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COMPLEX NUMBERS AND QUADRATIC EQUATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS4 Marks
Question
If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1,$ then find $\Big|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\Big|.$
✓

Answer

Let $\alpha=\text{a}+\text{ib}$ and $\beta=\text{x}+\text{iy}$ It is given that, $|\beta|=1$ $\therefore\ \sqrt{\text{x}^2+\text{y}^2}=1$ $\Rightarrow\text{x}^2+\text{y}^2=1\ .....(1)$ $\bigg|\frac{\beta-\alpha}{1-\bar{\alpha}{\beta}}\bigg|=\bigg|\frac{(\text{x}+\text{iy})-(\text{a}+\text{ib})}{1-(\text{a}-\text{ib})(\text{x}+{\text{iy}})}\bigg|$ $=\bigg|\frac{(\text{x}-\text{a})+\text{i}(\text{y}-\text{b})}{1-(\text{ax}+\text{aiy}-\text{ibx}+\text{by})}\bigg|$ $=\bigg|\frac{(\text{x}-\text{a})+\text{i}(\text{y}-\text{b})}{(1-\text{ax}-\text{by})+\text{i}(\text{bx}-\text{ay})}\bigg|$ $=\frac{\big|(\text{x}-\text{a})+\text{i}(\text{y}-\text{b})\big|}{\big|(1-\text{ax}-\text{by})+\text{i}(\text{bx}-\text{ay})\big|}\ \ \bigg[\Big|\frac{\text{z}_1}{\text{z}_2}\Big|=\frac{|\text{z}_1|}{|\text{z}_2|}\bigg]$ $=\frac{\sqrt{(\text{x}-\text{a})^2+(\text{y}-\text{b})^2}}{\sqrt{(1-\text{ax}-\text{by})^2+(\text{bx}-\text{ay})^2}}$ $=\frac{\sqrt{\text{x}^2+\text{a}^2-2\text{ax}+\text{y}^2+\text{b}^2-2\text{by}}}{\sqrt{1+\text{a}^2\text{x}^2+\text{b}^2\text{y}^2-2\text{ax}+2\text{abxy}-2\text{by}+\text{b}^2\text{x}^2+\text{a}^2\text{y}^2-2\text{abxy}}}$ $=\frac{\sqrt{(\text{x}^2+\text{y}^2)+\text{a}^2+\text{b}^2-2\text{ax}-2\text{by}}}{\sqrt{1+\text{a}^2(\text{x}^2+\text{y}^2)+\text{b}^2(\text{y}^2+\text{x}^2)-2\text{ax}-2\text{by}}}$ $=\frac{\sqrt{1+\text{a}^2+\text{b}^2-2\text{ax}-2\text{by}}}{\sqrt{1+\text{a}^2+\text{b}^2-2\text{ax}-2\text{by}}}\ \ [\text{Using (1)}]$ $=1$ $\therefore\ \bigg|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\bigg|=1$

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