If $$z_1$$ and $$z_2$$ are two complex numbers, then $$Re(z_1z_2)$$ is:

$$Re(z_1).Re(z_2)-Im(z_1).Im(z_2)$$

MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 2

For the quadratic equation

$ax^2 + bx + c = 0, a, b, c, \in Q$ , If

$D = 0$ then ...................
Choose the correct option in respect to the statements below.
(P) The roots of the equation are equal.
(Q) The roots of the equation are not equal.
(R) The roots of the equation are rational numbers.
(S) The roots of the equation has no roots.

0%
Statements $P$ and $R$ are correct
0%
Statements $Q$ and $R$ are correct
0%
Only statement $S$ is correct
0%
Only statement $P$ is correct

Explanation

For the quadratic equation

$ax^2 + bx + c = 0, a, b, c, \varepsilon Q$

Roots are given by

$\alpha=\dfrac{-b + \sqrt{b^2 - 4ac}}{2a}$ and

$\beta=\dfrac{-b - \sqrt{b^2 - 4ac}}{2a}$

$D = 0$ then

$\alpha=\dfrac{-b}{2a}=\beta$ i.e. the roots are equal and real.

Since

$a, b$ and

$c \in Q$ , the roots will be equal and rational.

Hence, statements P and R are correct.

What is the modulus of

$\cfrac { \sqrt { 2 } +i }{ \sqrt { 2 } -i }$ where

$i=\sqrt { -1 }$

0%
$3$
0%
$\dfrac{1}{2}$
0%
$1$
0%
None of the above

Explanation

$\left| \dfrac { \sqrt { 2 } +i }{ \sqrt { 2 } -i } \right| =\dfrac { \sqrt { ({ \sqrt { 2 } })^{2}+{ 1 }^{ 2 } } }{ \sqrt { (\sqrt { 2 })^{ 2 }+(-1)^{ 2 } } } =1$
Hence, option C is correct.

In the complex plane, what is the distance of

$4-2i$ from the origin?

0%
$2$
0%
$3.46$
0%
$4.47$
0%
$6$
0%
$12$

Explanation

Let

$z=4-2i$

In a complex plane, distance of

$z$ from any point origin is given by

$|z|=\sqrt{(x_1)^2+(y_1)^2}$

$\therefore$ Distance of

$z=4-2i$ from origin is given by

$|z|=|4-2i|=\sqrt{4^2+(-2)^2}=\sqrt{20}\approx 4.47$

Hence, the answer is

$4.47$ .

If the roots of an equation

$p{ x }^{ 2 }+qx+r=0$ are equal, then

0%
${ q }^{ 2 }=pr$
0%
${ q }^{ 2 }=4pr$
0%
${ p }^{ 2 }=4qr$
0%
$p=qr$

Explanation

For equal roots Discriminant has to be

$0$

${q}^{2}-4 pr = 0$

$\Rightarrow q^2 = 4pr$

In the complex plane, the number 4 + j3 is located in the

0%
first quadrant
0%
second quadrant
0%
third quadrant
0%
fourth quadrant

For a quadratic equation if

$D < 0$ then which of the following is true?

0%
Real roots do not exist
0%
Roots are real and equal
0%
Roots are rational and distinct
0%
Roots are real and distinct

Explanation

For a quadratic equation, if

$D < 0$ then real roots do not exist.

The roots of the equation

${ \left( z+\alpha \beta \right) }^{ 3 }={ \alpha }^{ 3 }$ represent the vertices of a triangle, one of whose sides is of length

0%
$\sqrt { 3 } \left| \alpha \beta \right|$
0%
$\sqrt { 3 } \left| \alpha \right|$
0%
$\sqrt { 3 } \left| \beta \right|$
0%
$None\ of\ these$

Explanation

$z_1=\alpha-\alpha \beta,z_2=\alpha\omega-\alpha \beta,z_3=\alpha\omega^2-\alpha \beta$

Let vertices be denoted by

$A,B,C$

Length of

$AB=|z_1-z_2|=|\alpha||1-\omega|=|\alpha|\left|1-\dfrac{-1-i\sqrt{3}}{2}\right|=\sqrt{3}|\alpha|$

Length of

$BC=|z_2-z_1|=|\alpha|\left|1-\dfrac{-1+i\sqrt{3}}{2}\right|=\sqrt{3}|\alpha|$

Length of

$BC=|z_3-z_1|=|\alpha||\omega|\left|\dfrac{-1+i\sqrt{3}}{2}\right|\left|1-\dfrac{-1+i\sqrt{3}}{2}\right|=\sqrt{3}|\alpha|$

Hence option B is correct.

