If $\mathrm{a},\mathrm{b},\mathrm{c},{\mathrm{a}}_1,{\mathrm{b}}_1\mathrm{and}{\mathrm{c}}_1$ are non-zero complex numbers satisfying $\frac{\mathrm{a}}{{\mathrm{a}}_1}+\frac{\mathrm{b}}{{\mathrm{b}}_1}+\frac{\mathrm{c}}{{\mathrm{c}}_1}=1+\mathrm{i}\mathrm{and}\frac{{\mathrm{a}}_1}{\mathrm{a}}+\frac{{\mathrm{b}}_1}{\mathrm{b}}+\frac{{\mathrm{c}}_1}{\mathrm{c}}=0$ where $\mathrm{i}=\sqrt{-1}$,  the value of $\frac{{\mathrm{a}}^2}{{\mathrm{a}}_1^2}+\frac{{\mathrm{b}}^2}{{\mathrm{b}}_1^2}+\frac{{\mathrm{c}}^2}{{\mathrm{c}}_1^2}$ is

$\text{ We have, }\frac{\mathrm{a}}{{\mathrm{a}}_1}+\frac{\mathrm{b}}{{\mathrm{b}}_1}+\frac{\mathrm{c}}{{\mathrm{c}}_1}=1+\mathrm{i}$
On squaring both sides, we get
$\begin{array}{l}\frac{{\mathrm{a}}^2}{{\mathrm{a}}_1^2}+\frac{{\mathrm{b}}^2}{{\mathrm{b}}_1^2}+\frac{{\mathrm{c}}^2}{{\mathrm{c}}_1^2}+2\left(\frac{\mathrm{ab}}{{\mathrm{a}}_1{\mathrm{b}}_1}+\frac{\mathrm{bc}}{{\mathrm{b}}_1{\mathrm{c}}_1}+\frac{\mathrm{ca}}{{\mathrm{c}}_1{\mathrm{a}}_1}\right)={\left(1+i\right)}^2\\ \Rightarrow \frac{{\mathrm{a}}^2}{{\mathrm{a}}_1^2}+\frac{{\mathrm{b}}^2}{{\mathrm{b}}_1^2}+\frac{{\mathrm{c}}^2}{{\mathrm{c}}_1^2}+2\frac{\mathrm{abc}}{{\mathrm{a}}_1{\mathrm{b}}_1{\mathrm{c}}_1}\left(\frac{{\mathrm{c}}_1}{\mathrm{c}}+\frac{{\mathrm{a}}_1}{\mathrm{a}}+\frac{{\mathrm{b}}_1}{\mathrm{b}}\right)=2i\\ \Rightarrow \frac{{\mathrm{a}}^2}{{\mathrm{a}}_1^2}+\frac{{\mathrm{b}}^2}{{\mathrm{b}}_1^2}+\frac{{\mathrm{c}}^2}{{\mathrm{c}}_1^2}+0=2\mathrm{i}\\ \therefore \frac{{\mathrm{a}}^2}{{\mathrm{a}}_1^2}+\frac{{\mathrm{b}}^2}{{\mathrm{b}}_1^2}+\frac{{\mathrm{c}}^2}{{\mathrm{c}}_1^2}=2\mathrm{i}\end{array}$