Question

If $\dfrac{{2{z_1}}}{{3{z_2}}}$is a purely imaginary number, then find the value of $\left| {\dfrac{{\left( {{z_1} - {z_2}} \right)}}{{\left( {{z_1} + {z_2}} \right)}}} \right|$.

Answer 1

As it is given that $\dfrac{{2{z_1}}}{{3{z_2}}}$is a purely imaginary, we can equate them to imaginary number and get the ratio $\dfrac{{{z_1}}}{{{z_2}}}$. Then we can simplify the expression with this ratio. Then we can take the modulus to find the value of the expression.

Complete step by step answer:

It is given that $\dfrac{{2{z_1}}}{{3{z_2}}}$is purely imaginary. It means it has only imaginary part and real part is 0. So we can write,
$\dfrac{{2{z_1}}}{{3{z_2}}} = 0 + ai$, where a is any real number.
On simplification, we get,
$\dfrac{{{z_1}}}{{{z_2}}} = \dfrac{{3ai}}{2}$ … (1)
Now we have the expression,
Let $I = \left| {\dfrac{{\left( {{z_1} - {z_2}} \right)}}{{\left( {{z_1} + {z_2}} \right)}}} \right|$.
On dividing both numerator and denominator with ${z_2}$, we get,
$I = \left| {\dfrac{{\left( {\dfrac{{{z_1}}}{{{z_2}}} - 1} \right)}}{{\left( {\dfrac{{{z_1}}}{{{z_2}}} + 1} \right)}}} \right|$
Using equation (1), we get
$I = \left| {\dfrac{{\left( {\dfrac{{3ai}}{2} - 1} \right)}}{{\left( {\dfrac{{3ai}}{2} + 1} \right)}}} \right|$
Multiplying both numerator and denominator with 2, we get,
$I = \left| {\dfrac{{\left( {3ai - 2} \right)}}{{\left( {3ai + 2} \right)}}} \right|$
By properties of modulus of complex numbers,$\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_1}} \right|}}$,
$\Rightarrow I = \dfrac{{\left| {3ai - 2} \right|}}{{\left| {3ai + 2} \right|}}$
We know that modulus of a complex number $z = a + ib$ is given by, $\left| z \right| = \sqrt {{a^2} + {b^2}}$
$\Rightarrow I = \dfrac{{\sqrt {{{\left( {3a} \right)}^2} + {{\left( { - 2} \right)}^2}} }}{{\sqrt {{{\left( {3a} \right)}^2} + {{\left( 2 \right)}^2}} }} \\\ = \dfrac{{\sqrt {9{a^2} + 4} }}{{\sqrt {9{a^2} + 4} }} \\\$
Cancelling the common terms, we get,
$I = 1$
$\Rightarrow \left| {\dfrac{{\left( {{z_1} - {z_2}} \right)}}{{\left( {{z_1} + {z_2}} \right)}}} \right| = 1$
Thus, the value of $\left| {\dfrac{{\left( {{z_1} - {z_2}} \right)}}{{\left( {{z_1} + {z_2}} \right)}}} \right|$is equal to 1.

Note: A complex number is number of the form $a + ib$ where a and b are real numbers and $i$ is an imaginary part which satisfies the equation ${i^2} = - 1$. In a complex number $a + ib$, $a$ is called the real part and $ib$ is called the imaginary part. Modulus of a complex number is the absolute value of the complex number. It is given by the equation $\left| z \right| = \sqrt {{a^2} + {b^2}}$ where $z = a + ib$