Put the following in the form of A + iB :

$\dfrac{(3 \, - \, 2i)(2 \, + \, 3i)}{(1 \, + \, 2i)(2 \, - \, i)}$

0%
$\dfrac{3}{4} \, + \, \dfrac{9}{4} \, i$
0%
$\dfrac{63}{25} \, - \, \dfrac{16}{25} \, i$
0%
$\dfrac{5}{4} \, + \, \dfrac{9}{4} \, i$
0%
$\dfrac{1}{4} \, + \, \dfrac{7}{4} \, i$

Explanation

$\cfrac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

$\implies \cfrac{6+9i-4i+6}{2-i+4i+2}$

$\implies \cfrac{12+5i}{4+3i}$

$\implies \cfrac{(12+5i)(4-3i)}{4^2+3^2}$

$\implies \cfrac{48-36i+20i+15}{25}$

$\implies \cfrac{63}{25}$

$-i \cfrac{16}{25}$

State true or false:

The following quadratic equations has real roots

$3a^2x^2 \, + \, 8abx \, + \, 4b^2 \, = \, 0, \, a ,b\, \neq \, 0$

0%
True
0%
False

Explanation

$3a^{2}x^{2}+8ab+4b^{2}=0$

Discriminant

$\rightarrow B^{2}-4AC=64a^{2}b^{2}-48a^{2}b^{2}$

$=16a^{2}b^{2}$

since it is greater than zero,

The roots are real

When will the quadratic equation

$ax^2+bx+c=0$ NOT have Real Roots?

0%
$b^2 - 4ac \ge 0$
0%
$b^2 - 4ac > 0$
0%
$b^2 - 4ac < 0$
0%
None of these

Explanation

We learnt that the quadratic equation

$ax^2+bx+c=0$ will NOT have Real Roots when

$b^2-4ac< 0$ . Option c is correct.

If the value of ' $b^2-4ac$ ' is greater than zero, the quadratic equation $ax^2+bx+c=0$ will have

0%
Two Equal Real Roots.
0%
Two Distinct Real Roots.
0%
No Real Roots.
0%
No Roots or Solutions.

Explanation

${ b }^{ 2 }-4ac>0$

$\therefore \triangle >0$

So, two equal real roots or only one real root.

$a, b, c$ are real and

$b^2- 4ac$ is perfect square then the roots of the equation

$ax^2+bx+c=0$ , will be:

0%
Rational & distinct
0%
Real & equal
0%
Irrational & distanct
0%
Imaginary & distinct

Explanation

Given that the discriminant

$=b^{2}-4ac$ is a perfect square

Let

$b^{2}-4ac =k^{2}$

$\Rightarrow$ Roots

$=\dfrac {-b\pm \sqrt {k^{2}}}{2a}$

$\Rightarrow$ roots

$=\dfrac {-b\pm k} {2a}$

As given

$a, b, c$ are real

$\Rightarrow$ Roots are rational and distinct.

If the value of ' $b^2-4ac$ ' is less than zero, the quadratic equation $ax^2+bx+c=0$ will have

0%
Two Equal Real Roots.
0%
Two Distinct Real Roots.
0%
No Real Roots.
0%
None of the above.

Explanation

${ b }^{ 2 }-4ac<0$

$\therefore \triangle <0$

So no real roots.

The number of solution of

$z^2 + \bar{z} = 0$ is

0%
$5$
0%
$4$
0%
$2$
0%
$3$

Explanation

Let

$z=x+iy$ .

Now,

$z^2+\overline{z}=0$

or,

$x^2-y^2+2ixy+(x-iy)=0$

or,

$(x^2-y^2+x)+i(2xy-y)=0$

Now comparing the real and imaginary part both sides we get,

$x^2-y^2+x=0$ .....(1) and

$2xy-y=0$ .....(2).

From (2) we get,

$x=\dfrac{1}{2}$ or

$y=0$ .

Now

$x=\dfrac{1}{2}$ gives from (1) we get,

$y=\pm \dfrac{\sqrt{3}}{2}$ .

And

$y=0$ gives from (1) we get,

$x=0, 1$ .

