Complex Numbers and Quadratic Equations Test 39

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Complex Numbers and Quadratic Equations Test 39

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Weekly Quiz Competition

Question 1
1 / -0

If $$z_1, z_2$$ be two non zero complex numbers satisfying the equation $$\displaystyle \left | \frac{z_1 + z_2}{z_1 - z_2} \right | = 1$$ then $$\displaystyle \frac{z_1}{z_2} + \left ( \frac{z_1}{z_2} \right )$$ is

A
zero

B
1

C
purely imaginary

D
2

Solution

let $$\frac{z_{1}}{z_{2}} =x$$
given $$|\frac{x+1}{x-1}| = 1$$
$$\Rightarrow |x+1|=|x-1|$$
Let $$x=a+ib$$
We get $$|a+1+ib|=|a-1+ib|$$
$$\Rightarrow a=0$$
Therefore $$x=ib$$ and $$\overline { x } =-ib$$
Therefore $$x+\overline { x } =ib-ib=0$$
So the correct option is $$A$$
Question 2
1 / -0

Interpret the following equations geometrically on the Argand plane.
$$1 < |z - 2 - 3 i| < 4$$

A
Annular

B
Straight line

C
A point

D
Ring

Solution

We have $$1 < |z - 2 - 3i| < 4$$
Represents a circle with centre at $$(2, 3)$$ and radius $$r \in (1, 4)$$
$$ |z - 2 - 3 i| > 1$$ represents the region in the plane outside the circle $$|z - 2 - 3i| = 1$$
Similarly $$|z - 2 - 3i| < 4$$ represents the region inside the circle $$|z - 2 - 3i| = 4$$
$$\therefore$$ $$1 < |z - 2 - 3i| < 4$$ represents the annular space between the two circles.
Question 3
1 / -0

Find the range of real number $$\alpha$$ for which the equation $$z + \alpha |z - 1| + 2i = 0; z= x + iy$$ has a solution. Find the solution.

A
$$\displaystyle x = 5/2, y = - 2$$

B
$$\displaystyle x = -2, y = 5/2$$

C
$$\displaystyle x = -5/2, y = 2$$

D
$$\displaystyle x = 2, y = -5/2$$

Solution

$$x+iy+\alpha(\sqrt{(x-1)^{2}+y^{2}})+2i=0$$
$$x+i(y+2)=\alpha(\sqrt{(x-1)^{2}+y^{2}}$$
$$x^{2}-(y+2)^{2}+i2x(y+2)=\alpha^{2}((x-1)^{2}+y^{2})$$
Real part
$$x^{2}-\alpha(x-1)^{2}-\alpha(y^{2})=0$$ and
Imaginary part
$$2x(y+2)=0$$
$$x(y+2)=0$$
Either $$x=0 $$or $$y=-2$$ ...if the above equations have a solution.
Considering y=-2,
Hence
$$\alpha\epsilon[-\dfrac{\sqrt{5}}{2},\dfrac{\sqrt{5}}{2}]$$
Hence
$$x=\dfrac{5}{2}$$ if $$|\alpha|\neq1$$.
Question 4
1 / -0

Let $$\displaystyle z=1+i\:b=(a,b)$$ be any complex number, $$\displaystyle a,b,\epsilon R$$ and $$\displaystyle \sqrt{-1}=i.$$ Let $$\displaystyle z\neq 0+0i,arg z=\tan^{-1}\left (\frac{Im\:z}{Re\:z}\right)$$ where $$\displaystyle -\pi<arg z\leq \pi$$

$$\displaystyle arg(\bar{z})+arg(-z)=\left\{\begin{matrix}\pi, \; if\: arg (z)<0 & \\ -\pi, \; if\: arg (z)>0 & \end{matrix}\right.$$

Let $$z$$ & $$w$$ be non-zero complex numbers such that they have equal modulus values and $$\displaystyle arg z- arg \bar{w} =\pi,$$ then z equals

