Solution

$z=x+iy$
amplitude $= tan^{-1}\frac{y}{x}$
$\Rightarrow$ amplitude $=\theta \pm 2k\pi$ where $\theta \epsilon \left [ -\pi ,\pi \right ]$ $\forall k\epsilon R$
since $k\epsilon R$
$\Rightarrow$ Amplitude of any complex number is many valued.
Question 9
1 / -0

The sum of two complex numbers $a + ib$ and $c+ id$ is purely imaginary if

A
$a + c = 0$

B
$a + d = 0$

C
$b + d = 0$

D
$b + c = 0$

Solution

It is given that
$z_{1}=a+ib$ and
$z_{2}=c+id$
$z_{1}+z_{2}=(a+c)+i(b+d)$
$z_{1}+z_{2}$ is purely imaginary. (Given)
Then the real part has to be $0$ .
Hence
$a+c=0$ .
Question 10
1 / -0

For $a > 0$ , arg $(ia) =$

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

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

C
$\pi$

D
$-\pi$

Solution

$z= 0 + ia$
$\therefore arg(z) = arg (ia) = \tan^{-1} \cfrac a0$
= $\cfrac {\pi}{2}$ ...(since a is greater than or equal to zero)

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Re-Attempt Weekly Quiz Competition

Complex Numbers and Quadratic Equations Test -7

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Complex Numbers and Quadratic Equations Test -7

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SHARING IS CARING

Question 1

000

111

222

333

Solution

Question 2

I

II

III

IV

Solution

Question 3

151515

15i15i15i

−15-15−15

−15i-15i−15i

Solution

Question 4

a+c=0a + c = 0a+c=0

b+d=0b + d = 0b+d=0

a+b=0a + b= 0a+b=0

b+c=0b + c = 0b+c=0

Solution

Question 5

Negative y-axis

Negative x-axis

The point (-2, 0)

The point (2, 0)

Solution

Question 6

positive rational number

purely imaginary

positive Integer

negative integer

Solution

Question 7

(1,113)\left ( 1,\dfrac{1}{13} \right )(1,131​)

(−513,1413)\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )(−135​,1314​)

(513,1413)\left ( \dfrac{5}{13},\dfrac{14}{13} \right )(135​,1314​)

(−513,−1413)\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )(−135​,−1314​)

Solution

Question 8

Double valued

Single valued

Many valued

Triple valued

Solution

Question 9

a+c=0a + c = 0a+c=0

a+d=0a + d = 0a+d=0

b+d=0b + d = 0b+d=0

b+c=0b + c = 0b+c=0

Solution

Question 10

π2\dfrac{\pi }{2}2π​

−π2-\dfrac{\pi }{2}−2π​

π\pi π

−π-\pi −π

Solution

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$0$

$1$

$2$

$3$

$15$

$15i$

$-15$

$-15i$

$a + c = 0$

$b + d = 0$

$a + b= 0$

$b + c = 0$

$\left ( 1,\dfrac{1}{13} \right )$

$\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )$

$\left ( \dfrac{5}{13},\dfrac{14}{13} \right )$

$\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )$

$a + c = 0$

$a + d = 0$

$b + d = 0$

$b + c = 0$

$\dfrac{\pi }{2}$

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

$\pi$

$-\pi$