Complex Numbers and Quadratic Equations Test -4

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Complex Numbers and Quadratic Equations Test -4

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Weekly Quiz Competition

Question 1
1 / -0

Find the modulus and the principal value of the argument of the number $$1-i$$

A
$$\displaystyle \sqrt{2},\pi/4$$

B
$$\displaystyle \sqrt{2},-\pi/4$$

C
$$\displaystyle \sqrt{2},-\pi/3$$

D
$$\displaystyle \sqrt{2},3\pi/4$$

Solution

Let $$\displaystyle 1=r\cos \theta ,-1=r\sin \theta$$
Squaring and adding these relations, we get $$\displaystyle r^{2}\left ( \cos ^{2}\theta +\sin ^{2}\theta \right )=1^{2}+\left ( -1 \right )^{2}=2$$.
Since, $$cos^2\theta+sin^2\theta=1, \Rightarrow r=\pm\sqrt{2}$$
$$r$$ cannot be negative
Hence. $$r=\sqrt{2}$$
Then $$\displaystyle \cos \theta =\dfrac{1}{\sqrt{2}},\sin \theta =\dfrac{-1}{\sqrt{2}},$$
The value of $$\displaystyle \theta $$ between-$$\displaystyle \pi $$ and $$\displaystyle \pi $$ which satisfies these equations is $$\displaystyle -\left ( \pi /4 \right )$$ Thus $$\displaystyle \left | 1-i \right |=r=\sqrt{2}$$ and $$arg (1-i)=-\displaystyle \left ( \pi /4 \right ).$$

Ans: B
Question 2
1 / -0

Check whether $$2x^2 - 3x + 5 = 0$$ has real roots or no.

A
The equation has real roots.

B
The equation has no real roots.

C
Data insufficient

D
None of these

Solution

Here the quadratic equation is $$2x^2 - 3x + 5 = 0$$
Comparing it with $$ax^2+bx+c=0$$, we get
$$a=2, b=-3, c=5$$
Therefore, discriminant, $$D=b^2-4ac$$
$$(-3)^2-4\times 2\times 5$$
$$=9-40$$
$$=-31$$
Here, $$D<0$$
Therefore the equation has no real roots.
Question 3
1 / -0

If i = $$\sqrt {-1}, then 1 + i^2 + i^3 -i^6 + i^8 $$ is equal to -

A
2- i

B
1

C
-3

D
-1

Solution

Given, $$i= \sqrt {-1}$$
$$\therefore 1+i^{ 2 }+i^{ 3 }-i^{ 6 }+i^{ 8 }=1-1-i+1+1$$
$$=2-i$$
Question 4
1 / -0

Solve $$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i,$$

A
$$\displaystyle x= -1, y= 2.$$

B
$$\displaystyle x= 2, y= -1.$$

C
$$\displaystyle x= 2, y= 1.$$

D
$$\displaystyle x= 1, y= 2.$$

Solution

$$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i.$$
Equating real and imaginary parts, we get
$$\displaystyle x+y= 1$$ and $$\displaystyle -x+y= -3.$$
Adding both equations we get $$y=-1$$
Substituting this value of $$y$$ in any of the 2 equations, we get $$x=2$$
$$\therefore \displaystyle x= 2, y= -1.$$

Ans: B
Question 5
1 / -0

If $$z = \displaystyle \frac{(3 + 4i)(5 - 7 i)}{(7 + 5i)(4 - 3i)}$$ then $$|z| = ?$$

A
5

B
4

C
1

D
None of these

Solution

Given, $$z = \displaystyle \frac{(3 + 4i)(5 - 7 i)}{(7 + 5i)(4 - 3i)}$$
$$\therefore |z| = \displaystyle \frac{|(3 + 4i) ||(5 - 7i)|}{|(7 + 5i)||(4 - 3i)|}$$

$$= \displaystyle \frac{\sqrt{9 + 16} \sqrt{25 + 49}}{\sqrt{49 + 25} \sqrt{16 + 9}} = 1$$
Question 6
1 / -0

($$i^{10}+1) (i^9 + 1)(i^8 +1).......(i+1)$$ equal to

A
-1

B
1

C
0

D
i

Solution

$$(i^{10}+1) (i^9 + 1)(i^8 +1).......(i+1)$$
$$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(i^2+1)(i+1)$$
$$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(-1+1)(i+1)[\because i^2=-1]$$
$$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(0)(i+1) =0$$
Question 7
1 / -0

The simplest form of the expression $$\dfrac {10 - \sqrt {-12}}{1 - \sqrt {-27}} $$ is

