Algebraic Operations on Complex Numbers MCQ Quiz - Objective Question with Answer for Algebraic Operations on Complex Numbers - Download Free PDF

Concept:

i^4k = 1, i^4k+1 = i, i^4k+2 = – 1, i^4k+3 = – i, where k ∈ ℤ

Calculation:

Given, Z1 = 4i40 - 5i35 + 6i17 + 2

= 4i4(10) - 5i(4(8)+3) + 6i(4(4)+1) + 2

= 4(1) - 5(- i) + 6(i) + 2

= 4 + 5i + 6i + 2

= 6 + 11i

Also, Z2 = -1 + i

∴ |Z1 + Z2|

= |6 + 11i - 1 + i|

= |5 + 12i|

= \(\sqrt{5^2+12^2}\)

= 13

∴ The value of |Z1 + Z2| is equal to 13.

The correct answer is Option 2.

Concept:

\(z = r(\cos \theta + i \sin \theta)\), where \(r = |z| = \sqrt{a^2 + b^2}\) 

\(\theta\) is the argument of z , i.e., the angle it makes with the positive real axis.

Explanation:

For 1 + i, a = 1 , b = 1

The modulus \(r = \sqrt{1^2 + 1^2} = \sqrt{2}\)

The argument \(\theta = \tan^{-1} \left( \frac{1}{1} \right) = \frac{\pi}{4}\)

So, we can write 1 + i as \(1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)\)

Now, using De Moivre's Theorem,

\(z^n = r^n \left( \cos (n \theta) + i \sin (n \theta) \right)\)

\((1 + i)^4 = (\sqrt{2})^4 \left( \cos \left( 4 \times \frac{\pi}{4} \right) + i \sin \left( 4 \times \frac{\pi}{4} \right) \right)\)

= \(4 \left( \cos \pi + i \sin \pi \right)\)

= \(4 \left( -1 + i \cdot 0 \right)\)

= -4

So, the correct answer is option 1.

Concept:

(i) ln(z)=ln(∣z∣)+iarg(z)

where ∣z∣ is the magnitude (absolute value) of z, and arg(z) is the argument (angle) of z, which is typically defined to lie in the interval (−π,π].

(ii) arg(z)=arg(x+iy)=tan^-1(\(\frac{y}{x}\))

Explanation:

Let y= iⁱ

Take Log_e both sides

log_ey = ilog_ei

log_ey = i(log_e|i|+i arg(i))

log_ey = i(log_e(1)+itan^-1(\(\frac{1}{0}\)))

log_ey = i(0+i(\(\frac{\pi}{2}+2n\pi\)))

log_ey = - (\(\frac{\pi}{2}+2n\pi\))

y = e^{\(-(\frac{\pi}{2}+2n\pi)\)}

Hence, y = e\(\frac{-(4n+1)\pi}{2}\) is answer.

Concept:

The square root of a complex number z = a + ib is

\(\sqrt{a+ib} = ±\left( \sqrt{|z|+a \over 2}+i{b\over|b|}\sqrt{|z|-a\over 2}\right)\)

Calculation:

Let z = x + iy

Given, z2(1 − z2) = 16

⇒ z⁴ - z² + 16 = 0

⇒ z² = \({1 ± \sqrt{(-1)^2-4(16)} \over 2}\)

⇒ z2 = \({1 ± 3 \sqrt 7 i \over 2}\), where |z²| = \(\sqrt{{1\over 4} + {63 \over 4} }= 4\)

⇒ z = \(\sqrt{{1\over 2} ± i{ 3\sqrt 7 \over 2}}\)

⇒ z = \( ±\left( \sqrt{4+{1 \over 2} \over 2}+i{(\pm 1)}\sqrt{4 - {1 \over 2}\over 2}\right)\)

⇒ z = \( ±\left( \sqrt{9\over 4}\pm i\sqrt{7\over 4}\right)\) 

⇒ z = \( ±\left({3 \over 2}\pm i{\sqrt{7}\over 2}\right)\)

⇒ |z| = \(\sqrt{\left({3 \over 2}\right)^2+\left({\sqrt{7}\over 2}\right)^2}\)

⇒ |z| = \(\sqrt{4} = 2\)

∴ The correct option is (3).

Formula used:

i² = -1 & i⁴ = 1

Calculation:

Given that, i¹⁹

This can be written as

(i⁴)⁴ × i² × i

By using the above formula 

(1) × (-1) × i = -i

∴ i¹⁹ = - i

Formula used:

i² = -1 & i⁴ = 1

Calculation:

Given that, i¹⁹

This can be written as

(i⁴)⁴ × i² × i

By using the above formula 

(1) × (-1) × i = -i

∴ i¹⁹ = - i

Concept:

The conjugate of the complex number a + bi is a - bi.

Property of iota power:

i² = -1

i⁴ = 1

Calculation:

Given: (x - iy)(3 + 5i) is the conjugate of -6 - 24i

According to the question, (x - iy)(3 + 5i) = -6 + 24i.

⇒ 3x + 5xi - 3yi - 5yi² = -6 + 24i

⇒ (3x + 5y) + (5x - 3y)i = -6 + 24i

Comparing the real and imaginary parts, we get:

3x + 5y = -6               ...(1)

5x - 3y = 24               ...(2)

Multiplying equation (1) by 3 and equation (2) by 5 and adding, we get:

9x + 25x = -18 + 120

34x = 102

x = 3

Using equation (1), we get:

y = -3.

