Question

# If we have two complex numbers $z_1$ and $z_2$ such that $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|+\left| {{z}_{1}}-{{z}_{2}} \right|$ then(A). $\operatorname{Im}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=0$ (B). $\operatorname{Re}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=0$(C). $\operatorname{Re}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\operatorname{Im}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)$(D). None of these

Hint: In the given expression, first of all subtract $\left| {{z}_{2}} \right|$ on both the sides then take the square on both the sides. After squaring on both the sides, you will find the relation between the angles of the two complex numbers then find the ratio of two complex numbers $z_1$ and $z_2$.

Complete step-by-step solution -
Let us assume that ${{z}_{1}}=\left| {{z}_{1}} \right|{{e}^{i{{\theta }_{1}}}}$&${{z}_{2}}=\left| {{z}_{2}} \right|{{e}^{i{{\theta }_{2}}}}$.
The relation between the complex numbers $z_1$ and $z_2$ which is given in the above problem is:
$\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|+\left| {{z}_{1}}-{{z}_{2}} \right|$
Subtracting$\left| {{z}_{2}} \right|$from both the sides we get,
$\left| {{z}_{1}} \right|-\left| {{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|$
Squaring on both the sides we get,
${{\left( \left| {{z}_{1}} \right|-\left| {{z}_{2}} \right| \right)}^{2}}={{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}$
${{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$
From the above equation, as L.H.S should be equal to R.H.S so the coefficient of $-2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|$must be equal on both the sides.
$1=\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$
In the above equation, $θ_1$ is the angle of a complex number $z_1$ from real axis in the argand plane and $θ_2$ is the angle of a complex number $z_2$ from real axis in the argand plane and $θ_1 – θ_2$ is the angle between $z_1$ and $z_2$ complex numbers,
The above equation in cosine will resolve to:
$θ_1 – θ_2$ = 0
Now, ratio of$z_1$ and $z_2$ is:
\begin{align} & \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{\left| {{z}_{1}} \right|{{e}^{i{{\theta }_{1}}}}}{\left| {{z}_{2}} \right|{{e}^{i{{\theta }_{2}}}}} \\ & \Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{\left| {{z}_{1}} \right|{{e}^{i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}}}{\left| {{z}_{2}} \right|} \\ & \Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|} \\ \end{align}
When we substitute $(θ_1 – θ_2)$ as 0 then $e^{i(0)}$ becomes 1 and the ratio of $z_1$ and $z_2$ is purely real.
So, we can say that$\operatorname{Im}\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=0$.
Hence, the correct option is (a).

Note: In the above steps, we have equated ${{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}$ to ${{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$. So, we are going to show the proof of this equation.
\begin{align} & {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}=\left( {{z}_{1}}-{{z}_{2}} \right)\left( \overline{{{z}_{1}}}-\overline{{{z}_{2}}} \right) \\ & \Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{z}_{1}}\overline{{{z}_{1}}}+{{z}_{2}}\overline{{{z}_{2}}}-{{z}_{1}}\overline{{{z}_{2}}}-{{z}_{2}}\overline{{{z}_{1}}} \\ \end{align}
We know that $z\overline{z}={{\left| z \right|}^{2}}$and
$z\overline{z}=\left| {{z}_{1}} \right|{{e}^{i{{\theta }_{1}}}}\left| {{z}_{2}} \right|{{e}^{-i{{\theta }_{2}}}}=\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|{{e}^{i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}}$
Substituting these values in the above equation we get,
\begin{align} & {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|{{e}^{i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}}-\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|{{e}^{-i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}} \\ & \Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\left( {{e}^{i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}}+{{e}^{-i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}} \right) \\ & \Rightarrow {{\left| {{z}_{1}}-{{z}_{2}} \right|}^{2}}={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}-2\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \\ \end{align}