A
$$-\dfrac {2}{7}$$

B
$$\dfrac {28}{3}$$

C
$$-\dfrac {2}{7} + i\sqrt {3}$$

D
$$1 + i \sqrt {3}$$

Solution

We need to find simplest form of $$\dfrac{10-\sqrt{-12}}{1-\sqrt{-27}}$$

$$=\dfrac{10-2i\sqrt{3}}{1-3i\sqrt{3}}$$

Taking conjugate, we get

$$=$$ $$\dfrac{10-2\sqrt{-3}}{1-3\sqrt{-3}}\times \dfrac{1+3i\sqrt{3}}{1+3i\sqrt{3}}$$

$$=$$ $$\dfrac{\frac{10-2\sqrt{-3}}{1-3\sqrt{-3}}}{1^2-(3i\sqrt{3})^2}$$

$$=$$ $$\dfrac{10+30i\sqrt{3}-2i\sqrt{3}-18i^2}{28}$$

$$=$$ $$\dfrac{10+28i\sqrt{3}+18}{28}$$

$$=$$ $$\dfrac{28(1+i\sqrt{3})}{28}$$

$$=$$ $$1+i\sqrt{3}$$
Question 8
1 / -0

If $$i^2$$ $$= -1$$, then find the odd one out of the following expressions.

A
$$-i^2$$

B
$$(-i)^2$$

C
$$i^4$$

D
$$(-i)^4$$

E
$$-i^6$$

Solution

$$i$$ is an imagiary number which has a value of $$i=\sqrt{-1}$$. Then
A. $$-i^2=-1*-1=1$$
B. $${(-i)}^2=i^2=-1$$
C. $$i^4={(i^2)}^2={(-1)}^2=1$$
D. $${(-i)}^4={(i^2)}^2={(-1)}^2=1$$
E. $$-i^6=-{(i^2)}^3=-1\times {(-1)}^3=-1\times (-1)=1$$

All are $$1$$ except option $$B$$ which is $$-1$$.
Question 9
1 / -0

When $$(3-2i)$$ is subtracted from $$(4 + 7i)$$, then the result is

A
$$1 + 5i$$

B
$$1 + 9i$$

C
$$7 + 5i$$

D
$$7 + 9i$$

Solution

We need to subtract $$(3-2i)$$ from $$(4+7i)$$
$$\therefore 4 + 7i - (3 - 2i)$$
$$=4 + 7i - 3 + 2i $$
$$= 1 + 9i$$
Question 10
1 / -0

If $$u = 3 - 5i$$ and $$v = -6 + i$$, then the value of $$(u+v)^2$$ is

A
$$-7 + 24i$$

B
$$9 + 8i$$

C
$$9 + 16i$$

D
$$25 - 24i$$

Solution

$$u=3-5i$$
$$v=-6+i$$
$$\therefore (u+v)^2=[3-5i+[-6+i]$$
                      $$=[-3-4i]^2$$
                      $$=[-3]^2+[-4i]^2+2*[-3]*[-4i]$$
                      $$=9+16i^2+24i$$
                      $$=9+16[-1]+24i$$
                      $$=-7+24i$$
Option A is correct.

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Complex Numbers and Quadratic Equations Test -4

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Complex Numbers and Quadratic Equations Test -4

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Question 1

$$\displaystyle \sqrt{2},\pi/4$$

$$\displaystyle \sqrt{2},-\pi/4$$

$$\displaystyle \sqrt{2},-\pi/3$$

$$\displaystyle \sqrt{2},3\pi/4$$

Solution

Question 2

The equation has real roots.

The equation has no real roots.

Data insufficient

None of these

Solution

Question 3

2- i

1

-3

-1

Solution

Question 4

$$\displaystyle x= -1, y= 2.$$

$$\displaystyle x= 2, y= -1.$$

$$\displaystyle x= 2, y= 1.$$

$$\displaystyle x= 1, y= 2.$$

Solution

Question 5

5

4

1

None of these

Solution

Question 6

-1

1

0

i

Solution

Question 7

$$-\dfrac {2}{7}$$

$$\dfrac {28}{3}$$

$$-\dfrac {2}{7} + i\sqrt {3}$$

$$1 + i \sqrt {3}$$

Solution

Question 8

$$-i^2$$

$$(-i)^2$$

$$i^4$$

$$(-i)^4$$

$$-i^6$$

Solution

Question 9

$$1 + 5i$$

$$1 + 9i$$

$$7 + 5i$$

$$7 + 9i$$

Solution

Question 10

$$-7 + 24i$$

$$9 + 8i$$

$$9 + 16i$$

$$25 - 24i$$

Solution

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