Question

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$, then $\arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right)$ is equal to
A. $- \pi$
B. $- \dfrac{\pi }{2}$
C. $0$
D. $\dfrac{\pi }{2}$
E. $\pi$

Answer 1

We will first assume ${z_1} = \cos {\theta _1} + \sin {\theta _1}$ and ${z_2} = \cos {\theta _2} + \sin {\theta _2}$ where ${\theta _1}$ is the argument of ${z_1}$ and ${\theta _2}$ is the argument of ${z_2}$. After that 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:
$\left| z \right| = \sqrt {{x^2} + {y^2}}$

Complete step-by-step solution:
Let the complex number ${z_1}$ be,
${z_1} = \cos {\theta _1} + \sin {\theta _1}$
where ${\theta _1}$ is the argument of ${z_1}$
Let the complex number ${z_2}$ be,
${z_2} = \cos {\theta _2} + \sin {\theta _2}$ where ${\theta _2}$ is the argument of ${z_2}$.
Now according to the question, it is given that,
$\Rightarrow \left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$
Substitute the values,
$\Rightarrow \left| {\cos {\theta _1} + i\sin {\theta _1} + \cos {\theta _2} + i\sin {\theta _2}} \right| = \left| {\cos {\theta _1} + i\sin {\theta _1}} \right| + \left| {\cos {\theta _2} + i\sin {\theta _2}} \right|$
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 the modulus of a complex number $z = x + iy$ is given by:
$\left| z \right| = \sqrt {{x^2} + {y^2}}$
Let us first consider the left-hand side.
Evaluating the modulus of the left-hand side we get,
$\Rightarrow \left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + \sin {\theta _2}} \right)} \right| = \sqrt {{{\left( {\cos {\theta _1} + \cos {\theta _2}} \right)}^2} + {{\left( {\sin {\theta _1} + \sin {\theta _2}} \right)}^2}}$
Simplify the term,
$\Rightarrow \left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + \sin {\theta _2}} \right)} \right| = \sqrt {{{\cos }^2}{\theta _1} + {{\cos }^2}{\theta _2} + 2\cos {\theta _1}\cos {\theta _2} + {{\sin }^2}{\theta _1} + {{\sin }^2}{\theta _2} + 2\sin {\theta _1}\sin {\theta _2}}$
Now using the following identity,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
Then, we get,
$\Rightarrow \left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + \sin {\theta _2}} \right)} \right| = \sqrt {2 + 2\cos {\theta _1}\cos {\theta _2} + 2\sin {\theta _1}\sin {\theta _2}}$
Taking 2 as common we get,
$\Rightarrow \left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + \sin {\theta _2}} \right)} \right| = \sqrt {2\left( {1 + \cos {\theta _1}\cos {\theta _2} + \sin {\theta _1}\sin {\theta _2}} \right)}$
We know that,
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
Using the above formula, we get,
$\Rightarrow \left| {\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + i\left( {\sin {\theta _1} + \sin {\theta _2}} \right)} \right| = \sqrt {2\left( {1 + \cos \left( {{\theta _1} - {\theta _2}} \right)} \right)}$.................….. (1)
Let us first consider the right-hand side.
Evaluating the modulus of the right-hand side we get,
$\Rightarrow \left| {\cos {\theta _1} + i\sin {\theta _1}} \right| + \left| {\cos {\theta _2} + i\sin {\theta _2}} \right| = \sqrt {{{\cos }^2}{\theta _1} + {{\sin }^2}{\theta _1}} + \sqrt {{{\cos }^2}{\theta _2} + {{\sin }^2}{\theta _2}}$
Now using the following identity,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
Then, we get,
$\Rightarrow \left| {\cos {\theta _1} + i\sin {\theta _1}} \right| + \left| {\cos {\theta _2} + i\sin {\theta _2}} \right| = \sqrt 1 + \sqrt 1$
Simplify the terms,
$\Rightarrow \left| {\cos {\theta _1} + i\sin {\theta _1}} \right| + \left| {\cos {\theta _2} + i\sin {\theta _2}} \right| = 2$................….. (2)
Now, equate the equations (1) and (2),
$\Rightarrow \sqrt {2\left( {1 + \cos \left( {{\theta _1} - {\theta _2}} \right)} \right)} = 2$
Square on both sides,
$\Rightarrow 2\left( {1 + \cos \left( {{\theta _1} - {\theta _2}} \right)} \right) = 4$
Divide both sides by 2,
$\Rightarrow 1 + \cos \left( {{\theta _1} - {\theta _2}} \right) = 2$
Simplify the terms,
$\Rightarrow \cos \left( {{\theta _1} - {\theta _2}} \right) = 1$
We know that,
$\cos 0 = 1$
Substitute the value in the expression,
$\Rightarrow {\theta _1} - {\theta _2} = 0$
Thus, $\arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right) = 0$

Hence, option (C) is the correct answer.

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 $\arg \left( z \right) = {\tan ^{ - 1}}\dfrac{y}{x}$.