For a non-zero complex number z , let arg ( z ) denote the principal argument with - π < a r g z ≤ π then, which of the following statement(s) is (are) FALSE?

A Arg - 1 - i = - 3 π 4 B Arg - 1 + i t = π - tan - 1 ⁡ t , t ≥ 0 - π + tan - 1 ⁡ t , t < 0 Discontinuous at t = 0 ( C ) a r g z 1 z 2 - a r g z 1 + a r g z 2 = a r g z 1 - a r g z 2 - a r g z 1 + a r g z 2 =0 D a r g z - z 1 z 2 - z 3 z - z 3 z 2 - z 1 = π ⇒ z - z 1 z 2 - z 3 z - z 3 z 2 - z 1 is real ⇒ z , z 1 , z 2 , z 3 are concyclic

Mathematics - Complex Number Question with Solution | TestHub

MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Hard2 minPYQ_2018

MathematicsHardmultiple choice

For a non-zero complex number $z$ , let arg $(z)$ denote the principal argument with $- π < a r g (z) \leq π$ then, which of the following statement(s) is (are) FALSE?

Options:(select one or more)

Answer:

A, B, D

Solution:

$(A)$ $Arg (- 1 - i) = - \frac{3 π}{4}$
$(B)$ Arg $(- 1 + i t) = \{\begin{matrix} π - \tan^{- 1} (t), & t \geq 0 \\ - π + \tan^{- 1} (t), & t < 0 \end{matrix}$
Discontinuous at $t = 0$
$(C)$ $a r g (\frac{z_{1}}{z_{2}}) - a r g (z_{1}) + a r g (z_{2})$
$= a r g z_{1} - a r g (z_{2}) - a r g (z_{1}) + a r g (z_{2}) =0$
$(D)$ $a r g (\frac{(z - z_{1}) (z_{2} - z_{3})}{(z - z_{3}) (z_{2} - z_{1})}) = π$
$\Rightarrow \frac{(z - z_{1}) (z_{2} - z_{3})}{(z - z_{3}) (z_{2} - z_{1})}$ is real
$\Rightarrow z, z_{1}, z_{2}, z_{3}$ are concyclic

Stream:JEE_ADVSubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)

⏱ 2mℹ️ Source: PYQ_2018

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