Complex Numbers and Quadratic Equations Test 41

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Complex Numbers and Quadratic Equations Test 41

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Weekly Quiz Competition

Question 1
1 / -0

Add and express in the form of a complex number $$a+bi$$
$$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$$

A
$$-4+9i$$

B
$$-5+9i$$

C
$$2+9i$$

D
$$-5+8i$$

Solution

Adding the given expression as follows:
$$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$$
$$=(2+3i)+(-4+5i)-(3-i)$$
$$=2+3i-4+5i-3+i$$
$$=-5+9i$$
Question 2
1 / -0

If $$\alpha $$ and $$\beta$$ are two different complex numbers with $$|\beta|=1$$, then $$\left | \dfrac{\beta -\alpha}{1-\bar{\alpha }\beta } \right |$$ is equal to.

A
$$0$$

B
$$1$$

C
$$\dfrac{1}{2}$$

D
$$-1$$

Solution

Let $$t = \left|\cfrac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right|$$
Multiply the numerator and denominator with $$\overline { \beta } $$
$$\Rightarrow t = \left|\cfrac { 1-\alpha \overline { \beta } }{ \overline { \beta } -\overline { \alpha } } \right| = \left|\cfrac { \overline { 1- \alpha \overline { \beta } } }{ \overline { \overline{\beta} -\overline{\alpha} } } \right| = \left|\cfrac { 1-\overline { \alpha } \beta }{ \beta -\alpha } \right|=\cfrac{1}{t}$$
$$\Rightarrow {t}^{2} = 1$$
$$\Rightarrow t = 1$$
Question 3
1 / -0

If $$z_ 1 = 2 \sqrt 2 (1 + i)$$ and $$z = 1 + i \sqrt 3$$, then $$z_1^2 z_2^3$$ is equal to

A
$$128 i$$

B
$$64 i$$

C
$$-64 i$$

D
$$- 128 i$$

E
$$256$$

Solution

Given, $$z_1=2\sqrt 2(1+i), z_2=1+i\sqrt 3$$
Therefore, $$z_1^2=8(1+i)^2$$
$$=8(1+2i-1)$$
$$=16i$$
and $$z_2^3=(1+i\sqrt 3)^3$$
$$=1+3\sqrt3i-9-3\sqrt3i$$
$$=-8$$
Thus, $$z_1^2z_2^3=-8(16i)$$ $$=-128i$$
Question 4
1 / -0

Evaluate in standard form: $$\dfrac {(2-3i)}{(2-2i)}$$, where $${i}^{2}=-1$$.

A
$$\dfrac {5}{4}-\dfrac {i}{4}$$

B
$$\dfrac {5}{4}+\dfrac {i}{4}$$

C
$$-\dfrac {5}{4}-\dfrac {i}{4}$$

D
$$-\dfrac {5}{4}+\dfrac {i}{4}$$

Solution

Consider $$\dfrac {(2-3i)}{(2-2i)}$$
Rationalise and solve considering that $$i^2=-1$$ as shown below:
$$\Rightarrow \dfrac { 2-3i }{ 2-2i } \times \dfrac { 2+2i }{ 2+2i } =\dfrac { 4+4i-6i-{ 6i }^{ 2 } }{ 4-{ 4i }^{ 2 } }$$
$$ =\dfrac { 4-2i+6 }{ 4+4 } $$
$$=\dfrac { -2i+10 }{ 8 } $$
$$=\dfrac { -i }{ 4 } +\dfrac { 5 }{ 4 }$$
Question 5
1 / -0

Express in the form of a complex number $$a+bi$$
$$-(7-i)(-4-2i)(2-i)$$

A
$$60-10i$$

B
$$70-10i$$

C
$$28-14i$$

D
$$70-15i$$

Solution

Multiplying the given expression as follows:
$$-[(7-i)(-4-2i)(2-i)]$$
$$=-[(-28-14i+4i+2i^2)(2-i)]$$
$$=-[(-28-10i-2)(2-i)]$$
$$=-[(-30-10i)(2-i)]$$
$$=-[-60+30i-20i+10i^2]$$
$$=-[-60-10+10i]$$
$$=-[-70+10i]$$
$$=70-10i$$
Question 6
1 / -0

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

A
Is strictly greater than $$\dfrac {5}{2}$$

B
Is strictly greater than $$\dfrac 32$$ but less than $$\dfrac {5}{2}$$

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

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

Solution

$$\left| z \right| \ge 2$$ is the region on or out side the circle whose center is $$\left( 0,0 \right)$$ and radius is $$2$$.