So the solution s are

$(0,0), (1,0), \left(\dfrac{1}{2},\pm \dfrac{\sqrt{3}}{2}\right)$ .

So we have

$4$ solutions.

$arg\left( -\cfrac { 3 }{ 2 } \right)$ equals

0%
$\cfrac{\pi}{2}$
0%
$-\cfrac{\pi}{2}$
0%
$0$
0%
$\pi$

Explanation

Let

$z = r(cos\theta +isin \theta )$

and

$r>=0$

$\therefore arg (\dfrac{-3}{2}) = \pi$

Where

$r = \frac{3}{2} \ and \ \theta = \pi$

Find the which of the complex number has greatest modulus.

0%
$7-5i$
0%
$\sqrt{3}+\sqrt{2}i$
0%
$-8+15i$
0%
$-3(1-i)$

Explanation

$|7-5i| = \sqrt {54 }$

$|\sqrt {3} +i \sqrt 2| = \sqrt 5$

$| -8+15i| = \sqrt {289}$

$|-3+3i| = 3\sqrt 2$

Thus option C has the greatest modulus.

The roots of the equation

$x^{2}+2\sqrt {3}x+3=0$ are

0%
real and unequal
0%
rational and equal
0%
irrational and equal
0%
irrational and unequal

Explanation

Calculating Discriminant of this quadratic,

$D=(2\sqrt3)^2-4.(3).(1)$

$\implies D=12-12=0$

Hence roots are real and equal.

For any complex number

$z$ the minimum value of

$|z|+|z-2013i|$ is...

0%
$2010$
0%
$2011$
0%
$2013$
0%
$2012$

Explanation

Q:- minimum value of

$\left | z \right |+\left | z-2013i \right |$

Solution For ZEG

$\Rightarrow 2013 \leqslant \left | z \right |+\left | z-2013i \right |$

$\Rightarrow$ minimum value of

$\left | z \right |+\left | z-2013i \right |$ is

2013

which is attained when

$z = i$

so, c is correct option.

$x^{2}+y^{2} \leqslant 1$ &

$y^2 \leqslant 1-x .$

1120336_1140024_ans_b1d27da3ec984cf389d2244ff874fc8e.jpg

Which of the following is true

0%
$(3 + \sqrt{-5})(3 - \sqrt{-5}) = 14$
0%
$(-2 + \sqrt{-3})(-3 + 2\sqrt{-3}) = -7\sqrt{3}i$
0%
$(2 + 3i)^2 = (-5 + 12i)$
0%
$(\sqrt{5} - 7i)^2 = -44 - 14\sqrt{5}i$

Explanation

(a)

$(3+\sqrt{-5})(3-\sqrt{-5})=14$

L.H.S

$= (3+\sqrt{-5})(3-\sqrt{-5})$

$=((3)^2-(\sqrt{-5})^2) ((a-b)(a+b)=a^2-b^2)$

$=9+5$

$=14 = R.H.S$

(b)

$(-3+\sqrt{-3})(-3+2\sqrt{-3})=-7\sqrt{3i}$

L.H.S

$=(-2+\sqrt{-3}) (-3+2\sqrt{-3})$

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

$=4-4\sqrt{3i}-3\sqrt{3i}-6=-7\sqrt{3i}=R.H.S$

(c)

$(2+3i)^2=(-5+12i)$

L.H.S

$= (2+3i)^2$

$= 4+9i^2+12i$

$=4-9+12i= -5+12i=R.H.S$

(d)

$(\sqrt{5}-7i)^2=-44- 14\sqrt{5}i$

L.H.S

$= (\sqrt{5}-7i)^2$

$=5-49-14\sqrt{5}i$

$= - 44- 14\sqrt{5}i = R.H.S$

$\therefore (a), (b), (c), (d)$ all are correct

Argument and modulus of

$\left[\dfrac {1+i}{1-i}\right]^{2013}$ are respectively ____

0%
$\dfrac {-\pi}{2}$ and $1$
0%
$\dfrac {\pi}{2}$ and $\sqrt {2}$
0%
$0$ and $\sqrt {2}$
0%
$\dfrac {\pi}{2}$ and $1$

