Imaginary Number i and its properties MCQ Quiz in বাংলা - Objective Question with Answer for Imaginary Number i and its properties - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Power of iota (i)

Depending upon the power of “i”, it can take the following values;

i4k+1 = i, i4k+2 = -1,  i4k+3 =  -i,  i4k = 1

Calculation:

We have to find the value of 1 + i2 + i4 + i6 +....+ i100

⇒ 1 + i⁽4 × 0 + 2) + i⁽4 × 1) + i(4 × 1 + 2) +....+ i(4 × 25)

⇒ 1 - 1 + 1 - 1 + ..... + 1 

⇒ 1 + 0

⇒ 1

∴ The value of 1 + i2 + i4 + i6 +....+ i100 is 1.

Mistake Points

Note that, there are 101 terms, starting from 1 (i.e. i⁰) to i¹⁰⁰ . 

i² = -1 and i¹⁰⁰ = 1. So i² to i¹⁰⁰ digit will cancel out, but the first digit (i.e. 1) will left.

Concept:

\(\rm i^2 = -1\)

\(\rm i^3 = -i\)

\(\rm i^4 = 1\)

Calculation:

\(\rm (\frac {1 + i} {1 - i})^x = 1\)

Take rationalization 

⇒ \(\rm (\frac {1 + i} {1 - i} \times \frac {1 + i} {1 + i})^x = 1\)

⇒ \(\rm [\frac {(1 + i)^2} {(1)^2 - (i)^2}]^x = 1\)                     \(\rm [ \because (a + b)^2 = a^2 + b^2 +2ab]\)

⇒ \(\rm[ \frac {(1 + i^2 + 2i)} {1 + 1}]^x = 1 \)                    \(\rm [\because i^2 = -1]\)

⇒ \(\rm (\frac {2i} {2})^x = 1 \)

⇒ \(\rm (i)^x = 1\)

x = 4n, where n is any positive integer

Concept:

Power of i:

1) i = \(\sqrt{-1}\)

2) i2 = -1

3) i3 = i × i2 = -i

4) i4 = (i2)2 = (-1)2 = 1

5) i4n = 1

Calculation:

To Find: Value of i12 + i13 + i14 + i15

As we know, i2 = -1

So, i¹² = (i4)³ = 1

i¹³ = (i4)³ × i = 1 × i = i

i¹⁴ = (i4)³ × i² = 1 × (-1) = -1

i¹⁵ = (i⁴)³ × i³ = 1 × (-i) = -i

Now, 

i12 + i13 + i14 + i15  = 1 + i - 1 - i = 0

Concept:

 i4k+1  = i = √-1  

 i4k+2  = i² = -1  

 i4k+3  = i³ = -i

 i4k = i⁴ = 1

where k is integer.

Calculation:

We have to find the value of (-√-1)4n + 3 + (i41 + i-257)9

⇒ (- i)4n + 3 + (i41 + i-257)9

⇒ (-1)^{4n + 3} × (i)4n + 3 + (i4×10 +1 + i4(-65)+3)9

⇒ -1 × (-i) + [(i²)²⁰ ×i¹ +  i4(-65) × i³]⁹ 

⇒ -1 × (-i) + (i + i³)⁹

⇒ i + (i - i)⁹

⇒ i + 0 = i

∴ The correct answer is option (3).

Concept:

Let z = a + ib, where a and b are real numbers. Then, 

|z| = √(a2 + b2)

|z̅| = |conjugate of z|

De Moivre’s Formula

For any real number x, we have (cos x + i sin x)ⁿ = cos(nx) + i sin(nx)

or (e^iθ)ⁿ = eiθn where n is a positive integer and “ i “ is the imaginary part, and i = √(-1).  

Calculation:

Given:

(1 + i√3)9 = a + ib

Multiplying and dividing 2 on right hand side,

⇒ (1 + i√3)9 = 2⁹ \(\displaystyle \left(\frac{1}{2}+i\frac{\sqrt3}{2}\right)^9 \)

⇒ (1 + i√3)9 = \(\displaystyle 2^9\left(cos\ \frac{\pi}{3}+i\ sin\ \frac{\pi}{3}\right)^9\)

Apply de moivre’s theorem

⇒ (1 + i√3)9 = \(\displaystyle 2^9\left(cos\ \frac{9\pi}{3}+i\ sin\ \frac{9\pi}{3}\right) \)

⇒ (1 + i√3)9 = \(\displaystyle 2^9\left(cos\ 3\pi + i\ sin\ 3\pi\right)\)

⇒ (1 + i√3)9 = \(\displaystyle 2^9\left(cos\ \pi + i\ sin\ \pi\right)\)

⇒ (1 + i√3)9 = 29 (-1 + 0)

⇒ (1 + i√3)9 = - 29

Imaginary part is 0.

