A) 3
B) 8
C) 9
D) \[\sqrt{8}\]
E) 4
Correct Answer: C
Solution :
Let\[\sqrt{8}+i=r(\cos \theta +i\sin \theta )\] Here,\[r=3\]and \[\theta ={{\tan }^{-1}}\left( \frac{1}{2\sqrt{2}} \right)\] Now, \[{{(\sqrt{8}+i)}^{50}}={{3}^{50}}{{(\cos \theta +i\sin \theta )}^{50}}\] But \[{{(\sqrt{8}+i)}^{50}}={{3}^{48}}(a+ib)\] \[\therefore \]\[a=3\text{ }cos\theta \]and\[b=3\text{ }sin\theta \] So, \[{{a}^{2}}+{{b}^{2}}={{(3\text{ }cos\theta )}^{2}}+{{(3\text{ }sin\theta )}^{2}}\] \[=9\]You need to login to perform this action.
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