Explanation

Let

$P={ [\cfrac { 1+i }{ 1-i } ] }^{ 2013 }={ [\cfrac { \sqrt { 2 } { e }^{ \cfrac { i\pi }{ 4 } } }{ \sqrt { 2 } { e }^{ \cfrac { -i\pi }{ 4 } } } ] }^{ 2013 }=({ e }^{ \cfrac { i2013\pi }{ 2 } })$

$P=\cos { (2013\cfrac { \pi }{ 2 } ) } +i\sin { (2013\cfrac { \pi }{ 2 } ) } =i\sin { (-\cfrac { \pi }{ 2 } ) }$

So, argument

$=-\cfrac { \pi }{ 2 }$

and modulus

$=1$

$z_1=\sqrt { 3 } -i,z_2=1+i\sqrt { 3 } ,$ then amp

$(z_1+z_2)=$

0%
$\dfrac { \pi }{ 12 }$
0%
$\dfrac { \pi }{ 15 }$
0%
$\dfrac { \pi }{ 6 }$
0%
$\dfrac { \pi }{ 4 }$

Explanation

$z_1=\sqrt { 3 } -i,z_2=1+i\sqrt { 3 }$

$$ z_1+z_2=(\sqrt { 3 }+1)+i(\sqrt{3}-1) $$

$z_1+z_2=(\sqrt { 3 }+1)+i(\sqrt{3}-1)$

$\left| z_1+z_2 \right|= \sqrt {(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}=2\sqrt{2}$

If the roots of

$2x^2+3x+p=0$ be equal, then the value of p is :

0%
$\dfrac{9}{8}$
0%
$\dfrac{6}{5}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{5}{4}$

Explanation

$D=(3^2-4\times 2\times p)=9-8p$

Since, roots are equal

$D=0$

$9-8p=0$

$p=\dfrac{9}{8}$

$2^{x^{2}} : 2^{2x}=8^{k}:1$ , then equation has only one solution if

0%
$k > \dfrac{1}{3}$
0%
$k = \dfrac{1}{3}$
0%
$k < \dfrac{-1}{3}$
0%
$k = \dfrac{-1}{3}$

Explanation

$2^{x^{2}}:2^{2^{x}}$

$\Rightarrow \frac{2^{x^{2}}}{2^{2^{x}}}=\frac{(2)^{3k}}{1}$

$\Rightarrow (2)^{x^{2}-2x}=(2)^{3k}$

Comparing powers we get

$\Rightarrow x^{2}-2x=3k$

$\Rightarrow x^{2}-2x\div 3k=0$

equation has only 1 solution

it has equal roots

D=0

$b^{2}=4ac$

$4=4\times 3k$

$[k=-\frac{1}{3}]$

1211739_1362990_ans_ef518b7cfefe4efc93fcba6b452b4559.jpg

$z_1=3+4i\\z_2=4-5i$ Then find

$z_1+z_2$

0%
7-i
0%
7+i
0%
7+9i
0%
None of these

Explanation

$z_1=3+4i\\z_2=4-5i\\z_1+z_2=3+4i+4-5i\\7-i$

Find the least positive value of n, if

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

0%
1
0%
2
0%
3
0%
4

Explanation

We have,

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

Therefore,

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

$i^n=1$

$\implies$ n is a multiple of 4.

The smallest positive value of n is 4.

The complex numbers

$z_1=8+9i, z_2=4-6i$ then

$z_1-z_2$

0%
$4+15i$
0%
$4-3i$
0%
$12+3i$
0%
$12-15i$

Explanation

$z_1=8+9i\\z_2=4-6i\\z_1-z_2=8+9i-4+6i=4+15i$

$z$ is a complex number such that

$|z|=1$ , then

$\left|\dfrac 1{\bar z}\right|$ is

0%
$0$
0%
$-1$
0%
$\sqrt{2}$
0%
$1$

Explanation

Let

$z=x+iy\\|z|=\sqrt{x^2+y^2}=1\\x^2+y^2=1\\\dfrac 1z=\dfrac1{x+iy}\\\dfrac{x-iy}{\sqrt{x^2+y^2}}=x-iy\\\left|\dfrac 1z\right|=\sqrt{x^2+y^2}=1$

$z_1=3+4i,z_2=2-i$ find

$z_2-z_1$

0%
-1-5i
0%
2-5i
0%
1+5i
0%
1-5i

Explanation

Given

$z_1=3+4i\\z_2=2-i\\z_2-z_1=2-i-3-4i=-1-5i$

$\alpha \epsilon \left( -1,1 \right)$ then roots of the quadratic equation

$\left( a-1 \right) { x }^{ 2 }+ax+\sqrt { 1-{ a }^{ 2 } } =0$ are

0%
real
0%
imaginary
0%
both equal
0%
none of these

Explanation

Given,

$\left(a-1\right)x^2+ax+\sqrt{1-a^2}=0$

$x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-a\pm \sqrt{a^2-4\left(a-1\right)\sqrt{1-a^2}}}{2\left(a-1\right)};\quad \:a\ne \:1$

Hence the roots of quadratic equation.

