$x: y=(a+i b):(c+i$ d ) হলে $\quad\left(c^{2}+d^{2}\right) x^{2}-2(a c+b d) x y+\left(a^{2}+b^{2}\right) y^{2}$=?

Solve: $x: y=(\mathrm{a}+i \mathrm{~b}):(\mathrm{c}+i \mathrm{~d})$$\Rightarrow \frac{x}{y}=\frac{a+i b}{c+i d} \Rightarrow \mathrm{cx}+i \mathrm{~d} x=\mathrm{ay}+i \mathrm{~b} y$

উভয় পক্ষ হতে বাত্তব ও কাল্পনিক অংশ সমীকৃত করে পাই,

$\begin{aligned} & \mathrm{cx}=\mathrm{ay} \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{a}}{\mathrm{c}} \text { and } \mathrm{dx}=\mathrm{by} \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{b}}{\mathrm{d}} \\ \therefore & \frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{a}}{\mathrm{c}}=\frac{\mathrm{b}}{\mathrm{d}} \\ & \text { (ii) } x: y=(\mathrm{a}+i \mathrm{~b}):(\mathrm{c}+i \mathrm{~d}) \\ \Rightarrow & \frac{x}{y}=\frac{a+i b}{c+i d} \Rightarrow \mathrm{c} x+i \mathrm{~d} x=\mathrm{a} y+i \mathrm{~b} y \\ \Rightarrow & \mathrm{c} x-\mathrm{ay}=i(\mathrm{~b} y-\mathrm{d} x) \\ \Rightarrow & (\mathrm{c} x-\mathrm{a} y)^{2}=i^{2}(\mathrm{~b} y-\mathrm{d} x)^{2} \\ \Rightarrow & \mathrm{c}^{2} x^{2}-2 \mathrm{c} x \cdot \mathrm{ay}+\mathrm{a}^{2} y^{2} \\ & \quad=-\left(\mathrm{b}^{2} y^{2}-2 \mathrm{~b} y \cdot \mathrm{d} x+\mathrm{d}^{2} x^{2}\right) \end{aligned}$

$\begin{array}{l} \Rightarrow c^{2} x^{2}-2 c x \cdot a y+a^{2} y^{2}+ \\ \quad\left(b^{2} y^{2}-2 b y \cdot d x+d^{2} x^{2}\right)=0 \\ \therefore \quad\left(c^{2}+d^{2}\right) x^{2}-2(a c+b d) x y+\left(a^{2}+b^{2}\right) y^{2}=0 \end{array}$