CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers and Quadratic Equations3 Marks
Question
If $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy},$ what is the value of $x^2 + y^2$ ?
✓
Answer
Given that$, \frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy}\ ...(\text{i})$
Taking conjugate on both sides$,$
$\frac{(\text{a}^2+1)^2}{2\text{a}+\text{i}}=\text{x}-\text{iy}\ ...(\text{ii})$
Multiplying $eq. (i)$ and $(ii)$ we have
$\frac{(\text{a}^2+1)^2(\text{a}^2+1)^2}{(2\text{a}-\text{i})(2\text{a}+\text{i})}=\text{x}^2+\text{y}^2$
$\Rightarrow\frac{(\text{a}^2+1)^4}{4\text{a}^2-\text{i}^2}=\text{x}^2+\text{y}^2$
$\Rightarrow\frac{(\text{a}^2+1)^4}{4\text{a}^2-\text{i}^2}=\text{x}^2+\text{y}^2$
Hence$,$ the value of $\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
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