A
$$-w$$

B
$$w$$

C
$$\displaystyle -\bar{w}$$

D
$$\displaystyle \bar{w}$$

Solution

Let $$\overline{w}=cos\theta+isin\theta=e^{i\theta}$$

Therefore

$$arg(z)=\pi+\theta$$

Hence $$z=e^{i(\pi+\theta)}$$

$$=cos(\pi+\theta)+isin(\pi+\theta)$$

$$=-cos\theta-isin\theta$$

$$=-\overline{w}$$
Question 5
1 / -0

If z be a complex number satisfying$$\displaystyle\ z^{4}+z^{3}+2z^{2}+z+1=0$$ then $$\displaystyle\ |z|$$ is

A
$$\displaystyle\ \frac{1}{2}$$

B
$$\displaystyle\ \frac{3}{4}$$

C
$$\displaystyle\ 1$$

D
None of these

Solution

$${ z }^{ 4 }+{ z }^{ 3 }+2{ z }^{ 2 }+z+1=0$$

$$\left[ { z }^{ 4 }+2{ z }^{ 2 }+1 \right] +\left[ { z }^{ 3 }+z \right] =0$$

$$\left[ { z }^{ 2 }+1 \right] \left[ { z }^{ 2 }+1+z \right] =0$$

$${ z }^{ 2 }+1=0\Longrightarrow z=\pm i$$

$$\left| z \right| =1$$
or
$${ z }^{ 2 }+z+1=0$$

$$z=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } =\cfrac { -1\pm \sqrt { 3i } }{ 2 } $$

$$\left| z \right| =\sqrt { \cfrac { 1 }{ 4 } +\cfrac { 3 }{ 4 } } =1$$
$$\therefore \boxed { \left| z \right| =1 } $$
Question 6
1 / -0

If $$z = x+iy$$ and $$w = \dfrac{(1-iz)}{(z-i)}$$, then $$|w| = 1$$ implies that, in the complex plane

A
$$z$$ lies on the imaginary axis

B
$$z$$ lies on the real axis

C
$$z$$ lies on the unit circle

D
None of these

Solution

Given $$|w|$$$$=1$$

$$1=|\dfrac{1-iz}{z-i}|$$

$$\Rightarrow |z-i|=|1-iz|$$.....(i)

Now let $$z=x+iy$$
$$\Rightarrow iz=-y+ix$$

Put in (i)

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

$$\Rightarrow (y+1)^{2}-(y-1)^{2}=0$$

$$\Rightarrow (2y)(2)=0$$

$$\Rightarrow y=0$$.

This is the equation of the x-axis.
Question 7
1 / -0

$$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }=$$

A
$$-2$$

B
$$2$$

C
$$-1$$

D
$$1$$

Solution

We have $$\displaystyle \frac { \sqrt { 3 } +i }{ 2 } =\frac { i\sqrt { 3 } +{ i }^{ 2 } }{ 2i } $$ (Here we multiplied the numerator and denominator by $$(-i)$$ on the LHS)
We get: $$-i\left( \frac { -1+\sqrt { 3 } i }{ 2 } \right) =i\omega $$
Simliarly applying the same idea we get the steps below:
and $$\displaystyle \frac { i-\sqrt { 3 } }{ 2 } =\frac { { i }^{ 2 }-i\sqrt { 3 } }{ 2i } =-i\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) =-i{ \omega }^{ 2 }$$
Hence $$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }={ \left( -i\omega \right) }^{ 6 }+{ \left( -i{ \omega }^{ 2 } \right) }^{ 6 }\\ \\ ={ i }^{ 6 }\left( { \omega }^{ 6 }+{ \omega }^{ 12 } \right) =-1\left( 1+1 \right) =-2$$
Question 8
1 / -0

Let $$\displaystyle\ z_{1}= a+ib, z_{2}= p+iq$$ be two unimodular complex numbers such that $$\displaystyle\ Im(z_{1}z_{2})=1$$. If$$\displaystyle\ \omega_{1}= a+ip, \omega_{2}=b+iq$$ then

