A) \[\frac{1}{2}(\sqrt{{{a}^{2}}+1}+a)\]
B) \[\frac{1}{2}(\sqrt{{{a}^{2}}+2}+a)\]
C) \[\frac{1}{2}(\sqrt{{{a}^{2}}+4}+a)\]
D) None of these
Correct Answer: C
Solution :
Let\[z=r\]\[(\cos \theta +i\sin \theta )\]. Then \[\left| z+\frac{1}{z} \right|=a\Rightarrow {{\left| z+\frac{1}{z} \right|}^{2}}={{a}^{2}}\] Þ \[{{r}^{2}}+\frac{1}{{{r}^{2}}}+2\cos 2\theta ={{a}^{2}}\] ??(i) Differentiating w.r.t. \[\theta \]we get \[2r\frac{dr}{d\theta }-\frac{2}{{{r}^{3}}}\frac{dr}{d\theta }-4\]sin \[2\theta =0\] Putting \[\frac{dr}{d\theta }=0,\]we get \[\theta =0,\frac{\pi }{2}\] \[r\] is maximum for \[\theta =\frac{\pi }{2},\] therefore from (i) \[{{r}^{2}}+\frac{1}{{{r}^{2}}}-2={{a}^{2}}\,\,\,\,\Rightarrow r-\frac{1}{r}=a\,\Rightarrow r=\frac{a+\sqrt{{{a}^{2}}+4}}{2}\]
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