Question

If $\arg \left[ {\dfrac{{{z_1}}}{{{z_2}}}} \right] = \dfrac{\pi }{2}$, then find the value of $\left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right|$.

Answer 1

Here, we have to find the value of the given expression of a complex number. We will use the triangle inequality to prove that the division of two complex numbers is purely imaginary. Then by using the division of two complex numbers and the absolute value of a complex number, we will find the value of the given expression of a complex number.

Formula Used:
We will use the following formula:

1. Triangle Inequality: $2{\mathop{\rm Re}\nolimits} \left( {{z_1}{z_2}} \right) = {z_1}\overline {{z_2}} + \overline {{z_1}} {z_2}$
2. Absolute value of a complex number is given by the formula $\left| {x + iy} \right| = \sqrt {{{\left( x \right)}^2} + {{\left( y \right)}^2}}$

Complete step by step solution:
We are given that $\arg \left[ {\dfrac{{{z_1}}}{{{z_2}}}} \right] = \dfrac{\pi }{2}$.
Let ${z_1}$ and${z_2}$ be two complex numbers.
Now, we will prove that $\dfrac{{{z_1}}}{{{z_2}}}$ is purely imaginary. So,
${\left| {{z_1} + {z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} + 2{\mathop{\rm Re}\nolimits} \left( {{z_1}{z_2}} \right)$
We know that the triangle inequality $2{\mathop{\rm Re}\nolimits} \left( {{z_1}{z_2}} \right) = {z_1}\overline {{z_2}} + \overline {{z_1}} {z_2}$.
By using the triangle inequality, we get
$\Rightarrow {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} = {\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} + {z_1}\overline {{z_2}} + \overline {{z_1}} {z_2}$
By cancelling similar terms from both the sides, we get
$\Rightarrow {z_1}\overline {{z_2}} = - \overline {{z_1}} {z_2}$
By rewriting the equation, we get
$\Rightarrow \dfrac{{\overline {{z_1}} }}{{\overline {{z_2}} }} = - \dfrac{{{z_1}}}{{{z_2}}}$
$\Rightarrow \left( {\overline {\dfrac{{{z_1}}}{{{z_2}}}} } \right) = - \dfrac{{{z_1}}}{{{z_2}}}$
So, $\dfrac{{{z_1}}}{{{z_2}}}$ is purely imaginary.
Let us consider $\dfrac{{{z_1}}}{{{z_2}}} = ki$
Now, we will find the value of $\left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right|$ .
Taking common ${z_2}$ from both numerator and denominator, we get
$\left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right| = \dfrac{{{z_2}\left| {\dfrac{{{z_1}}}{{{z_2}}} + 1} \right|}}{{{z_2}\left| {\dfrac{{{z_1}}}{{{z_2}}} - 1} \right|}}$
By cancelling the similar terms from numerator and denominator, we get
$\Rightarrow \left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right| = \dfrac{{\left| {\dfrac{{{z_1}}}{{{z_2}}} + 1} \right|}}{{\left| {\dfrac{{{z_1}}}{{{z_2}}} - 1} \right|}}$
By substituting $\dfrac{{{z_1}}}{{{z_2}}} = ki$ in the above equation, we get
$\Rightarrow \left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right| = \dfrac{{\left| {ki + 1} \right|}}{{\left| {ki - 1} \right|}}$
Absolute value of a complex number is given by the formula $\left| {x + iy} \right| = \sqrt {{{\left( x \right)}^2} + {{\left( y \right)}^2}}$
Now, by using the Absolute value of a complex number formula, we get
$\Rightarrow \left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right| = \dfrac{{\sqrt {1 + {k^2}} }}{{\sqrt {1 + {k^2}} }}$
By dividing the terms, we get
$\Rightarrow \left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right| = 1$

Therefore, the value of $\left| {\dfrac{{{z_1} + {z_2}}}{{{z_1} - {z_2}}}} \right|$ is 1.

Note:
We know that a Complex number is defined as a number written in the form$a + bi$ where$a$ is any real number and $bi$ where$b$ is any real number and $i$ is the imaginary unit. The absolute value of a complex number is defined as the distance between the origin and any point in the complex plane. Absolute value is also known as Modulus. The modulus of a complex number is always equal to the product of the square root of the complex number and its conjugate complex number.