A
$$\displaystyle\ Re(\omega_{1}\omega_{2})=1$$

B
$$\displaystyle\ Im(\omega_{1}\omega_{2})=1$$

C
$$\displaystyle\ Rm(\omega_{1}\omega_{2})=0$$

D
$$\displaystyle\ Im(\omega_{1}\bar{\omega_{2}})=0$$

Solution

$$z_1 = a + ib$$ and $$z_2 = p + iq$$
Given that $$Im(z_1z_2) = 1$$, so $$aq + bp = 1$$
Now, $$\omega_1 = a + ip, \omega_2 = b + iq$$
$$\Rightarrow \omega_1\omega_2 = ab - pq + i(aq + bp)$$
$$\Rightarrow Im(\omega_1\omega_2) = 1$$
Question 9
1 / -0

The root of $$(x + a) (x + b) - 8k = (k - 2)^2$$ are real and equal, when $$a,b,c$$ $$\epsilon$$ R, then

A
$$a + b = 0$$

B
$$a =b$$

C
$$k = -3$$

D
$$k = 0$$

Solution

The equation:
$$(x + a) (x + b) - 8k = (k - 2)^2$$
$$x^2 + x (a + b) - 8k + ab = k^2 + 4 - 4k$$
$$x^2 + x (a + b) - k^2 - 4k - 4 + ab = 0 $$

The roots of the equation are real and equal.
Thus, discriminant = $$0$$
$$D = 0 $$
$$(a + b)^2- 4(1)(ab- k^2- 4k - 4)= 0$$
$$a^2 +b^2 + 2ab - 4ab + 4k^2 + 16 k + 16 = 0 $$
$$(a - b)^2 + 4(k +2)^2 = 0 $$
Thus, both $$(a -b) =0$$ and $$(k + 2) = 0$$
Hence, $$a = b$$ and $$k = -2$$
Question 10
1 / -0

If $$\displaystyle y^{2}< x$$ and $$\displaystyle x\in \left ( -\infty ,0 \right )$$ then $$y$$ must

A
be positive

B
be negative

C
have no real value

D
None of the above

Solution

Given $$ x $$ is a negative value
And $$ {y}^{2} < x $$, means it is also a negative value. Ideally, no real value has its square as negative.

Thus option $$C$$ is correct.

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Complex Numbers and Quadratic Equations Test 39

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Complex Numbers and Quadratic Equations Test 39

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SHARING IS CARING

Question 1

zero

1

purely imaginary

2

Solution

Question 2

Annular

Straight line

A point

Ring

Solution

Question 3

$$\displaystyle x = 5/2, y = - 2$$

$$\displaystyle x = -2, y = 5/2$$

$$\displaystyle x = -5/2, y = 2$$

$$\displaystyle x = 2, y = -5/2$$

Solution

Question 4

$$-w$$

$$w$$

$$\displaystyle -\bar{w}$$

$$\displaystyle \bar{w}$$

Solution

Question 5

$$\displaystyle\ \frac{1}{2}$$

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

$$\displaystyle\ 1$$

None of these

Solution

Question 6

$$z$$ lies on the imaginary axis

$$z$$ lies on the real axis

$$z$$ lies on the unit circle

None of these

Solution

Question 7

$$-2$$

$$2$$

$$-1$$

$$1$$

Solution

Question 8

$$\displaystyle\ Re(\omega_{1}\omega_{2})=1$$

$$\displaystyle\ Im(\omega_{1}\omega_{2})=1$$

$$\displaystyle\ Rm(\omega_{1}\omega_{2})=0$$

$$\displaystyle\ Im(\omega_{1}\bar{\omega_{2}})=0$$

Solution

Question 9

$$a + b = 0$$

$$a =b$$

$$k = -3$$

$$k = 0$$

Solution

Question 10

be positive

be negative

have no real value

None of the above

Solution

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