Test: Algebra of Complex Numbers (May 14) - JEE MCQ

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:
(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 (y/x)
Explanation:
Let y= iⁱ
Take Log_e both sides
log_ey = ilog_ei
log_ey = i(log_e|i|+i arg(i))

Concept:
If z = x + iy, then

• ̅ z = x - iy
• |z|² = x² + y² 

Calculation:


and z² = (2i)² = 4 (-1) = - 4
So, z2 + z ̅z = - 4 + (2i).(- 2i)
⇒  z2 + z ̅z = - 4 - (- 4) = 0
∴ The correct answer is option (1).

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

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 

∴ The square root of 

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,

On comparing both side
a = 1/5 and b = 1/5
Hence, a/b will be 1.

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.