Concept:

z = x + iy

when x > 0, y > 0 then point lie in quadrant I.

Calculation:

Given

z1 = \(\rm 2\sqrt{5} + 5i\) and z2 = \(\rm \sqrt{5} + i\)

z1 - z2 = \(\rm \sqrt{5} + 4i\) 

which is represented by a point in quadrant I

Formula used:

(a + b)² = a² + b² + 2ab

(a + ib) (a - ib) = a² - i²b² = a² + b²

i² = -1

Concept Used:

if y = x² then we can say that x is the square root of y

Calculation:

We have the algebric expression \(\rm 2a+2\sqrt{a^2 + b^2}\)

⇒ {(a + ib) + (a - ib)} + \(\rm 2\sqrt{(a + ib)(a - ib)}\)

⇒ \(\rm {(\sqrt{(a + ib)})^{2} + ((\sqrt{(a - ib)})^{2} + 2\sqrt{(a + ib)(a - ib)}}\)

⇒ \(\rm (\sqrt{a + ib} + \sqrt{a - ib})^{2}\)

∴ The square root of  \(\rm 2a+2\sqrt{a^2 + b^2}\) is  \(\rm (\sqrt{a + ib} + \sqrt{a - ib})\).

Concept:

Whenever a complex number is a root of a polynomial with real coefficients, its complex conjugate is also a root of that polynomial.

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

e.g., z = a + ib then conjugate of z is given by z̅ = a - ib

Quadratic equation in the form of roots: x2 – (sum of roots) x + product of roots = 0

(a + b)(a - b) = (a2 - b2)

i2 = -1

Calculation:

Since complex roots exists in conjugate pairs.

∴ if 3 + 4i is one root of the equation x² + px + q = 0 then other root will be:

3 – 4i 

Quadratic equation obtained by these roots is:

x² - (3 + 4i + 3 - 4i)x + (3 + 4i)(3 - 4i) = 0

x² - 6x + 9 + 16 = 0

x2 - 6x  +25 = 0

Comparing this equation with the x2 + px + q = 0;

p = -6 and q = 25

CONCEPT:

• i² = - 1
• If x + iy = a + ib then x = a and y = b where a, b, x and y are real numbers.

CALCULATION:

Given: i9 + i19 = x + iy

⇒ (i²)⁴ ⋅ i + (i²)⁹ ⋅ i = x + iy

As we know that, i2 = - 1

⇒ i - i = 0 = 0 + i 0 = x + i y 

As we know that, if x + iy = a + iv then x = a and y = b

⇒ x = 0 and y = 0

Hence, correct option is 2.

Concept:

If two complex number are equal that means real part and imaginary parts are equal.

Calculation:

 a = (x - 4) + 5i and b = 2 + i(y - 1)

⇒ x - 4 = 2 and y - 1 = 5

⇒ x = 6 and y = 6

Concept:

Complex number z = x + iy, on graph represent x = x-coordinate y = y-coordinate where Y-axis is imaginary axis and X-axis is real axis. 

Calculation:

Let, z = x + iy ⇒ z̅ = x - iy

iz̅ - iz = 5 will be

⇒ i (x - iy) - i (x + iy) = 5

⇒ xi - i²y - ix - i²y = 5

⇒ 2y = 5    (∵ i²  = -1)

⇒ y = 5/2 

x - coordinate of z₂ will be the same as of the z₁ as the point z₂ is the reflection of the z₁ about the line y = 5/2

∴ x-coordinate of z₂ = 1

Distance |Cz₁| = (√ [(1 - 1)² + (5/2 - 1)²)

⇒ Distance |Cz1| = 3/2

Now, y-coordinate of z₂ = 5/2 + 3/2 = 8/2

⇒y-coordinate of z₂ = 4

∴ z₂ = 1 + 4i

Hence, option (1) is correct.

Formula used:

(a + b)(a - b) = a² - b²

i² = -1

Calculation:

Given that,

\(\frac{1}{3\ + i}\ + \frac{i}{3\ - i} \ = a \ + ib\)

\(\Rightarrow \frac{3 \ - \ i}{(3\ + i)(3 \ - \ i)}\ + \frac{i(3\ + i)}{(3\ - i)(3 \ + \ i)} \ = a \ + ib\)

\(\Rightarrow \frac{(3\ - \ i\ + \ 3i \ + \ i^2)}{3^2\ -\ i^2 }\ = \ a \ +\ ib\)         (Using above identity)

\(\Rightarrow \ \frac{2 \ + \ 2i}{10}\ = a\ +\ ib\)                     (∵ i² = -1)

\(\Rightarrow \ \frac{1}{5}\ +\ \frac{1}{5}i\ =\ a\ +\ ib\)

On comparing both side

a = 1/5 and b = 1/5

Hence, a/b will be 1.

Calculation:

\([ (\frac{1}{5}+\frac{7i}{5})-(6+\frac{i}{5})]-\rm (\frac{-4}{5}+i)\)

= \(\rm (\frac{1}{5}-6+\frac{7i}{5}-\frac{i}{5})-\rm (\frac{-4}{5}+i)\)

= \(\rm (\frac{-29}{5}+\frac{6i}{5})-\rm (\frac{-4}{5}+i)\)

= \(\rm (\frac{-29}{5}+\frac{4}{5}+\frac{6i}{5}-i)\)

= \(\rm (\frac{-25}{5}+\frac{i}{5})\)

= \(\rm (-5+\frac{i}{5})\)