$z_1=4+i,z_2=4-i$ find

$z_1z_2$

0%
$17$
0%
$16$
0%
$17-i$
0%
$16i$

Explanation

$z_1=4+i\\z_2=4-i\\z_1z_2=(4+i)(4-i)\\16-i^2=16+1=17$

$z_1=9+8i\ \ \ |z|=$

0%
$\sqrt {145}$
0%
$\sqrt {163}$
0%
$\sqrt {117}$
0%
$\sqrt {137}$

Explanation

$z=9+8i\\|z|=\sqrt {a^2+b^2}\\\\|z|=\sqrt {9^2+8^2} =\sqrt {81+64}=\sqrt {145}$

$(x+iy)(2-3i)=4+i\left ( \dfrac{1}{2} \right )$ then

$x + y =$

0%
$\dfrac{3}{2}$
0%
$\dfrac{1}{2}$
0%
$0$
0%
$\dfrac{2}{3}$

Explanation

$(x+iy)(2-3i)=4+i\dfrac{1}{2}$

$2x+2iy-3ix+3y=4+\dfrac{i}{2}$

$2x+3y+i(2y-3x)=4+\dfrac{i}{2}$
Hence

$2x+3y=4$ ................................(i)
And

$2y-3x=\dfrac{1}{2}$ .............................(ii)

Solving both the equations give us

$y=1$

$x=\dfrac{1}{2}$

Hence

$x+y$

$=1+\dfrac{1}{2}$

$=\dfrac{3}{2}$ .

$z_1$ and

$z_2$ are two complex numbers, then

$Re(z_1z_2)$ is:

0%
$Re(z_1)Re(z_2)$
0%
$Re(z_1).Re(z_2)-Im(z_1).Im(z_2)$
0%
$Im(z_1).Re(z_2)$
0%
$Re(z_1).Im(z_2)$

Explanation

$z_{1}=a_{1}+ib_{1}; z_{2}=a_{2}+ib_{2}$

$z_{1}z_{2}=(a_{1}+ib_{1})(a_{2}+ib_{2})$

$=a_{1}a_{2}+(a_{1}b_{2})i+(a_{2}b_{1})i-b_{1}b_{2}$

$=a_{1}a_{2}-b_{1}b_{2}+(a_{1}b_{2}+a_{2}b_{1})i$

$\therefore Re(z_{1}z_{2})=a_{1}a_{2}-b_{1}b_{2}$

$=Re(z_{1})\cdot Re(z_{2})-Im(z_{1})\cdot Im(z_{2})$

$\left| \dfrac { (3+i)(2-i) }{ 1+i } \right|=$

0%
$\sqrt{5}$
0%
$5\sqrt{2}$
0%
$\sqrt{10}$
0%
$5$

Explanation

$= | \dfrac{(3+i)(2-i)}{1+i}|$

$=|\dfrac{6-3i+2i+1}{1+i}|$

$=|\dfrac{7-i}{1+i}|$

$=|\dfrac{(7-i)(1-i)}{2}|$

$=|\dfrac{7-7i-i-1}{2}|$

$=|\dfrac{6-8i}{2}|$

$=|3-4i|$

$=\sqrt{3^{2}+4^{2}}$

$=\sqrt{25}$

$=5$

The real part of

$\left ( \dfrac{1+i}{3-i} \right )^2=$

0%
$1$
0%
$16$
0%
$16\omega ^2$
0%
$\displaystyle \frac{-3}{25}$

Explanation

$\displaystyle \frac{1+i}{3-i}=\frac{(1+i)(3+i)}{(3-i)(3+i)}=\frac{3-1+4i}{3^{2}+(1)^{2}}=\frac{2+4i}{10}=\frac{1+2i}{5}$

