Question

If ${\mathrm{z}}_1$ and ${\mathrm{z}}_2$ are two complex numbers such that $\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|+\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|,$ then

• A. $\mathrm{Im}\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=0$
• B. $\mathrm{Re}\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=0$
• C. $\mathrm{Re}\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=0=\mathrm{Im}\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)$
• D. None of these

Answer 1

The correct option is A

$\mathrm{Im}\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=0$

Explanation for the correct option

Finding the value of $\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|+\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|$:

The given two complex numbers are ${\mathrm{z}}_1$ and ${\mathrm{z}}_2$

The given equation is $\left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|+\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|...\left(\mathrm{i}\right)$

Let write above equation $\left|{\text{z}}_1\right|-\left|{\text{z}}_2\right|=\left|{\text{z}}_1-{\text{z}}_2\right|$

consider squaring on both sides,

$\begin{array}{l}{\left(\left|{\text{z}}_1\right|-\left|{\text{z}}_2\right|\right)}^2={\left|{\text{z}}_1-{\text{z}}_2\right|}^2\\ {\left|{\text{z}}_1\right|}^2+{\left|{\text{z}}_2\right|}^2-2\left|{\text{z}}_1\right|\left|{\text{z}}_2\right|={\left|{\text{z}}_1\right|}^2+{\left|{\text{z}}_2\right|}^2-2\left|{\text{z}}_1\right|\left|{\text{z}}_2\right|\cos \left({\mathrm{\theta }}_1-{\mathrm{\theta }}_2\right)...\left(\mathrm{ii}\right)\end{array}$

$\begin{array}{l}\cos \left({\mathrm{\theta }}_1-{\mathrm{\theta }}_2\right)=0...\left(\mathrm{iii}\right)\end{array}$

Substitute $\left(\mathrm{iii}\right)$ in $\left(\mathrm{ii}\right)$, we get

$\begin{array}{l}{\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2-2\left|{\mathrm{z}}_1\right|\left|{\mathrm{z}}_2\right|={\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2-2\left|{\mathrm{z}}_1\right|\left|{\mathrm{z}}_2\right|.0\\ {\left|{\mathrm{z}}_1\right|}^2+{\left|{\mathrm{z}}_2\right|}^2-2\left|{\mathrm{z}}_1\right|\left|{\mathrm{z}}_2\right|=0\end{array}$

From $\left(\mathrm{i}\right)$, we get

$\Rightarrow \arg \left({\mathrm{z}}_1\right)-\arg \left({\mathrm{z}}_2\right)$

$\Rightarrow \frac{\left({\mathrm{z}}_1\right)}{\left({\mathrm{z}}_2\right)}$ is purely real, implying that the imaginary component is zero.

Finally, $\mathrm{Im}\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=0$

Therefore, the correct option is $\mathrm{a}$