∴  Imaginary part, b = 0.

Concept:

The value i or the concept of i is used in explaining and expressing complex numbers. Complex numbers are numbers with a real and imaginary part. The imaginary part is defined with the help of i. Basically, “i” is the imaginary part which is also called iota.

Value of i is √-1 A negative value inside a square root signifies an imaginary value. All the basic arithmetic operators are applicable to imaginary numbers. On squaring an imaginary number, we obtain a negative value.

i² = -1, i³ = -i, i⁴ = 1

Calculation:

Given

i = √-1

⇒ i² = -1

⇒ i-2n 

⇒ (i²)^-n

⇒ (-1)^-n

Since, n = 0, 1, 2, 3, ...

When n = even then i^-2n = 1

When n = odd then i-2n = -1

∴ Total possible values of the i-2n for n ϵ Z is two which are - 1 and 1.

Concept:

Power of i:

• i = \(\sqrt{-1}\)
• i2 = -1
• i3 = i × i2 = -i
• i4 = (i2)2 = (-1)2 = 1
• i4n = 1

Calculation:

To Find: Value of i3 + i6 + i9

As we know, i2 = -1

So,i3 =  i2 × i= -1 × i = -i

i6 = i3 × i3 = -i × -i = i² = -1

i⁹ = i3 × i3 × i3  = -i × -i × -i = -i³ = -(-i) = i 

Now,

i3 + i6 + i9 = (-i) + (-1) + i = -1 

Concept:

Property of iota:

i²  = -1

Power of i: 

For any integer k, i^4k = 1, i^{4k + 1} = i, i^{4k + 2} = -1 and i^{4k + 3} = -i

Calculation:

i⁴⁰ + i⁴¹ + i⁴² + i⁴³

= \(\rm i^{4 \times 10} + i^{(4 \times 10 + 1)} + i^{(4 \times 10 + 2)} + i^{(4 \times 10 + 3)}\)

= 1 + i - 1 - i

= 0

Concept

The cube roots of unity can be defined as the numbers which when raised to the power of 3 gives the result as 1.

Formulae

• ω2 = \( \displaystyle -\frac{1}{2}- i\frac{\sqrt{3}}{2}\)  and ω = \( \displaystyle -\frac{1}{2}+ i\frac{\sqrt{3}}{2}\)
• 1 + ω + ω2 = 0
• ω3 = 1

Calculation

Given:

z + z⁻¹ = 1

multiplying both sides by z

⇒ z + z⁻¹ = 1

⇒ \( z+\frac 1z-1=0\)

⇒ z² + 1 = z

⇒ z² − z + 1 = 0

By solving, z = \(\displaystyle \frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm i\sqrt{3}}{2}\)

We know that z = \(\displaystyle \frac{1+ i\sqrt{3}}{2}\) = -ω²   and z = \(\displaystyle \frac{1- i\sqrt{3}}{2}\) = ω are the roots of the equation.

⇒ ω2 = \( \displaystyle -\frac{1}{2}- i\frac{\sqrt{3}}{2}\)  and ω = \( \displaystyle -\frac{1}{2}+ i\frac{\sqrt{3}}{2}\)

Now z2 − z + 1 = [z - (-ω)][z -(-ω2)] = 0 where - ω and - ω2 are the roots of the equation.

Sum of roots = \(\displaystyle -\frac{b}{a}\) = - ω - ω² = 1

We know that 1 + ω + ω2 = 0 and ω³ = 1

z100 + z-100 = (- ω)¹⁰⁰ + (- ω)-100

⇒ z100 + z-100 = ω100 + ω-100

⇒ z100 + z-100 = (ω³)³³.ω + \(\displaystyle \frac{1}{(ω^3)^{33}.ω}\)

⇒ z100 + z-100 = (1)33.ω + \(\displaystyle \frac{ω^3}{(1)^{33}.ω}\)         [ω3 = 1]

⇒ z100 + z-100 = ω + ω²

⇒ z100 + z-100 = -1            [1 + ω + ω2 = 0]

∴ z100 + z-100 is −1.

Concept:

The imaginary unit i is defined as i = √-1.

i4n + 1 = i, i4n + 2 = -1, i4n + 3 = -i, i4n = 1, etc.

Calculation:

Let z = i-25 + i²⁵

= i^-24 i^-1 + i²⁴i

= (i⁴)^-6 × \(\rm \frac 1 i\) + (i⁴)⁶ i

= 1 × \(\rm \frac 1 i\) + i                 (∵ i⁴ⁿ = 1)

= \(\rm \frac {1 + i^2} i = \rm \frac {1 -1} i =0\)