Minimum $$\left| z+\dfrac { 1 }{ 2 } \right|$$ is distance of $$z$$ which lie on circle $$\left| z \right| =2$$ from $$\left( -\dfrac { 1 }{ 2 } ,0 \right)$$

Thus minimum $$\left| z+\dfrac { 1 }{ 2 } \right| =$$ Distance between $$\left( -\dfrac { 1 }{ 2 } ,0 \right)$$ to $$\left( 0,0 \right)$$

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

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

Apply radical rule $$\sqrt [ n ]{ a^{ n } } =a,$$ assuming $$a\ge \: 0$$

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

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

$$=\dfrac { 2\cdot \: 2 }{ 2 } -\dfrac { 1 }{ 2 } $$

$$=\dfrac { 2\cdot \: 2-1 }{ 2 } $$

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

Hence, option B is correct.
Question 7
1 / -0

What is $${ i }^{ 1000 }+{ i }^{ 1001 }+{ i }^{ 1002 }+{ i }^{ 1003 }$$ equal to (where $$i=\sqrt { -1 } $$)?

A
$$0$$

B
$$i$$

C
$$-i$$

D
$$1$$

Solution

$$i^{1000} + i^{100} + i^{1002} + i^{1003}$$
$$i^{2} = -1, i^{3} = -1, i^{4} = 1$$
$$i^{1000} = (i^{4})^{250} = 1\ i^{1002} = i^{2} - i^{1000} = -1$$
$$i^{1001} = i\cdot i^{1000} = i\ i^{1003} = i^{3}\cdot i^{1000} = -i$$
$$\therefore i^{1000} + i^{1001} + i^{1002} + i^{1003} = 1 - 1 + i - i = 0$$
Question 8
1 / -0

A complex number z is said to be unimodular if $$|z| =1. $$. Suppose $$z_1$$ and $$z_2$$ are complex numbers such that $$\frac{z_1-2z_2}{2-z_1\overline {z}_2}$$ is unimolecolar and $$z_2$$ is not unimodular. Then the point $$z_1$$ lies on a:

A
straight line parallel to y-axis

B
circle of radius 2.

C
Circle of radius $$\sqrt2$$.

D
Staright line parallel to x-axis.

Solution
Question 9
1 / -0

Let $$z_1$$ = 18 + 83i, $$z_2$$ = 18 + 39i, ana $$z_3 $$= 78 + 99i. where i = $$\sqrt-1$$. Let z be a unique comlpex number with the properties that $$\dfrac{z_3 - z_1}{z_2 - z_1}$$ $$\cdot$$ $$\dfrac{z - z_2}{z - z_3}$$ is a real number and the imaginary part of the size z is the greatest possible.

A
$$Re (z) = 56$$

B
$$Re (z) = 61$$

C
$$Re (z) = 54$$

D
$$Re (z) = 59$$

Solution

Using the property that

if $$\dfrac{z_3-z_1}{z_2-z_1}.\dfrac{z-z_2}{z-z_3}$$is a real number then the four points are concyclic i.e. are present on a circle

since it is told the imaginary part is maximum so,it the top point of the circle
so we take the perpendicular bisector of any two line and take its intersection.
As shown in the figure,

Equation of perpendicular bisector of $$z_1z_2$$
$$y=61$$

Equation of perpendicular bisector of $$z_3z_2$$
$$y+x=117$$

Solving the equation we get,
$$x=56$$

So, $$Re(z)=56$$
Question 10
1 / -0

If $$z=\sqrt{20i-21}+\sqrt{21+20i}$$, then the principal value of arg 'z' can be

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

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

C
$$-\dfrac{3\pi}{4}$$

D
$$-\dfrac{\pi}{4}$$

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Re-Attempt Weekly Quiz Competition

Complex Numbers and Quadratic Equations Test 41

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Complex Numbers and Quadratic Equations Test 41

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SHARING IS CARING

Question 1

$$-4+9i$$

$$-5+9i$$

$$2+9i$$

$$-5+8i$$

Solution

Question 2

$$0$$

$$1$$

$$\dfrac{1}{2}$$

$$-1$$

Solution

Question 3

$$128 i$$

$$64 i$$

$$-64 i$$

$$- 128 i$$

$$256$$

Solution

Question 4

$$\dfrac {5}{4}-\dfrac {i}{4}$$

$$\dfrac {5}{4}+\dfrac {i}{4}$$

$$-\dfrac {5}{4}-\dfrac {i}{4}$$

$$-\dfrac {5}{4}+\dfrac {i}{4}$$

Solution

Question 5

$$60-10i$$

$$70-10i$$

$$28-14i$$

$$70-15i$$

Solution

Question 6

Is strictly greater than $$\dfrac {5}{2}$$

Is strictly greater than $$\dfrac 32$$ but less than $$\dfrac {5}{2}$$

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

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

Solution

Question 7

$$0$$

$$i$$

$$-i$$

$$1$$

Solution

Question 8

straight line parallel to y-axis

circle of radius 2.

Circle of radius $$\sqrt2$$.

Staright line parallel to x-axis.

Solution

Question 9

$$Re (z) = 56$$

$$Re (z) = 61$$

$$Re (z) = 54$$

$$Re (z) = 59$$

Solution

Question 10

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

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

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

$$-\dfrac{\pi}{4}$$

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