$\displaystyle \therefore \left ( \frac{1+2i}{5} \right )^{2}$

$\displaystyle =\frac{1^{2}-2^{2}+4i}{25}$

$\displaystyle =\frac{-3+4i}{25}$

$\displaystyle \therefore$ Real part

$=\dfrac{-3}{25}$

$z=-1+3i$ then

$z^2+2z+10=$

0%
$0$
0%
$1$
0%
$–1$
0%
$2$

Explanation

$z^{2}+2z+10$

$=(z+1)^{2}+10-1$

$=(z+1)^{2}+9$

$=(-1+3i+1)^{2}+9$

$=(3i)^{2}+9$

$=-9+9$

$=0$

Hence,

$z^{2}+2z+10=0$ where

$z=-1+3i$ .

In the argand diagram, the complex number z is in the fourth quadrant, then

$\overline{z}$ ,

$-z$ ,

$\overline{-z}$ are respectively are in quardrants

0%
1,3,2
0%
1,2,3
0%
3,2,1
0%
2,1,3

Explanation

Let

$z = x +iy$
As

$z$ is in the fourth quadrant,

$x >0$ and

$y < 0$

$\overline{z} = x - iy$
In this case, the real part would be positive and the imaginary part would be positive. Hence, this would lie in the first quadrant.

$-z = -x - iy$
In this case, the real part would be negative and the imaginary part would be positive. Hence this would lie in the second quadrant.

$\overline{-z} = -x+iy$
In this case, the real part would be negative and the imaginary part would also be negative. Hence, this would lie in the third quadrant.

The value of

$1+(1+i)+(1+i)^2+(1+i)^3=$

0%
$0$
0%
$5i$
0%
$4i$
0%
$3i$

Explanation

The given series is a G.P with a common ratio of

$(1+i)$ .
Hence

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

$=\dfrac{1-(1+i)^{4}}{1-(1+i)}$

Now

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

$=\sqrt{2}(e^{i\frac{\pi}{4}})$
Hence

$\dfrac{1-(1+i)^{4}}{-i}$

$=\dfrac{1-(\sqrt{2}(e^{i\frac{\pi}{4}}))^{4}}{-i}$

$=\dfrac{1-4(e^{i\pi})}{-i}$

$=\dfrac{1-4(-1)}{-i}$

$=\dfrac{5}{-i}$

$=5i$

$\left ( 5+3i \right )(x+iy)=3-4i$ then

$34x =$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$(5+3i)(x+iy)=3-4i$
L.H.S.

$=(5x-3y)+(5y+3x)i=3-4i$

$5x-3y=3$

$5y+3x=-4$
Solving the equations simultaneously,

$x=\dfrac{3}{34} ; y=\dfrac{-29}{34}$

$\therefore 34x=34\times \dfrac{3}{34}$

$=3$

$z_1$ ,

$z_2$ are the complex numbers such that

$|z_1+z_2|=|z_1|+|z_2|$ then arg

$z_1 -$ arg

$z_2$ is

0%
$-\pi$
0%
$\dfrac{-\pi }{2}$
0%
0
0%
$\dfrac{\pi }{2}$

Explanation

Let

$z_{1}=x_{1}+iy_{1}$

$z_{2}=x_{2}+iy_{2}$

$z_{1}+z_{2}=(x_{1}+x_{2})+i(y_{1}+y_{2})$ ...(i)

Now

$|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$

$\sqrt{(x_{1}+x_{2})^{2}+(y_{1}+y_{2})^{2}}=\sqrt{x_{1}^{2}+y_{1}^{2}}+\sqrt{x_{2}^{2}+y_{2}^{2}}$

Squaring both sides give us

$(x_{1}+x_{2})^{2}+(y_{1}+y_{2})^{2}=(x_{1})^{2}+(x_{2})^{2}+(y_{1})^{2}+(y_{2})^{2}+2\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}$

$(x_{1})^{2}+(x_{2})^{2}+(y_{1})^{2}+(y_{2})^{2}+2x_{1}x_{2}+2y_{1}y_{2}=(x_{1})^{2}+(x_{2})^{2}+(y_{1})^{2}+(y_{2})^{2}+2\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}$

$2x_{1}x_{2}+2y_{1}y_{2}=2\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}$

$(x_{1}x_{2}+y_{1}y_{2})^{2}=x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}$

$x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+2x_{1}x_{2}y_{1}y_{2}=x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}$

$2x_{1}x_{2}y_{1}y_{2}=x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}$
Or

$x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}-2x_{1}x_{2}y_{1}y_{2}=0$
Or

$(x_{1}y_{2}-x_{2}y_{1})^{2}=0$
Or

$x_{1}y_{2}-x_{2}y_{1}=0$

$x_{1}y_{2}=x_{2}y_{1}$
Or

$\dfrac{x_{1}}{y_{1}}=\dfrac{x_{2}}{y_{2}}$

Hence

$z_{1}$ and

$z_{2}$ are colinear.

