Question

For a non-zero complex number $z$, let $\arg \left(z\right)$ denote the principal argument with $-\pi <\arg \left(z\right)\le \pi$. Then which of the following statement(s) is (are) $\text{FALSE}$?

• A. $\arg (-1-i)=\frac{\pi }{4},$ where $i=\sqrt{-1}$
• B. The function $f:R\to (-\pi ,\pi ]$, defined by $f\left(t\right)=\arg (-1+it)$ for all $t\in R$, is continuous at all points of $R$, where $i=\sqrt{-1}$
• C. For any two non-zero complex numbers $z_1$ and $z_2$, $\arg \left(\frac{z_1}{z_2}\right)-\arg \left(z_1\right)+\arg \left(z_2\right)$ is an integer multiple of $2\pi$.
• D. For any three given distinct complex numbers $z_1,z_2$ and $z_3$, the locus of the point $z$ satisfying the condition $\arg \left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right)=\pi$,
  lies on a straight line.

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\arg (-1-i)=\frac{\pi }{4},$ where $i=\sqrt{-1}$
B The function $f:R\to (-\pi ,\pi ]$, defined by $f\left(t\right)=\arg (-1+it)$ for all $t\in R$, is continuous at all points of $R$, where $i=\sqrt{-1}$
D For any three given distinct complex numbers $z_1,z_2$ and $z_3$, the locus of the point $z$ satisfying the condition $\arg \left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right)=\pi$,

lies on a straight line.
$\arg (-1-i)=\frac{\pi }{4}-\pi =-\frac{3\pi }{4}$
$f\left(t\right)=\arg (-1+it)=\{\begin{array}{c}\pi -{\tan }^{-1}t,t\ge 0\\ -{\tan }^{-1}t-\pi ,t<0\end{array}$
Clearly, $f$ is discontinuous at $t=0$.

$\arg \left(\frac{z_1}{z_2}\right)-\arg \left(z_1\right)+\arg \left(z_2\right)$
$=\arg \left(z_1\right)-\arg \left(z_2\right)+2n\pi -\arg \left(z_1\right)+\arg \left(z_2\right)$
$=2n\pi$

$\arg \left(\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\right)=\pi$
$\Rightarrow \arg \left(\frac{z-z_1}{z_2-z_1}\right)+\arg \left(\frac{z_2-z_3}{z-z_3}\right)=\pi$
$\Rightarrow z,z_1,z_2,z_3$ are concyclic.
Hence, locus of $z$ is a circle.