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Question: For a non-zero complex number \(z\), let \(\text{arg}\left( \text{z} \right)\) denotes the principal...

For a non-zero complex number zz, let arg(z)\text{arg}\left( \text{z} \right) denotes the principal argument with −π<arg⁡(z)≤π-\pi <\arg \left( z \right)\le \pi .$$$$ Then, which of the following statement(s) is (are) FALSE?
A.arg⁡(−1 − i) = π4\arg \left( -1\text{ }-\text{ }i \right)\text{ }=\text{ }\dfrac{\pi }{4} where \text{i}~=\text{ }\surd -\text{1}~$$$$$ B. The function f\text{ }:\text{ }R\to \left( -\pi ,\text{ }\pi ~ \right],definedbydefined byf\left( t \right)\text{ }=\text{ }\arg \left( -1\text{ }+\text{ }it \right)forallfor allt\in R,iscontinuousatallpointsofR,where, is continuous at all points of R, where i=\sqrt{-1}$$$$$

C. For any two non-zero complex numbers z1{{z}_{1}} and z2{{z}_{2}}, arg⁡(z1z2)−arg⁡(z1)+arg⁡(z2)\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)-\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right) is an integer multiple of 2\pi $$$$$ D. For any three given distinct complex numbers {{z}{1}},, {{z}{2}}andand{{z}{3}}thelocusofthepointzsatisfyingtheconditionthe locus of the point z satisfying the condition\arg \left( \dfrac{\left( z-{{z}{1}} \right)\left( {{z}{2}}-{{z}{3}} \right)}{\left( z-z \right)\left( {{z}{2}}-{{z}{1}} \right)} \right)=\pi ,liesonastraightline., lies on a straight line.$$$

Explanation

Solution

Use the identities and complex numbers to test the truth value of each option. The argument or principal argument of any non-zero complex number is the smallest angle it makes at the origin with positive real axis (xx-axis is the real axis and yy-axis is the imaginary axis) when represented in the argand plane. You will also need the identity of concyclic points and operation between arguments of complex numbers.

Complete step-by-step answer:
Checking the truth value of A: It is given that $\arg \left( -1-i \right)=\dfrac{\pi }{4}$ . We know that the argument of any complex number $a+ib$ is given by ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)$. So $\arg \left( -1-i \right)={{\tan }^{-1}}\left( \dfrac{-1}{-1} \right)=\dfrac{\pi }{4}$. So A is true.
Checking the truth value of B: The given function is $f\left( t \right)=\arg \left( -1+it \right)$. When $t$ approaches from left hand side that is $t<0$ then $f\left( t \right)=\arg \left( -1+it \right)=-\pi +{{\tan }^{-1}}t$. When $t$ approaches from right hand side that is $t\ge 0$ $f\left( t \right)=\arg \left( -1+it \right)=\pi +{{\tan }^{-1}}t$. So $f$ is not continuous .
Checking the truth value of C: The given expression is $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)-\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)$ on expanding transforms to $\arg \left( \dfrac{{{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 $ which is a integral multiple of $\pi $. So C is true.
Checking the truth value of D: The given expression is $\arg \left( \dfrac{\left( z-{{z}_{1}} \right)\left( {{z}_{2}}-{{z}_{3}} \right)}{\left( z-z \right)\left( {{z}_{2}}-{{z}_{1}} \right)} \right)=\pi $. We know from the complex number theory that given expression is possible when all the points ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ lie on one circle. So they can not lie on a straight line. So D is false.

So, the correct answer is “Option A and C”.

Note: While solving the problems in the argand plane involving the argument of complex numbers we need to take note of the quadrant where the complex numbers are present because that will determine the argument. .