The simplified value of

$\displaystyle \frac{1-i}{1+i}$ is:

0%
$i$
0%
$-i$
0%
$1$
0%
$-2i$

Explanation

Let

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

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

$=\dfrac{(1-i)^{2}}{1^{2}-(i)^{2}} \quad \dots (a^2-b^2=(a+b)(a-b))$

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

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

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

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

$=-i$

The minimum value of

$|\mathrm{z}|+|\mathrm{z}-1|+|\mathrm{z}-2|$ is

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

Let

$z = x+iy$

$f(z) = |z| + |z-1| +|z-2|$

$f(x,y) = \sqrt{x^2 +y^2} +\sqrt { (x-1)^2+ y^2} +\sqrt {(x-2)^2 +y^2}$

For minimum value,

$y= 0$

$f(x) = |x| + |x-1| +|x-2|$ for all

$x$

The function can be expressed as follows

$f(x) = 3x -3$ ,

$x \geq 2$

$f(x) = x+1,$

$1\leq x<2$

$f(x) = 3- x$ ,

$0\leq x\leq 1$

$f(x) = -3x +3$ ,

$x <0$

Hence,

$f(min) = f(1) = 2$

$\alpha$ and

$\beta$ are real then

$\left| \dfrac { \alpha +i\beta }{ \beta +i\alpha } \right|=$

0%
Lies betwen 0 and 1
0%
= 1
0%
>1
0%
2

Explanation

We know that

$|\dfrac{z_{1}}{z_{2}}|$

$=\dfrac{|z_{1}|}{|z_{2}|}$
Hence

$|\dfrac{\alpha+i\beta}{\beta+i\alpha}|$

$=\dfrac{|\alpha+i\beta|}{|\beta+i\alpha|}$

$=\dfrac{\sqrt{\alpha^{2}+\beta^{2}}}{\sqrt{\beta^{2}+\alpha^{2}}}$

$=1$ .

$m_1$ ,

$m_2$ ,

$m_3$ and

$m_4$ respectively denote the moduli of the complex numbers

$1 + 4i, 3 + i, 1 – i \ and\ 2 – 3i$ then the correct order among the following is :

0%
$m_1$ < $m_2$ < $m_3$ < $m_4$
0%
$m_4$ < $m_3$ < $m_2$ < $m_1$
0%
$m_3$ < $m_2$ < $m_4$ < $m_1$
0%
$m_3$ < $m_1$ < $m_2$ < $m_4$

Explanation

$m_{1}=|1+4i|=\sqrt{1+16}=\sqrt{17}$ .

$m_{2}=|3+i|=\sqrt{9+1}=\sqrt{10}$

$m_{3}=|1-i|=\sqrt{1+1}=\sqrt{2}$

$m_{4}=|2-3i|=\sqrt{4+9}=\sqrt{13}$
Hence

$m_{1}>m_{4}>m_{2}>m_{3}$

The principal argument of

$z=-3+3i$ is:

0%
$\dfrac{\pi }{4}$
0%
$-\dfrac{\pi }{4}$
0%
$\dfrac{3\pi }{4}$
0%
$-\dfrac{3\pi }{4}$

Explanation

Given,

$z=-3+3i$

$\Rightarrow z=3(-1+i)$

$=3\left (\dfrac{-1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}\right)$

$=3\sqrt2\left (\cos\left (\dfrac{3\pi}{4}\right)+i\sin\left (\dfrac{3\pi}{4}\right)\right)$

$=3\sqrt2e^{i\frac{3\pi}{4}}$

Hence, principal argument of

$z$ is

$\dfrac{3\pi}{4}$ .

Assertion (A): The principal amplitude of complex number

$x + ix$ is

$\cfrac{\pi }{4}$ .
Reason (R): The principal amplitude of a complex number

$x + iy$ is

$\cfrac{\pi }{4}$ if

$y = x$ .

