Question

If ${z_1}$ and ${z_2}$ are two nonzero complex numbers such that $\left| {{z_1} + {z_2}}\right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$. Find the value of $\arg {z_1} - \arg {z_2}$

Answer 1

We will first assume ${z_1} = \cos {\theta _1} + i\sin {\theta _1}$ and ${z_2} = \cos{\theta _2} + i\sin {\theta _2}$ where ${\theta _1}$is the argument of${z_1}$ and${\theta _2}$ is the argument of ${z_2}$and then put these values in the given condition and find the modulus and then finally find the relation between ${\theta _1}$and ${\theta _2}$.
The modulus of a complex number $z = x + iy$ is given by:-
$|z| = \sqrt {{x^2} + {y^2}}$

Complete step-by-step answer:
Let the complex number ${z_1}$be:-
${z_1} = \cos {\theta _1} + i\sin {\theta _1}$ where ${\theta _1}$is the argument of${z_1}$ ${\theta_2}$ is the argument of ${z_2}$
And, the complex number ${z_2}$ be:-
${z_2} = \cos {\theta _2} + i\sin {\theta _2}$where ${\theta _2}$ is the argument of ${z_2}$
Now according to the question it is given that,
$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$
Hence, putting in the respective values we get:-
\Rightarrow$$$\left| {\cos {\theta _1} + i\sin {\theta _1} + \cos {\theta _2} + i\sin {\theta _2}} \right| = \left| {\cos{\theta _1} + i\sin {\theta _1}} \right| + \left| {\cos {\theta _2} + i\sin {\theta _2}} \right|$$ Simplifying it further we get:- \Rightarrow\left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + \sin {\theta _2}}\right)} \right| = \left| {\cos {\theta _1} + i\sin {\theta _1}} \right| + \left| {\cos {\theta _2} + i\sin {\theta_2}} \right|$$ Now we have to find the modulus of each of the terms on the left hand side as well as the right hand side of the equation. Now we know that modulus of a complex number $$z = x + iy$$ is given by:- $$|z| = \sqrt {{x^2} + {y^2}} $$ Hence let us first consider the left hand side:- Evaluating the modulus of left hand side we get:- $\Rightarrow\left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + i\sin {\theta _2}}\right)} \right| = \sqrt {{{\left( {\cos {\theta _1} + \cos {\theta _2}} \right)}^2} + {{\left( {\sin {\theta _1} +\sin {\theta _2}} \right)}^2}} $Applying the following identity:-${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$We get:-$\left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + i\sin {\theta _2}}\right)} \right| = \sqrt {{{\cos }^2}{\theta _1} + {{\cos }^2}{\theta _2} + 2\cos {\theta _1}\cos {\theta _2} +{{\sin }^2}{\theta _1} + {{\sin }^2}{\theta _2} + 2\sin {\theta _1}\sin {\theta _2}} $Now using the following identity:-${\cos ^2}\theta + {\sin ^2}\theta = 1 We get:- $\Rightarrow$$$\left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + i\sin {\theta _2}}\right)} \right| = \sqrt {2 + 2\cos {\theta _1}\cos {\theta _2} + 2\sin {\theta _1}\sin {\theta _2}}
Taking 2 as common we get:-
\Rightarrow$$$\left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + i\sin {\theta _2}}\right)} \right| = \sqrt {2\left( {1 + \cos {\theta _1}\cos {\theta _2} + \sin {\theta _1}\sin {\theta _2}}\right)} $$ Using the following identity:- $$\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$$ We get:- \Rightarrow\left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + i\sin {\theta _2}}\right)} \right| = \sqrt {2\left( {1 + \cos \left( {{\theta _1} - {\theta _2}} \right)} \right)} $$……………………..(1) Now let us consider right hand side:- $$RHS = \left| {\cos {\theta _1} + i\sin {\theta _1}} \right| + \left| {\cos {\theta _2} + i\sin {\theta _2}}\right|$$ Now we know that modulus of a complex number $$z = x + iy$$ is given by:- $$|z| = \sqrt {{x^2} + {y^2}} $$ Hence applying this formula we get:- $\RightarrowRHS = \sqrt {\left( {{{\cos }^2}{\theta _1} + {{\sin }^2}{\theta _1}} \right)} + \sqrt {\left( {{{\cos}^2}{\theta _2} + {{\sin }^2}{\theta _2}} \right)} $Now using the following identity:-${\cos ^2}\theta + {\sin ^2}\theta = 1 We get:- $\Rightarrow$$$RHS = \sqrt 1 + \sqrt 1
Simplifying it further we get:-
\Rightarrow$$$RHS = 2$$………………….. (2) Equating equations 1 and 2 we get:- \Rightarrow\sqrt {2\left( {1 + \cos \left( {{\theta _1} - {\theta _2}} \right)} \right)} = 2$$ Squaring both the sides we get:- $\Rightarrow2\left( {1 + \cos \left( {{\theta _1} - {\theta _2}} \right)} \right) = 4 Simplifying it further we get:- $\Rightarrow$$$1 + \cos \left( {{\theta _1} - {\theta _2}} \right) = 2
Hence, we get:-
\Rightarrow$$$\cos \left( {{\theta _1} - {\theta _2}} \right) = 1$$ Now we know that, $$\cos 0 = 1$$ Therefore, \Rightarrow$$${\theta _1} - {\theta _2} = 0$Now since,${\theta _1}$is the argument of${z_1}$and${\theta _2}$is the argument of${z_2}$$

Hence, $\arg {z_1} - \arg {z_2} = 0$

Note: Students should take note that the argument of the complex number is the angle measured from the positive real axis to the line segment.
For a complex number $z = x + iy$
The argument is given by:-
$\arg \left( z \right) = {\tan ^{ - 1}}\dfrac{y}{x}$