Test: Geometry Of Complex Numbers - JEE MCQ

When x > 0 & y < 0, the quadrant contains positive X-axis and negative Y-axis
Therefore, the point lies in the fourth quadrant. 

Concept:
If a number ω = a + ib is purely imaginary, then

• a = 0 
• and ω =

Calculation:
Given, z is a complex number such that is purely imaginary,
Let = a + ib, then a = 0
And

z̅  = x - iy 
Real Part of = Real Part of = Real Part of
⇒Real Part of

⇒ x² - x - y² + y = 2x² + 2y² - 4y + 2
⇒ x² + 3y² + x - 5y + 2 = 0
The above equation does not represent a circle or Straight line.
So No Option is Correct.
Option (2) is correct only when we use Z in place of Z̅ in the numerator of 

P(-x,-y) lies on the 3rd or 4th quadrant
So, according to the question, points lies in 3rd quadrant
i.e. 3(π/2)

z = -i
Comlex number z is of the form x + iy
x = 0, y = -1
arg(z) = π − tan−1|y/x|
⇒ π − tan−1|-1/0|
= π - (π/2)
⇒  π - π/2
⇒ π/2

√(-1)² + (√3)² 

= 2

z = a + ib
a = -1, b = -√3
(-1, -√3) lies in third quadrant.
Arg(z) = -π + tan^-1(b/a)
= - π + tan^-1(√3)
= - π + tan^-1(tan π/3)
= - π + π/3
= -2π/3

x = |z| cos θ and y = |z| sin θ satisfies infinite values of θ and for any infinite values of θ is the value of Arg z. Thus, for any unique value of θ that lies in the interval - π < θ ≤ π and satisfies the above equations x = |z| cos θ and y = |z| sin θ is known as the principal value of Arg z or Amp z and it is denoted as arg z or amp z.
 
We know that, cos (2nπ + θ) = cos θ and sin (2nπ + θ) = sin θ (where n = 0, ±1, ±2, ±3, .............), then we get,
 
Amp z = 2nπ + amp z where - π < amp z ≤ π