A) \[\operatorname{Im}(z)=0\]
B) \[\operatorname{Re}(z)=0\]
C) \[amp(z)=\pi \]
D) None of these
Correct Answer: B
Solution :
Let\[z=r(\cos \theta +i\sin \theta )\]. Then \[\left| z+\frac{1}{z} \right|=1\,\,\Rightarrow {{\left| z+\frac{1}{z} \right|}^{2}}\]=1 Þ \[{{\left| r(\cos \theta +i\sin \theta )+\frac{1}{r}(\cos \theta -i\sin \theta ) \right|}^{2}}=1\]. Þ \[{{\left( r+\frac{1}{r} \right)}^{2}}{{\cos }^{2}}\theta +{{\left( r-\frac{1}{r} \right)}^{2}}{{\sin }^{2}}\theta =1\] Þ \[{{r}^{2}}+\frac{1}{{{r}^{2}}}+2\cos 2\theta =1\] Since \[|z|=r\] is maximum, therefore \[\frac{dr}{d\theta }=0\] Differentiating (i) 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 \[\sin 2\theta =0\]Þ \[\theta =0\]or \[\frac{\pi }{2}\] Þ z is purely imaginary or purely real. (\[\because \theta =0\] is not given)
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