0%
Both A and R are true and R is the correct explanation of A
0%
A is true R is false
0%
A is false, R is true
0%
Both A and R are false

Explanation

Assertion :

$arg(x+ix) = arg \cfrac{x}{x}$

$=\cfrac {\pi}{4}$

Reason : if

$y=x$ then

$arg(x+iy) = \cfrac {\pi}{4}$

$\therefore$ Both Assertion and reason are true and Reason is the correct explanation of Assertion.

Hence, option A is correct.

The area of the triangle formed by the three complex numbers

$1 + i$ ,

$i - 1$ ,

$2i$ in the Argand diagram is:

0%
$\dfrac{1}{2}$
0%
$1$
0%
$\sqrt{2}$
0%
$2$

Explanation

The three vertices are

$A (1,1) , B (1,-1)$ and

$C (0,2)$

Base

$AB = 2$

Height

$= 1 -0 = 1$

Area

$= \dfrac {1}{2} \times$ Base

$\times$ Height

Area

$= 1$

The modulus of

$(1 + i) (3 + 4i) =$

0%
$\sqrt{50}$
0%
$\sqrt{25}$
0%
$10$
0%
$10 \sqrt{2}$

Explanation

$z_{1},z_{2}$ being two complex numbers,

$|z_{1}.z_{2}|=|z_{1}|.|z_{2}|$

$\therefore |(1+i)(3+4i)|$

$=|1+i|\cdot|3+4i|$

$=\sqrt{1^{2}+1^{2}}\times \sqrt{3^{2}+4^{2}}$

$=\sqrt{2}\sqrt{25}$

$=\sqrt{50}$

$z_1$ ,

$z_2$ ,

$z_3$ are complex numbers such that

$\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =\left| \dfrac { 1 }{ { z }_{ 1 } } +\dfrac { 1 }{ { z }_{ 2 } } +\dfrac { 1 }{ { z }_{ 3 } } \right| =1$ , then

$|z_1+z_2+z_3|$ is:

0%
Equal to 1
0%
Less than 1
0%
Greater than 3
0%
Equal to 3

Explanation

Let

$z_1 = cis \theta_1$ ,

$z_2 = cis\theta_2$ and

$z_3 =cis\theta_3$

We have,

$\left | \dfrac{1}{z_1} + \dfrac{1}{z_2} +\dfrac{1}{z_3} \right| = 1$

$| cis(-\theta_1) + cis(-\theta_2) + cis (-\theta_3) | = 1$

Hence,

$| (\cos \theta_1 + \cos \theta_2 + \cos \theta_3 ) - i ( \sin \theta_1 + \sin \theta_2 + \sin \theta_3 ) | = 1$

Hence,

$| (\cos \theta_1 + \cos \theta_2 + \cos \theta_3 ) + i ( \sin \theta_1 + \sin \theta_2 + \sin \theta_3 ) | =1$

Hence,

$|z_1+ z_2+ z_3 | =1$

$(x+iy)(2+cos\theta+isin\theta)=3$ then

$x^{2}+y^{2}-4x+3$ is

0%
$0$
0%
$1$
0%
$3$
0%
$4$

Explanation

$(x+iy)(2+cos\theta+isin\theta)=3$

$2x+x(cos\theta+isin\theta)+i2y+iy(cos\theta+isin\theta)=3$

$2x+xcos\theta-ysin\theta+i(xsin\theta+2y+ycos\theta)=3$

Therefore, equating the real and imaginary parts of LHS and RHS gives us

$2x+xcos\theta-ysin\theta=3$

$xsin\theta+ycos\theta+2y=0$

$\Rightarrow xcos\theta-ysin\theta=3-2x$

Squaring both sides give us

$x^{2}cos^{2}\theta+y^{2}sin^{2}\theta-2xycos\theta.sin\theta=(2x-3)^{2}$ ...(i)

And

$xsin\theta+ycos\theta=-2y$

$x^{2}sin^{2}\theta+y^{2}cos^{2}\theta+2xysin\theta.cos\theta=4y^{2}$ ...(ii)

Adding both i and ii gives us

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

$x^{2}+y^{2}=4y^{2}+4x^{2}-12x+9$

$3x^{2}+3y^{2}-12x+9=0$

$x^{2}+y^{2}-4x+3=0$

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 2 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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